# Ver 3537 E1.1 Analysis of Circuits (2014) E1.1 Circuit Analysis. Problem Sheet 1 (Lectures 1 & 2)

Save this PDF as:

Size: px
Start display at page:

Download "Ver 3537 E1.1 Analysis of Circuits (2014) E1.1 Circuit Analysis. Problem Sheet 1 (Lectures 1 & 2)"

## Transcription

1 Ver 3537 E. Analysis of Circuits () Key: [A]= easy... [E]=hard E. Circuit Analysis Problem Sheet (Lectures & ). [A] One of the following circuits is a series circuit and the other is a parallel circuit. Explain which is which. (a) (b). [B] Find the power absorbed by by each of the subcircuits A and B given that the voltage and current are V and A as shown. 3. [B] For each of the four circuits below, find the power absorbed by the voltage source (P V ), the power absorbed by the current source (P I ) and the total power absorbed (P V + P I ). (a) (b) (c) (d). [B] Determine the voltage V X in the following circuit. 5. [B] Determine the current I X in the following circuit. 6. [B] What single resistor is equivalent to the three resistor sub-circuit shown below? Problem Sheet Page of

2 Ver 3537 E. Analysis of Circuits () 7. [B] What single resistor is equivalent to the three resistor sub-circuit shown below? 8. [C] What single resistor is equivalent to the five resistor sub-circuit shown below? 9. [A] If a resistor has a conductance of 8 µs, what is its resistance?. [B] Determine the voltage across each of the resistors in the following circuit and the power dissipated in each of them. Calculate the power supplied by the voltage source.. [B] Determine the current through each of the resistors in the following circuit and the power dissipated in each of them. Calculate the power supplied by the current source.. [B] Determine R so that Y = X. 3. [B] Choose R and R so that Y =.X and R + R = MΩ.. [D] You have a supply of resistors that have the values {,, 5, 8,, 7, 33, 39, 7, 56, 68, 8} n Ω for all integer values of n. Thus, for example, a resistor of 39 Ω is available and the next higher value is 7 Ω. Show how, by combining two resistors in each case, it is possible to make networks whose equivalent resistance is (a) 3 kω, (b) kω and (c) as close as possible to 3.5 kω. Determine is the worst case percentage error that might arise if, instead of combining resistors, you just pick the closest one available. Problem Sheet Page of

3 Ver 7 E. Analysis of Circuits () E. Circuit Analysis Problem Sheet - Solutions. Circuit (a) is a parallel circuit: there are only two nodes and all four components are connected between them. Circuit (b) is a series circuit: each node is connected to exactly two components and the same current must flow through each.. For subcircuit B the voltage and current correspond to the passive sign convention (i.e. the current arrow in the opposite direction to the voltage arrow) and so the power absorbed by B is given by V I = W. For device A we need to reverse the direction of the current to conform to the passive sign convention. Therefore the power absorbed by A is V I = W. As must always be true, the total power absorbed by all components is zero. 3. The power absorbed is positive if the voltage and current arrows go in opposite directions and negative if they go in the same direction. So we get: (a) P V = +, P I =, (b) P V = +, P I =, (c) P V =, P I = +, (a) P V =, P I = +. In all cases, the total power absorbed is P V + P I =.. We can find a path (shown highlighted below) from the bottom to the top of the V X arrow that passes only through voltage sources and so we just add these up to get the total potential difference: V X = ( 3) + (+) + (+9) = +8 V. 5. If we add up the currents flowing out of the region shown highlighted below, we obtain I X 5 + =. Hence I X = A. 6. The three series resistors are equivalent to a single resistor with a value of = 8 kω. 7. The three series resistors are equivalent to a single resistor with a value of /+ /5+ / =.7 =.588 kω. 8. We can first combine the parallel k and 3 k resistors to give 3 +3 =. k. This is then in series with the k resistor which makes 5. k in all. Now we just have three resistors in parallel to give a total of /+ /5+ = /5..39 =.78kΩ. 9. The resistance is 8 6 = 5 kω Solution Sheet Page of

4 Ver 7 E. Analysis of Circuits (). [Method ]: The resistors are in series and so form a potential divider. The total series resistance is 7 k, so the voltages across the three resistors are 7 = V, 7 = V and 7 = 8 V. The power dissipated in a resistor is V R, so for the three resistors, this gives = mw, = 8 mw and 8 = 6 mw. [Method ]: The total resistance is is 7 k so the current flowing in the circuit is 7 = ma. The voltage across a resistor is IR which, in for these resistors, gives = V, = V and = 8 V. The power dissipated is V I which gives = mw, = 8 mw and 8 = 6 mw. The current through the voltage source is ma, so the power it is supplying is V I = = 8 mw. This is, inevitably, equal to the sum of the power disspipated by the three resistors: = 8.. [Method ]: The resistors are in parallel and so form a current divider: the ma will divide in proportion to the conductances: ms,.5 ms and.5 ms. The total conductance is.75 ms, so the three resistor currents are = ma,.75 = 6 ma and.75 = 3 ma. The power dissipated in a resistor is I R which gives = mw, 6 = 7 mw and 3 = 36 mw. [Method ]: The equivalent resistance of the three resistors is /+ /+ = / 7 kω. Therefore the voltage across all components in the parallel circuit is 7 = V. The current through a resistor is V R which gives = ma, = 6 ma and = 3 ma. The power dissipated in a resistor is V I which gives = mw, 6 = 7 mw and 3 = 36 mw. The power supplied by the current source is = 5 mw which as expected equals The resistors form a potential divider, so Y X = R. So we want + R = k. R + = R + = 6 3. The resistors form a potential divider, so Y X = R R R +R. So we want R +R = and R + R = MΩ. Substituting one into the other and cross-multiplying gives R = MΩ R = MΩ. Substituting this into the simpler of the two initial equations gives R = = 9 MΩ.. (a) 3 k =.5 k +.5 k = 3.3 k 33 k, (b) k = 3.9 k +, (c) 3.88 k = 3.9 k 33 k. To make an exhaustive search for creating a resistance of R, you need to consider two possibilities: (i) for two resistors in series, the largest of the two resistors must be in the range [ R, R] or (ii) for two resistors in parallel, the smallest resistor must be in the range [R, R]. In both cases there are at most four possibilities, so you need to consider up to eight possibilities in all. So, for example, for R = 3.5 k, we would consider the following possibilities: (i).5 k +.8 k = 3.3 k,.8 k +.8 k = 3.6 k,. k +. k = 3. k,.7 k +.8 k = 3.5 k and (ii) 3.9 k 33 k = 3.88 k,.7 k 5 k = k, 5.6 k k = 3.59 k, 6.8 k 6.8 k = 3. k. The choice with least error is the one given above. Since we are interested in % errors, we need to consider the ratio between resistor values. The largest ratio between successive resistors is the series is 5 =.5 (this includes the wraparound ratio of 8 =.). The worst-case percentage error will arise if our target resistance is the mean of these two values, 3.5. The percentage error in choosing either one is then =.%. Solution Sheet Page of

5 Ver 7 E. Analysis of Circuits (5) Key: [A]= easy... [E]=hard E. Circuit Analysis Problem Sheet (Lectures 3 & ). [B] Calculate V X and I X in the following circuit using (a) nodal analysis and (b) simplifying the circuit by combining parallel resistors.. [B] Calculate V X and I X in the following circuit using (a) nodal analysis and (b) simplifying the circuit by combining parallel resistors. 3. [C] Calculate V X in the following circuit using (a) nodal analysis and (b) superposition.. [C] Calculate V X in the following circuit. 5. [C] Calculate V X in the following circuit. 6. [C] Calculate V X in the following circuit. Problem Sheet Page of 3

6 Ver 7 E. Analysis of Circuits (5) 7. [C] Calculate V X in the following circuit. The value of the dependent current source is 99 time the current flowing through the V voltage source. 8. [C] In the following circuit calculate V X in terms of V and I using (a) nodal analysis and (b) superposition. 9. [C] Calculate V X and I X in the following circuit using (a) nodal analysis and (b) superposition.. [C] Determine an expression for I X in terms of V in the following circuit. Determine the value of V that will make I X =.. [C] Calculate V X in the following circuit using (a) nodal analysis and (b) superposition.. [C] Calculate V X in the following circuit which includes a dependent voltage source. Problem Sheet Page of 3

7 Ver 7 E. Analysis of Circuits (5) 3. [C] Find the equivalent resistance of the network shown below.. [D] Prove that if V AB =, then R = kω in the following circuit. The circuit is used to detect small changes in R from its nominal value of kω. Find an expression for V AB as a function of R. If changes in V AB of mv can be detected, what is the smallest detectable change in R. 5. [D] Calculate V X in the following circuit. You can either use nodal analysis directly or else simplify the circuit a little to reduce the number of nodes. 6. [D] Calculate V X in the following circuit which includes a floating dependent voltage source. Problem Sheet Page 3 of 3

8 Ver 686 E. Analysis of Circuits (5) E. Circuit Analysis Problem Sheet - Solutions Note: In many of the solutions below I have written the voltage at node X as the variable X instead of V X in order to save writing so many subscripts.. [Nodal analysis] KCL at node V X gives V X + V X + V X = which simplifies to 7V X 56 = from which V X = 8. [Parallel resistors] We can merge the Ω and Ω resistors to make one of + = 3 Ω as shown below. Now we have a potential divider, so V X = = 8 V. In both cases, we can now calculate I X = V X = A. Note that when we merge the two resistors, I X is no longer a distinct current on the diagram. Original Simplified. [Nodal Analysis] KCL at node V x gives 5 + V X + V X = which simplifies to + 5V X = from which V X =. [Parallel Resistors] We can combine the Ω and Ω resistors to make one of below. Now we have V X = 5.8 = V. + = 5 Ω as shown Original Simplified In both cases, we can now calculate I X = V X = A. In this question, you have to be a bit careful about the sign used to represent currents. Whenever you use Ohm s law, you must be sure that you use the passive sign convention (with the current arrow in the opposite direction to the voltage arrow); this is why the current through the.8 Ω resistor is 5 A rather than +5 A. Solution Sheet Page of 5

9 Ver 686 E. Analysis of Circuits (5) 3. [Nodal Analysis] KCL at node V x gives V X 6 + V X + = which simplifies to 5V X + = from which V X =. [Superposition] (i) If we set the current source to zero, then the Ω resistor connected to it plays no part in the circuit and we have a potential divider giving V X = 6 5 =.. (ii) We now set the voltage source to zero and then simplify the resultant circuit as shown below. Being careful with signs, we now get V X =.8 = 3.. Adding these two values together gives a final answer of V X =. Original I = V = V = simplified. We first label the nodes; we only need two variables because of the floating voltage source. KCL at X gives X X (Y 3) X Y = which gives X 9Y =. KCL at the supernode {Y, Y 3}gives Y 5 + Y X (Y 3) X (Y 3) + + = which gives 9X + 9Y = 97. Solving these two simultaneous equations gives X = and Y = We first label the nodes; there are only two whose voltage is unknown. Working in ma and kω, KCL at X gives X 3 + X Y 6 + = which gives 3X Y =. KCL at Y gives Y X 6 + Y + Y 6 = which gives X + 7Y =. Solving these simultaneous equations gives X = 8 and Y =. 6. We first label the nodes; since the two nodes having unknown voltages are joined by a fixed voltage source, we only need one variable. We write down KCL for the supernode {X, X 5} (shaded in the diagram) which gives (X 5) (X 5) + X 3 + X 9 = which simplifies to X = 35 or X = 75 V. Solution Sheet Page of 5

10 Ver 686 E. Analysis of Circuits (5) 7. There is only one node with an unknown voltage, namely X. However, there is a dependent current source, so we need to express its value in terms of the node voltages: 99I = 99 X 5 where we are expressing currents in ma. So now we can apply KCL to node X to obtain X + X X = which simplifies to 5X = 35 from which X = [Nodal Analysis] Using KCL at node X gives X V + X I = which we can rearrange to give X = V + I. [Superposition] If we set I = then we have a voltage divider in which X = V (see middle diagram). If we set V = then (see right diagram) we can combine the two parallel resistors as + = Ω and it follows that X = I. By superposition, we can add these two expression together to give X = V + I. Original I = V = 9. [Nodal Analysis] We can easily see that the A current flowing through the leftmost resistor means the top left node has a voltage of 8 (although actually we do not need to calculate this because of the isolating effect of the current source). Using KCL at the supernode {X, X} gives + X + X = which we can rearrange to give X = or X = 6. It follows that I X = 6 = 3 A. [Superposition] If we set I = then we have a voltage divider (since the resistors are in series) in which X = = and I X = A (see middle diagram). If we set V = then (see right diagram) we can combine the two parallel resistors as + = Ω and it follows that X = and, by current division, that I X = A. By superposition, we can add these two expression together to give X = 6 V and I X = 3 A. Original I = V = Solution Sheet Page 3 of 5

11 Ver 686 E. Analysis of Circuits (5). [Nodal Analysis] We first label the unknown node as X. Now, KCL at this node gives 6+ X 3 + X V 6 = which rearranges to give 3X = V + 36 from which X = 3 V +. Now I X = X 3 so I X = 9 V +. [Superposition] If we set I = (see middle diagram) then I X = V 9. If we set V = then (see right diagram) we have a current divider in which the 6 A current divides in proportion to the conductances. So I X = 6 /3 /3+ = 6 /6 3 =. Adding these results together gives I X = 9 V +. If V = 36 then I X =. Original I = V =. [Nodal Analysis] KCL at node X gives X + X + X ( 3) 3 = which rearranges to give 8X = from which X = 3. [Superposition] If we set the left source to zero (see middle diagram) then, the two parallel Ω resistors are equivalent to Ω and so we have a potential divider and X = 3 =.75. If we set the other source to zero (see right diagram) we can combine the Ω and 3 Ω parallel resistors to obtain =. Ω. We again have a potential divider giving X = +. = Adding these together gives X = = 3 V. Original U = U =. KCL at node Y gives Y + Y X 5 = which rearranges to X +6Y =. We also have the equation of the dependent voltage source: X = 6Y. We can conveninetly eliminate 6Y between these two to give X = and so X =. 3. We first pick a ground reference at one end of the network and label all the other nodes. The equivalent resistance is now V I. We assume that we know V and then calculate I. KCL at node A gives A V 5 + A B 5 + A 5 = from which 3A B = V or B = 3A V. KCL at node B gives B V 5 + B A 5 + B 5 = from which B 5A = V. Substituting B = 3A V into this equation gives 33A 5A = V or A = 8 V which in turn gives B = 3A V = 8 8V. The current I is the sum of the currents throught the rightmost two 5 Ωresistors: I = A 5 + B 5 = 7V. So the equivalent resistance is V I = 7 Ω. Solution Sheet Page of 5

12 Ver 686 E. Analysis of Circuits (5). If V AB = then no current flows through the k resistor, so the two vertical resistor chains form potential dividers. In the leftmost chain, V A = + = 5 V. Since V AB =, V B = V A = 5 = R +R which implies that R = k. KCL at nodes A gives A + A + A B = which gives A B = or B = A 5. KCL at node B now gives B + B R + B A = into which we can substitute the expression for B to get A 5 + A 5 R + A 5 = from which A = +5R +R. Substituting this into B = A 5 gives B = 5R 5R +R and hence A B = +R. If we can detect a value of A B = mv =. then the corresponding value of R is the solution to 5R +R =. which gives 5.R = from which R = Ω which is a change of.8 Ω or.%. 5. We label the nodes as shown below (using X instead of V X for ease of writing). Note that when we have labelled the upper node of the floating voltage source as Y we can label the lower node as Y + 3 and do not need another variable. KCL at node X gives X 9 + X Z 3 + X Y = which gives 3X 3Y Z =. KCL at node Z gives Z + Z X Z Y = which gives X 3Y +8Z = 5. Finally, KCL at the supernode {Y, Y + 5} gives Y X + Y +5 Z = from which X 3Y + Z = 3. Solving these three simultaneous equations gives X =, Y = and Z = 8. Original Simplified An alternative approach is to notice that the three rightmost components are in series and so you can reorder them without affecting the rest of the circuit to give the simplified circuit shown above. Now, we only have two unknowns and hence only two simultaneous equations to solve. KCL at node X W X W X gives X 9 + gives W X 6 + W +5 X give X = and W = W +5 6 = which gives 6X 3W = 87. KCL at the supernode {W, W +5} = from which 3X 6W = 75. These equations are easily solved to 6. Since the floating voltage source is a dependent voltage source, we need to label its two ends with separate variables (see below). We now write down a KCL equation for the supernode shown shaded: Y 8 + Y + X X 6 = which simplifies to X + 8Y = 6. We also need to express the voltage source value in terms of nodal voltages: Y X = 8I = 8 X 6 = 3 X which rearranges to give Y = 7 3 X. Substituting this in the previous equation gives X X = 6 which simplifies to 5X = 6 from which X =. Solution Sheet Page 5 of 5

13 Ver 379 E. Analysis of Circuits () Key: [A]= easy... [E]=hard E. Circuit Analysis Problem Sheet 3 (Lectures 5, 6, 7 & 8). [B] Calculate the Thévenin and Norton equivalent networks at the terminals A and B for each of the following. (a) (b). [B] Use nodal analysis to calculate an expression for A in Fig. in terms of I and then rearrange this to give I in terms of A. Show how these expressions are related to the Thévenin and Norton equivalent networks at the terminals A and B. Fig. Fig. 3 Fig. 3. [B] Determine X in Fig. 3 when (a) U = +5 V and (b) U = 5 V. Assume that the diode has a forward voltage drop of.7 V.. [B] In Fig., calculate I and the power dissipation in the resistor and in the diode. Assume that the diode has a forward voltage drop of.7 V. 5. [B] Find the gains X U and Y U in the following circuits: (a) (b) 6. [C] Calculate the Thévenin and Norton equivalent networks at the terminals A and B in Fig. 6 in two ways (a) by combining resistors to simplify the circuit and (b) by using nodal analysis to express A in terms of I. Fig. 6 Problem Sheet 3 Page of 5

14 Ver 379 E. Analysis of Circuits () 7. [C] Find the current I in Fig. 7 in two ways: (a) by nodal analysis and (b) by combining the leftmost three components into their Thévenin equivalent. Fig. 7 Fig [C] For what value of R in Fig. 8 will the power dissipation in R be maximized. Find the power dissipation in R in this case. 9. [C] Find the Thévenin equivalent of the circuit between nodes A and B in two ways: (a) by performing a sequence of Norton Thévenin transformations and (b) using superposition to find the open-circuit voltage and combining resistors to find the Thévenin resistance.. [C] State whether the feedback in the following circuits is positive or negative: (a) (b) (c) (d). [C] Find the gain X U in the following circuit: Problem Sheet 3 Page of 5

15 Ver 379 E. Analysis of Circuits (). [C] Find expressions for X, Y and Z in terms of U, U and U 3. (a) (b) (c) 3. [C] Choose values of R and R in Fig. 3 so that X = U 3U. Fig. 3 Fig.. [C] Find an expression for Y in Fig. in terms of U and U. 5. [C] In the circuit diagram, the potentiometer resistance between the slider and ground is a kω where a. Find the gain of the circuit, X U as a function of a. What is the range of gains that the circuit can generate as a is varied. 6. [C] By replacing the rightmost three components in Fig. 6 by their Thévenin equivalent, find X when (a) U = V and (b) U = 5 V. Assume that the diode has a forward voltage drop of.7 V. Determine the value of U at which the diode switches between operating regions. Fig. 6 Fig [C] In the block diagram of Fig. 7, F, G, H represent the gains of the blocks. Find the overall gain Y X. Fig. 7 Problem Sheet 3 Page 3 of 5

16 Ver 379 E. Analysis of Circuits () 8. [D] The circuit of Fig. 8 includes a dependent current source whose value is proportional to the current J. Find the Thévenin equivalent of the circuit by two methods: (a) use nodal analysis including the current I and express the voltage V AB as a function of I and (b) assume I = and find (i) the open circuit voltage, V AB as a function of U and (ii) the short-circuit current (with A joined to B) as a function of U. Hence find the Thévenin equivalent of the circuit. 9. [D] Choose resistor values in Fig. 9 so that (a) X = U U + 6 U and (b) the Thévenin resistance between X and ground is 5 Ω. Why would the question be impossible if it had asked for X = U U + 3 U? Fig. 9 Fig.. [D] Choose resistor values in Fig. so that (a) the Thévenin between X and ground is 5 Ω and (b) it is possible to set X to,, 3 or V by setting the switches appropriately.. [D] State whether the feedback in the following circuits is positive or negative: (a) (b) Problem Sheet 3 Page of 5

17 Ver 379 E. Analysis of Circuits (). [D] Find an expression for Z in Fig. in terms of U and U. Fig. Fig [D] The circuit of Fig. 3 is called a Howland current source. Show that the circuit has negative feedback. Use nodal analysis to determine the current I and show that it does not depend on R.. [D] A non-linear device has a characteristic Y = X for inputs in the range X. To improve its linearity, the device is placed in a feedback loop using an op-amp with a gain of A as shown in Fig.. Determine an expression for Y in terms of U and A. Simplify the expression by using the Taylor series approximation: v + w ( v + w v w valid for v w. Estimate how large A 8v ) must be to ensure that Y U=.5 =.5 Y U= to within % of Y U=. Fig. 5. [D] The diodes in Fig. 5 have a characteristic I = k exp V V T where V T = 5 mv. (a) For the circuit of in Fig. 5(a), find an expression for U in terms of X assuming that X >. (b) For the circuit of in Fig. 5(b), find and expression for Y in terms of X assuming that X >. Fig. 5(a) Fig. 5(b) 6. [C] We normally assume that the current at the inputs to an opamp are negligible. However this is not always true and this question investigates the effect of non-zero input bias currents. If I B = na, find expressions for X U and Y U in the circuits below. Explain the advantage of circuit (b) and derive a general principal that should be followed when using opamps having non-negligible input bias currents, I B. (a) (b) Problem Sheet 3 Page 5 of 5

18 Ver 9 E. Analysis of Circuits (6) E. Circuit Analysis Problem Sheet 3 - Solutions. (a) Thévenin voltage equals the open circuit voltage is V (from potential divider). To obtain the Thévenin/Norton we set the voltage source to (making it a short circuit) and find the resistance of the network to be =.8 Ω. From this the Norton current is.8 = 5 A. This may also be found directly from the short-circuit current of the original circuit. (b) The open-circuit voltage is 8 V (since the A current flows anticlockwise). To obtain the Thévenin/Norton resistance, we set the current source to zero (zero current implies an open circuit), so the resultant network has a resistance of Ω. The Norton current is 8 = A; this may also be found by observing that the short-circuit current (flowing into node A) is + A.. KCL at node A gives A 5 + A I = from which 5A I = which we can rearrange to give A = +.8I = V T h + R T h I. We can also rearrange to give I = A = I Nor + R Nor A. 3. (a) When U = 5, the diode is on and so X = U.7 =.3. To check our assumption about the diode operating region, we need to calculate the current through the diode; this equals.3 ma which is indeed positive. (b) When U = 5, the diode is off, so the current through the resistor is zero and X =. As a check, the voltage across the diode is U X = 5 which confirms our assumption that it is off.. As in part (a) of the previous question, I =.3 ma. The power dissipation in the diode is therefore V D I =.7.3 = 3. mw. The power dissipation in the resistor is I R = 8.5 mw. 5. Circuit (a) is an inverting amplifier with gain X U = =. Circuit (b) is a non-inverting amplifier with gain Y U = + = +. Another way to see this is to notice that, since the opamp inputs draw no current, the potential divider means that the -ve opamp input is at Y and, since the negative feedback ensures the opamp terminals are at the same voltage, U = Y. 6. [Method - circuit manipulation] To calculate the Thévenin equivalent, we want to determine the open-circuit voltage and the Thévenin resistance. To determine the open-circuit voltage, we assume that I = and calculate V AB. Since I =, we can combine the Ω and 6 Ω resistors to give 7 Ω and then combine this with the 6 Ω resistor in parallel to give 3 Ω. We now have a potential divider so the voltage at point X is 63 /3. This is then divided by the Ω and 6 Ω resistors 3+ = 98 /3 3 to give an open-circuit voltage of = 8 V. The Thévenin/Norton resistance can be found by short-circuiting the voltage source to give 3 Ω in parallel with 6 Ω which equals Ω. This is then in series with Ω(to give 3 Ω ) and finally in parallel with 6 Ω to give Ω. [Method - Nodal Analysis]. We can do KCL at node X (see diagram below) to get X X 6 + X A = which simplifies to 9X 6A = 6 or 3X A =. We now do KCL at A but include an additional input current I as shown in the diagram. This gives A X + A 6 I = from which 7A 6X = 6I. Substituting for 6X = A + 8 gives 3A = 8 + 6I or A = 8 + I. This gives the Thévenin voltage as 8 and the Thévenin/Norton resistance as Ω.Hence the Norton current is A. 7. (a) X gives X + X + X = from which 7X = 56 X = 8 I = X = ma. (b)finding the Thévenin equivalent of the left three components: we consider the two resistors as a potential divider to give V T h = 5 =. V. Setting the source to zero (short circuit) gives V R T h = = 8 Ω. Hence I = T h R T h + =..8 = ma. Solution Sheet 3 Page of 5

19 Ver 9 E. Analysis of Circuits (6) 8. From question 7, the left three components have a Thévenin equivalent: V T h =. V and R T h = 8 Ω. It follows that the maximum power will be dissipated in R when R = R T h = 8 Ω (see notes page 5-8). Since the voltage across R will then be V T h the power dissipation will be R T h VT h = 39. mw. 9. (a) As shown in the sequence below, we first combine the current source with the Ωresistor, then combine the four series components into a single Thévenin equivalent and finally find the Thévenin equivalent of the simple network (using parallel resistors for R T h and a potential divider for V T h ). (b) Setting the voltage source to zero gives us the first diagram. Combining (6 + ) = 3 so X =. 3 = 3. V. it follows (potential divider) that A = 3. 6 = V. Now setting the current source to zero gives the second diagram and we have a potential divider giving V AB = = 3 V. Superposition now gives us V AB = V T h = + 3 = V. To find R T h we set both sources to zero and find the resultant resistance of (6+) = = 5 Ω.. (a) Negative, (b) Positive, (c) Negative, (d) Positive. In simple circuits like these, you can just see which terminal the output feeds back to.. The best way to think of this circuit is as a potential divider with gain Y U = 3 followed by a non-inverting opamp circuit with gain X Y = + 6 = 7. The combined gain is then X U = 3 7 = 7 3. Solution Sheet 3 Page of 5

20 Ver 9 E. Analysis of Circuits (6). (a) You can either recognise this a a standard inverting summing amplifier with gain X = ( U + U ) = U U or else apply KCL at the +ve input terminal with the assumption that negative feedback will ensure that this terminal is at the same voltage as the ve terminal U i.e. V. This gives: + U + X = from which X = U U. (b) The network connected to the +ve terminal is a weighted averaging circuit (page 3-7 of the notes) so V + = 3 U + 3 U. The opamp circuit itself is a non-inverting amplifier with a gain of + 5 = 6. So, Y = 6 ( 3 U + 3 U ) = U + U. ((c) [Superposition method] Following the method of part (b) above, if U 3 =, we have Z = 5 5 U + 5 U ) = U + U. If, on the other hand, U = U =, then V + = and so we have an inverting amplifier with a gain of =. Hence Z = U 3. Combining these gives Z = U + U U 3. [Nodal analysis method] The top two resistors are a weighted average circuit so V + = 5 U + 5 U. Now, assuming that V = V +, we do KCL at V to give 5 U+ 5 U U3 U + U U 3 Z = giving Z = U + U U U+ 5 U Z = from which 3. We can use superposition. If U =,then V + = and we have an inverting amplifier with a gain X U = 6 R. The question tells us that this must equal 3 so we must have R =. Now, if U =, 6 the circuit consists of a potential divider with a gain of R +6 followed by a non-inverting amplifier with a gain of + 6 R =. The combined gain must equal (from the question) so the potential divider must have a gain of which means R = 6 kω.. The first opamp is non-inverting with a gain X U = + 5 =.. We can use superposition to find Y Y : If U =,then X = and we have a non-inverting amplifier with a gain of U = + 5 = 6. If, on the other hand, U =, then we have an inverting amplifier and Y X = 5 = 5. It follows that Y U = Y X X U = 5. = 6. Combining both portions of the superposition, Y = 6U 6U = 6(U U ). This is therefore a differential amplifier (whose output is (U U )) that draws almost no current from either if its inputs. A better circuit is given in question. 5. The potentiometer is a potential divider and so V + = au. Assuming that V = V +, we can do KCL at V to get au U + au X = from which X = (a ) U. For a =, the gain is and when a =, the gain is +. So the circuit can generate any gain between these two extremes. 6. The Thévenin voltage is 3 V (potential divider) and the Thévenin resistance is 3 =.75 Ω as shown in the diagram below. Note that if you drawn the Thévenin voltage source the other way around (with + at the bottom) then the Thévenin voltage will be +3 V; this is an equally valid solution. (a) If U = V, the diode is forward biassed and X gives X ( 3).75 + X (.7) 3 = from which X =.5. The current through the diode is X ( 3).75 = 63 ma which is > confirming our guess about the diode operating region. (b) If U = 5 V, the diode is off and so X = 3 (since no current flows through the 75 Ω resistor). The diode forward voltage is ( U) X =. This is <.7 confirming our operating region guess. The diode switches regions when both operating region equations are true: I D = and V D =.7. The current equation implies X = 3 while the voltage equation (and the zero current through the 3 k resistor) implies X = U.7. Combining these gives U =.3. Extreme care with signs is needed in this question. Solution Sheet 3 Page 3 of 5

21 Ver 9 E. Analysis of Circuits (6) 7. There is only one feedback loop in this circuit: from W back to the input adder. We can write W = F G (X W ) from which W = F G +F GX. Then Y = F (X W ) + HW. From the first equation (X W ) = W F G so we can substitute this in to get Y = W G + HW = ( G + H) F G F +F GH +F GX = +F G X. 8. (a) [Nodal analysis including I ] We have J = U A, so the current source value (expressed in terms of node voltages) is 9J =.9 (U A). The KCL A is A U.9(U A) I = from which 5 (A U) = I which gives A = I + U = R T h I + V T h. So the Thévenin voltage is U and the Thévenin resistance is kω. (b) In this method, we assume I = and calculate the open-circuit voltage and short-circuit current. For the open-circuit voltage, we do KCL at A and obtain A U.9(U A) = from which 5 (A U) = so A = V T h = U. For the short-circuit current, we join A and B and the current is then V T h R T h = U +.9U =.5U. Hence R T h = kω. 9. From the notes (page 3-7) X = UG+UG+U3G3 G +G +G 3. The equations are made much easier to solve because we know that the Thévenin resistance must be 5 Ω. The Thévenin resistance is just the parallel combination R R R 3 = G +G +G 3 = ms. Hence X = U U + 6 U = 5 (U G + U G + U 3 G 3 ) where the first expression is given in the question and the second comes from substituting for G + G + G 3. Identifying the coefficients in this equation gives G = 3, G = 5 and G 3 = from which R = 3, R = 5 and R 3 =. In a weighted average circuit, the coefficients must sum to ; thus = but so the latter set of coefficients is inadmissible.. From the notes (page 3-7) X = UG+U3G3+5G G +G +G 3+G. where U and U 3 can equal or 5. As in the previous question, we are told that G + G + G 3 + G = ms. Hence X = 5 (U G + U 3 G 3 + 5G ). The only way that we can obtain the required voltages is if switching U causes X to change by V and switching U 3 causes X to change by V (or vice versa). Thus we need X = 5 U + 5 U 3 + ; it is easy to see that this satisfies the required output voltages. Now equating coefficients, we get 5G = 5, 5G 3 = 5 and 5G = from which R = G = 5, R 3 = G 3 = 5 and R = G = 5. Finally, G =. G G 3 G =. so R = G = 5.. We need to determine how the differential input to the opamp (V + V )depends on X. We are not interested in how it depends on U so it is convenient to set U =. So, with this assumption, we get potential divider equations: (a) V + = + X = 3 X and V = + X = 3 X. Hence V + V = 3X which, since the coefficient is positive means positive feedback. (b) V + = + X = 3 X and V = + X = 3 X. Hence V + V = 3X which, since the coefficient is negative means negative feedback.. We can split the circuit up into two independent parts because the opamp outputs are voltage sources whose voltage is not affected by how many other things are connected to them. Note that all three opamps have negative feedback and so we can assume that V + = V. For the first part, we have A = U and B = U. A therefore gives U X 5 + U U = which gives X = 6U 5U. Similarly, B gives U Y 5 + U U = which gives Y = 6U 5U. The second part of the circuit is a differential amplifier (see page 6-9 of the notes) for which Z = 6 (Y X). Substituting in the expressions for X and Y gives Z = 3 ((6U 5U ) (6U 5U )) = 3 (U U ) = 33 (U U ). Solution Sheet 3 Page of 5

22 Ver 9 E. Analysis of Circuits (6) 3. We can verify that the opamp has negative feedback since (if we set the V source to zero), we have V = Y but V + = αy where α = R + R <. So, we can assume that V + = V = X. V gives X ( ) + X Y = which gives X Y = X. V + gives X + X Y + I =. Substituting X Y = X gives X + X + I = which simplifies to I = 5. Notice that this does not depend on R, so we have constructed a current source.. From the block diagram, we can deduce Y = X = AU AY from which Y + AY AU =. Solving this quadratic equation gives Y (U) =.5 ( A + A + AU ). Notice that only one of the two roots will result in Y being positive. ( Provided ( that )) A U, we can use the Taylor series approximation to give Y (U).5 A + A + U A U A = U U A. From this, Y () = A and Y (.5) = A =.5Y () + A. We would like the error,.5 A to be % of Y (), i.e..5 A. ( A). Solving this gives A (a) The terminal of the opamp is a virtual earth so the current through the resistor is X R. All the current flows through the diode whose voltage is U. Therefore we have X U R = k exp V T from which U = V T ln X kr. (b) The second opamp is a non-inverting amplifier with a gain of 3 so W = 3U = 3V T ln X kr. For the third opamp, the current thorough the diode is Y W R = k exp V T = k exp ( ( 3 ln kr) X = k X 3. kr) From this we find that Y = k R X 3. This we have made a circuit that cubes its input voltage (times a scale factor). 6. (a) Despite the current I B, it is still the case that V + = and so, because of the negative feedback, P = also. P gives U +I B + Y = from which U +I B Y = or Y = U +I B. Substituting I B =. ma gives Y = U +. so there is an output error of mv. (b) This time, the current I B flows through the 8 kω resistor so V + = 8I B. As before, we assume that Q = V + also. Then Q gives Q U +I B + Q Y = from which U +I B Y +5Q =. Substituting Q = 8I B gives U + I B Y I B = which simplifies to Y = U. Thus in this circuit, the bias currents do not cause any error. The moral is that when designing opamp circuits, you should try to make the Thévenin resistance seen by the two input terminals the same. If you achieve this, and if the bias currents are the same at both inputs (usually approximately true) there will be no resultant errors. Solution Sheet 3 Page 5 of 5

23 Ver 55 E. Analysis of Circuits (5) Key: [A]= easy... [E]=hard E. Circuit Analysis Problem Sheet (lectures 9 & ) Note: A dimensioned sketch should show the values on the x and y axes corresonding to significant places on the corresponding graph.. [B] For each of the following waveforms, determine the corresponding phasor in both the form a + jb and r θ. (a) 8 cos ωt. (b) 3 cos ωt + sin ωt. (c) cos ( ωt + π ). (d) 8 sin ωt. (e) cos ωt. (f) sin ( ωt π ). (g) 8 cos ( ωt + π ) + 5 sin ( ωt π ).. [B] For each of the following phasors, determine the corresponding waveform in both the form a cos ωt + b sin ωt and a cos (ωt + θ). (a), (b), (c) 3j, (d) j, (e) j, (f) 3 j, (g) e j π, (h) e j π [B] For each of the following cases say which of the two waveforms or phasors leads the other: (a) sin ωt and cos ωt. (b) sin (ωt + π) and cos ωt. (c) sin (ωt π) and cos ωt. (d) ( + j) and ( + j). (e) ( + j) and ( j). (f) ( + j) and ( j). (g) and 35.. [B] Draw a dimensioned sketch of the waveform of i in the circuit of Fig. (a) when v has the waveform shown in Fig. (b). Fig. (a) Fig. (b) 5. [B] For each of the circuits shown in Fig. 5(a)-(d) determine the average value of y(t) when x(t) = + cos ωt for some non-zero frequency ω. Fig. 5(a) Fig. 5(b) Fig. 5(c) Fig. 5(d) 6. [B] Find the value of a single inductor equivalent to the circuit shown in Fig. 6. Problem Sheet Page of 3

24 Ver 55 E. Analysis of Circuits (5) Fig. 6 Fig [B] Find the value of a single capacitor equivalent to the circuit shown in Fig. 7 given that each of the capacitors has a value of µf. 8. [B] Find the average value of v in the circuit of Fig. 8 if u(t) = + 3 cos ωt. Fig. 8 Fig [B] Find the average value of v in the circuit of Fig. 9 if u(t) = 8 cos ωt.. [B] Find the complex impedance of the circuit shown in Fig. for (a) ω =, (b) ω =, (c) ω = and (d) ω =. Fig. Fig. (a) Fig. (b) Fig. (c). [B] The components in Fig. are labelled with their impedances. Calculate both the complex impedance and the complex admittance for each of the three networks.. [B] The components in Fig. are labelled with their impedances. Determine the values of a parallel inductor and resistor that will have the same overall impedance at (a) khz and (b) khz. Hint: first calculate the admittance of the original network. Fig. Fig. 3(a) Fig. 3(b) 3. [C] Draw a dimensioned sketch of the waveform of v(t) in the circuit of Fig. 3(a) when i has the waveform shown in Fig. 3(b). Problem Sheet Page of 3

25 Ver 55 E. Analysis of Circuits (5). [C] The three current i, i, i 3 in Fig. are equal to 5 cos ( ) ( ) ωt + 3π, cos ωt + π and 8 cos ωt but not necessarily in that order. Determine which current is which and find both the phasor I and time-waveform i(t). Fig. Fig. 7(a) Fig. 7(b) 5. [D] Draw a dimensioned sketch of the waveform of i in the circuit of Fig. 5 when v has the waveform shown in Fig. (b) given that at time t =, (a) i() = and (b) i() = A. 6. [D] Draw a dimensioned sketch of the waveform of v in the circuit of Fig. 6(a) when i has the waveform shown in Fig. 6(b) given that at time t =, (a) v() = and (b) v() = 5 V. Fig. 5 Fig. 6(a) Fig. 6(b) Fig [D] In the circuit of Fig. 7(a), the voltage v has the periodic waveform shown in Fig. 7(b) with a period of µs and an amplitude of V. (a) State the duty cycle of v. (b) Determine the average value of x. (c) Determine the average value of i R. (d) Determine the average value of i L. (e) Assuming that x is constant (at its average value), draw a dimensioned sketch of the waveform of i L (t) and determine its maximum and minimum values. (f) Assuming that x is constant (at its average value), determine the average, positive peak and negative peak of the powers absorbed by each of R, L and C. 8. [D] In the circuit of Fig. 8, the output logic levels from the inverter are V and 5 V and the inverter has a maximum output current of ± ma. The inverter senses a low voltage when its input is <.5 V. If x changes from logic to logic, determine the delay until z changes. Ignore the inverter input currents and any delays inside the inverters themselves. Problem Sheet Page 3 of 3

26 Ver 667 E. Analysis of Circuits (5) E. Circuit Analysis Problem Sheet - Solutions. (a) 8, (b) 3 j = 5.93 ( 53 ), (c). + j. =.79 (5 ), (d) j8 = 8.57 (9 ), (e) = 3. (8 ), (f), (g). + j. = 3.79 (5 ).. (a) cos ωt, (b) cos ωt = cos (ωt + π), (c) 3 sin ωt = 3 cos ( ) ( ) ωt + π, (d) sin ωt = cos ωt π, (e) cos ωt sin ωt =. cos ( ) ωt + 3π, (f) 3 cos ωt + sin ωt = 5 cos (ωt.93), (g) sin ωt = cos ( ) ( ) ωt + π, (h) 3.6 cos ωt + sin ωt = cos ωt π (a) cos ωt by π, (b) sin (ωt + π) by π, (c) sin (ωt π) by π : note that sin (ωt + π) and sin (ωt π) are actually the same waveform, (d) ( + j) by.3 rad, (e) ( + j) by π, (f) ( j) by π, (g) by. Because angles are only defined to within a multiple of 36, you always need to be careful when comparing them. To find out which is leading, you need to take the difference in phase angles and then add or subtract multiples of 36 to put the answer into the range ±8. Note that a sine wave is defined for all values of t (not just for t > ) and so there is no such thing as the first peak of a sine wave.. i = C dv (see Fig. ) dt. dv dt is 3 V/s for the first ms and 6 V/s for the next ms. So i = +5 or 3 ma. C t (ms) Fig. 5. The average value of x(t) is X = (note that we use capital letters for quantities that do not vary with time). For averages (or equivalently for DC or ω = ) capacitors act as open circuit and inductors as short circuits; this gives the simplified circuits shown below. So this gives (a) Y = 3, (b) Y = X =, (c) Y = X =, (d) Y = X =. Fig. 5(a) Fig. 5(b) Fig. 5(c) Fig. 5(d) 6. + = =. + = 8. 8 = 6 mh. 7. C 9 and C are short-circuted and play no part in the circuit. We can merge series and parallel capacitors as follows: C,5 =.5, C 7,8 =, C,3 =. Now merge C,5 with C 6 to give C,5,6 =.5 and merge this with C 7,8 = to give C,5,6,7,8 = 6 7. Now merge this with C,3 = to give C,3,,5,6,7,8 = 7. Finally merge this with C = to give C = 7 µf. 8. To determine average values, we can treat C as open circuit and L as short circuit. The original circuit simplifies to that shown in Fig. 8. So we have a simple potential divider and v = ū = V where the overbar denotes average value. Fig. 8 Fig. 9 Solution Sheet Page of 3

27 L Ver 667 E. Analysis of Circuits (5) 9. To determine average values, we can treat C as open circuit and L as short circuit. The original circuit simplifies to that shown in Fig. 9 and so v = ū = 8 V where the overbar denotes average value.. Z = R S + R P jωl. (a) Z = R S =, (b) Z = + j = + j = , (c) Z = + j = + j =. 86, (d) Z =.. We denote the impedance by Z and the admittance by Y = Z. (a) Z =. +.9j and Y =.5.33j, (b) Z = 3j and Y =.6 +.j, (c) Z = and Y =.5. Notice that the imaginary part of the impedance (the reactance) is positive for inductive circuits and negative for capacitive circuits, but that the imaginary part of the admittance (the susceptance) has the opposite sign. In part (c), the impedances of the inductor and the capacitor have cancelled out leaving an overall impedance that is purely real; because the impedances are frequency dependent, this cancellation will only happen at one particular frequency which is called the network s resonant frequency. I strongly advise you to learn how to do these complex arithmetic manipulations using the built-in capabilities of the Casio fx-99.. (a) ω = 683 so the impedance is Z = + 3j. Taking the reciprocal gives Y = Z =.9.89jmS. Since parallel admittances add, the parallel component values must be R P =.9 = 87Ω and L P =.89ω = 55. mh. For case (b), we now have ω = 683 and, following the same argument, we get R P = 98.8 kω and L P = 5.5 mh. As the frequency goes up, the series resistor becomes a less significant part of the total impedance and its effect becomes less. This means that L P becomes approximately equal to the original inductance and R P becomes larger. 3. v = L di Fig. 3) dt. di dt is 3 A/s for the first ms and 6 A/s for the next ms. So v = +6 or mv. (see. If we define the voltage phase to be φ will be i = φ +, φ π < i < φ +, i 3 = φ + π as the CIVIL mnemonic reminds us. Thus i 3 will have the most positive phase shift. It follows that i = cos ( ) ωt + π, i = 8 cos ωt and i 3 = 5 cos ( ) ωt + 3π and that φ = π. As phasors these are I =. + j. = 5, I =.8, I 3 = j3.5 = Adding these together gives I =.7 + j.95 = 5.3 (8 ). So i(t) =.7 cos ωt.95 sin ωt = 5 cos (ωt +.3) A. t 5. i(t) = i() + L v(τ)dτ where v(τ) = 3τ for τ ms and v(τ) = 36 6τ for τ= ms τ 6 ms (obtain this formula by finding the straight line equalling at τ = ms and at τ = 6 ms). So, for the first segment, i = L 5t which reaches A at t =.. For the second segment, i = L ( 36t 3t ) + c. To find c, we force i = at t =.. This gives c = 36 A. When t =.6 we then get i = 8 A. For part (b), we just add A to the curve. (see Fig. 5) 6. v = C idt where i = 3t or i = t. So, for the first segment, v = C.5t which reaches.8 V at t =.. For the second segment, v = C ( 36 3 t 3t ) + a. To find a, we force v =.8 at t =.. This gives a =. V. When t =.6 we then get v(t) = 7. V. For part (b), we just subtract 5 V from the curve. (see Fig. 6) 3 5 C - 6 t (ms) -5 6 t (ms) - 6 t (ms) Fig. 5 Fig. 6 Fig (a) The duty cycle is.5 = 5%, (b) x = v = = 5 V, (c) ī R = x R = 5 ma, (d) Since ī C =, ī L = ī R = 5 ma, (e) The voltage across the inductor is v x = L di L dt. So when v =, di L dt = 5 L = 7.5 ka/s. So the total change in i L over the µs interval is 7.5 ma. It follows that i L varies from its average value of 5 maby ±3.75 maand has minimum and maximum values of.5 and 8.75 ma (see Fig. 7(a)). (f) Average powers are P R = 5 mw, P L = P C =. Max powers are P R = 5 mw, P L = v L i L = = 3.5 mw, P C = v C i C = = 8.75 mw. Min powers are P R = 5 mw, P L = v L i L = = 3.75 mw (see Fig. 7(b)), P C = v C i C = = 8.75 mw (see Fig. 7(c)). Note that during the time that it is positive Solution Sheet Page of 3

28 Ver 667 E. Analysis of Circuits (5) (.5 < t <.5 ms), the average value of i C is ı C =.375 ma and so the total rise in v C will be v C = ı C t C =.375 = 75 mv (i.e. ±38mVaround its mean) which is small compared to its mean value of 5 V; this justifies the assumption that it is constant t (µs) t (µs) t (µs) Fig. 7(a) Fig. 7(b) Fig. 7(c) 8. When x changes from low to high, y will change from high to low. The maximum current is ma so dy dt = i 3.5 C = 5 MV/s. So the time to fall from 5 V to.5 V is 5 6 = 7 ns. Solution Sheet Page 3 of 3

29 Ver 78 E. Analysis of Circuits (6) Key: [A]= easy... [E]=hard E. Circuit Analysis Problem Sheet 5 (Lectures, & 3) Note: A sketch should show the values on the x and y axes corresonding to significant places on the corresponding graph.. [C] For each of the circuits in Fig. (i)-(vi), (a) Find the transfer function Y X (jω). (b) Find expressions for the low and high frequency asymptotes of H (jω). (c) Sketch the straight line approximation to the magnitude response, H (jω), indicating the frequency (in rad/s) and the gain of the approximation (in db) at each of the corner frequencies. Fig. (i) Fig. (ii) Fig. (iii) Fig. (iv) Fig. (v) Fig. (vi). [C] Sketch a straight line approximation for the phase response, H (jω), of the circuit in Fig. (v), indicating the frequency (in rad/s) and phase (in rad) at each of the corner frequencies. k(jω) 3. [B] A C-weighting filter in audio engineering has the form H (jω) = where a = 9 (jω+a) (jω+b) and b = rad/s. Calculate k exactly so that H (jω) = at ω = π rad/s.. [C] Design circuits with each of the magnitude responses given in Fig. (i)-(iii) using, in each case, a single capacitor and appropriate resistors. Fig. (i) Fig. (ii) Fig. 5. [B] Express each of the following transfer functions in a standard form in which the numerator and denominator are factorized into linear and quadratic factors of the form (jω+p) and ) ((jω) + qjω + r where p, q and r are real with an additional numerator factor of the form A(jω) k if required. Quadratic terms should be factorized if possible. (a) ω jω 3 ω +ω (b) (+ω ) ( ω )+jω (c) (jω) +jω+ (jω) +jω+ (d) jω+6(jω) +5 Problem Sheet 5 Page of 3

30 Ver 78 E. Analysis of Circuits (6) 6. [B] Without doing any algebra, determine the low and high frequency asymptotes of the following transfer functions: (a) ω jω 3 +ω (b) (jω)3 +3 (jω) + (c) jω((jω)6 +3)(5(jω) 3 +jω+3) ((jω) 5 +)((jω) 5 +5) (d) jω+6(jω) 7. [D] A circuit has a transfer function whose low and high frequency asymptotes are A (jω) α and B (jω) β respectively. What constraints can you place on α and β if you know that the magnitude of the transfer function is less than G at all frequencies where G is a fixed real-valued constant. 8. [C] For each of the following transfer functions, sketch the straight line approximation to the magnitude response and determine the gain of this approximation at ω = rad/s: (a) (b) (+jω /5) (+ jω /), (c) 3jω(+ jω /5) (+ jω /)(+ jω /)(+ jω /5). 5(+ jω /5) (+ jω /)(+ jω /), ( 9. [C] The frequency response of the circuit in Fig. 9 is given by jω p ) where the corner ( jω P ) +ζ( jω p )+ frquency, p = ζrc rad/s. Using capacitors of value C = nf, design a filter with ζ =.5 and a p corner frequency of π = khz. Determine the value of the transfer function at ω π = Hz, khz and khz. Gain (db) A B C Frequency (krad/s) Fig. 9 Fig. (a) Fig. (b). [D] A high-pass Butterworth filter of order N consists of N cascaded copies of Fig. 9 all having the same corner frequency, but with the k th stage having ζ k = cos (( ) ) k N π where k =,..., N. ( cascaded means you connect the output of one stage to the input of the next). Design a th order high-pass Butterworth filter with a corner frequency of khz. Write an expression for its transfer function and sketch its magnitude response using just the high and low asymptotes. Butterworth filters are widely used because they have a very smooth magnitude response without any peaks; the transfer function satisfies H(jω) = ( ω p ) N ( ω p ). N +. [C] The circuit of Fig. (a) is a high-pass filter whose magnitude response is marked A in Fig. (b). Using the filter transformations described in lectures, design filters with the magnitude responses marked B and C on the graph. Relative to A, these are respectively shifted up in frequency by a factor of 5 and reflected in the axis ω =,.. [C] (a) Find the resonant frequency, ω r, at which the impedance of the network in Fig. (i) is real. (b) Determine the Q of the circuit at ω r. (c) Find R P and L P in the circuit of Fig. (ii) so that the two networks have the same impedance at ω r. Fig. (i) Fig. (ii) Fig. 3 Problem Sheet 5 Page of 3

### Design Engineering MEng EXAMINATIONS 2016

IMPERIAL COLLEGE LONDON Design Engineering MEng EXAMINATIONS 2016 For Internal Students of the Imperial College of Science, Technology and Medicine This paper is also taken for the relevant examination

### Sinusoidal Steady-State Analysis

Chapter 4 Sinusoidal Steady-State Analysis In this unit, we consider circuits in which the sources are sinusoidal in nature. The review section of this unit covers most of section 9.1 9.9 of the text.

### Sinusoidal Steady-State Analysis

Sinusoidal Steady-State Analysis Almost all electrical systems, whether signal or power, operate with alternating currents and voltages. We have seen that when any circuit is disturbed (switched on or

### E1.1 Analysis of Circuits ( ) Revision Lecture 1 1 / 13

RevisionLecture 1: E1.1 Analysis of Circuits (2014-4530) Revision Lecture 1 1 / 13 Format Question 1 (40%): eight short parts covering the whole syllabus. Questions 2 and 3: single topic questions (answer

### Notes on Electric Circuits (Dr. Ramakant Srivastava)

Notes on Electric ircuits (Dr. Ramakant Srivastava) Passive Sign onvention (PS) Passive sign convention deals with the designation of the polarity of the voltage and the direction of the current arrow

### AC Circuit Analysis and Measurement Lab Assignment 8

Electric Circuit Lab Assignments elcirc_lab87.fm - 1 AC Circuit Analysis and Measurement Lab Assignment 8 Introduction When analyzing an electric circuit that contains reactive components, inductors and

### CIRCUIT ANALYSIS II. (AC Circuits)

Will Moore MT & MT CIRCUIT ANALYSIS II (AC Circuits) Syllabus Complex impedance, power factor, frequency response of AC networks including Bode diagrams, second-order and resonant circuits, damping and

### EECE 2150 Circuits and Signals, Biomedical Applications Final Exam Section 3

EECE 2150 Circuits and Signals, Biomedical Applications Final Exam Section 3 Instructions: Closed book, closed notes; Computers and cell phones are not allowed You may use the equation sheet provided but

### Dynamic circuits: Frequency domain analysis

Electronic Circuits 1 Dynamic circuits: Contents Free oscillation and natural frequency Transfer functions Frequency response Bode plots 1 System behaviour: overview 2 System behaviour : review solution

### RLC Series Circuit. We can define effective resistances for capacitors and inductors: 1 = Capacitive reactance:

RLC Series Circuit In this exercise you will investigate the effects of changing inductance, capacitance, resistance, and frequency on an RLC series AC circuit. We can define effective resistances for

### Circuit Theorems Overview Linearity Superposition Source Transformation Thévenin and Norton Equivalents Maximum Power Transfer

Circuit Theorems Overview Linearity Superposition Source Transformation Thévenin and Norton Equivalents Maximum Power Transfer J. McNames Portland State University ECE 221 Circuit Theorems Ver. 1.36 1

### EE 321 Analog Electronics, Fall 2013 Homework #3 solution

EE 32 Analog Electronics, Fall 203 Homework #3 solution 2.47. (a) Use superposition to show that the output of the circuit in Fig. P2.47 is given by + [ Rf v N + R f v N2 +... + R ] f v Nn R N R N2 R [

### Notes for course EE1.1 Circuit Analysis TOPIC 4 NODAL ANALYSIS

Notes for course EE1.1 Circuit Analysis 2004-05 TOPIC 4 NODAL ANALYSIS OBJECTIVES 1) To develop Nodal Analysis of Circuits without Voltage Sources 2) To develop Nodal Analysis of Circuits with Voltage

### Phasors: Impedance and Circuit Anlysis. Phasors

Phasors: Impedance and Circuit Anlysis Lecture 6, 0/07/05 OUTLINE Phasor ReCap Capacitor/Inductor Example Arithmetic with Complex Numbers Complex Impedance Circuit Analysis with Complex Impedance Phasor

### 3.1 Superposition theorem

Many electric circuits are complex, but it is an engineer s goal to reduce their complexity to analyze them easily. In the previous chapters, we have mastered the ability to solve networks containing independent

### AC Circuits Homework Set

Problem 1. In an oscillating LC circuit in which C=4.0 μf, the maximum potential difference across the capacitor during the oscillations is 1.50 V and the maximum current through the inductor is 50.0 ma.

### Lectures 16 & 17 Sinusoidal Signals, Complex Numbers, Phasors, Impedance & AC Circuits. Nov. 7 & 9, 2011

Lectures 16 & 17 Sinusoidal Signals, Complex Numbers, Phasors, Impedance & AC Circuits Nov. 7 & 9, 2011 Material from Textbook by Alexander & Sadiku and Electrical Engineering: Principles & Applications,

### D is the voltage difference = (V + - V - ).

1 Operational amplifier is one of the most common electronic building blocks used by engineers. It has two input terminals: V + and V -, and one output terminal Y. It provides a gain A, which is usually

### Basics of Network Theory (Part-I)

Basics of Network Theory (PartI). A square waveform as shown in figure is applied across mh ideal inductor. The current through the inductor is a. wave of peak amplitude. V 0 0.5 t (m sec) [Gate 987: Marks]

### 0 t < 0 1 t 1. u(t) =

A. M. Niknejad University of California, Berkeley EE 100 / 42 Lecture 13 p. 22/33 Step Response A unit step function is described by u(t) = ( 0 t < 0 1 t 1 While the waveform has an artificial jump (difficult

### D.M. Brookes P. Georgiou

EE1-01 IMPERIAL COLLEGE LONDON DEPARTMENT OF ELECTRICAL AND ELECTRONIC ENGINEERING EXAMINATIONS 2017 ExamHeader: EEE/EIE PART I: MEng, BEng and ACGI ANALYSIS OF CIRCUITS Tuesday, 6 June 10:00 am Time allowed:

### Time Varying Circuit Analysis

MAS.836 Sensor Systems for Interactive Environments th Distributed: Tuesday February 16, 2010 Due: Tuesday February 23, 2010 Problem Set # 2 Time Varying Circuit Analysis The purpose of this problem set

### LCR Series Circuits. AC Theory. Introduction to LCR Series Circuits. Module. What you'll learn in Module 9. Module 9 Introduction

Module 9 AC Theory LCR Series Circuits Introduction to LCR Series Circuits What you'll learn in Module 9. Module 9 Introduction Introduction to LCR Series Circuits. Section 9.1 LCR Series Circuits. Amazing

### Experiment Guide for RC Circuits

Guide-P1 Experiment Guide for RC Circuits I. Introduction 1. Capacitors A capacitor is a passive electronic component that stores energy in the form of an electrostatic field. The unit of capacitance is

### Designing Information Devices and Systems II Spring 2016 Anant Sahai and Michel Maharbiz Midterm 2

EECS 16B Designing Information Devices and Systems II Spring 2016 Anant Sahai and Michel Maharbiz Midterm 2 Exam location: 145 Dwinelle (SIDs ending in 1 and 5) PRINT your student ID: PRINT AND SIGN your

### RC, RL, and LCR Circuits

RC, RL, and LCR Circuits EK307 Lab Note: This is a two week lab. Most students complete part A in week one and part B in week two. Introduction: Inductors and capacitors are energy storage devices. They

### ELEC 202 Electric Circuit Analysis II Lecture 10(a) Complex Arithmetic and Rectangular/Polar Forms

Dr. Gregory J. Mazzaro Spring 2016 ELEC 202 Electric Circuit Analysis II Lecture 10(a) Complex Arithmetic and Rectangular/Polar Forms THE CITADEL, THE MILITARY COLLEGE OF SOUTH CAROLINA 171 Moultrie Street,

### Lecture #3. Review: Power

Lecture #3 OUTLINE Power calculations Circuit elements Voltage and current sources Electrical resistance (Ohm s law) Kirchhoff s laws Reading Chapter 2 Lecture 3, Slide 1 Review: Power If an element is

### Department of Electrical Engineering and Computer Sciences University of California, Berkeley. Final Exam Solutions

Electrical Engineering 42/00 Summer 202 Instructor: Tony Dear Department of Electrical Engineering and omputer Sciences University of alifornia, Berkeley Final Exam Solutions. Diodes Have apacitance?!?!

### Simultaneous equations for circuit analysis

Simultaneous equations for circuit analysis This worksheet and all related files are licensed under the Creative Commons Attribution License, version 1.0. To view a copy of this license, visit http://creativecommons.org/licenses/by/1.0/,

### I. Impedance of an R-L circuit.

I. Impedance of an R-L circuit. [For inductor in an AC Circuit, see Chapter 31, pg. 1024] Consider the R-L circuit shown in Figure: 1. A current i(t) = I cos(ωt) is driven across the circuit using an AC

### Chapter 33. Alternating Current Circuits

Chapter 33 Alternating Current Circuits 1 Capacitor Resistor + Q = C V = I R R I + + Inductance d I Vab = L dt AC power source The AC power source provides an alternative voltage, Notation - Lower case

### Electric Circuits I Final Examination

EECS:300 Electric Circuits I ffs_elci.fm - Electric Circuits I Final Examination Problems Points. 4. 3. Total 38 Was the exam fair? yes no //3 EECS:300 Electric Circuits I ffs_elci.fm - Problem 4 points

Network Topology-2 & Dual and Duality Choice of independent branch currents and voltages: The solution of a network involves solving of all branch currents and voltages. We know that the branch current

### ECE2262 Electric Circuits. Chapter 6: Capacitance and Inductance

ECE2262 Electric Circuits Chapter 6: Capacitance and Inductance Capacitors Inductors Capacitor and Inductor Combinations Op-Amp Integrator and Op-Amp Differentiator 1 CAPACITANCE AND INDUCTANCE Introduces

### UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS

UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS 1.0 Kirchoff s Law Kirchoff s Current Law (KCL) states at any junction in an electric circuit the total current flowing towards that junction is equal

### Problem Set 4 Solutions

University of California, Berkeley Spring 212 EE 42/1 Prof. A. Niknejad Problem Set 4 Solutions Please note that these are merely suggested solutions. Many of these problems can be approached in different

### ECE145A/218A Course Notes

ECE145A/218A Course Notes Last note set: Introduction to transmission lines 1. Transmission lines are a linear system - superposition can be used 2. Wave equation permits forward and reverse wave propagation

### Chapter 2. Engr228 Circuit Analysis. Dr Curtis Nelson

Chapter 2 Engr228 Circuit Analysis Dr Curtis Nelson Chapter 2 Objectives Understand symbols and behavior of the following circuit elements: Independent voltage and current sources; Dependent voltage and

### EIT Review 1. FE/EIT Review. Circuits. John A. Camara, Electrical Engineering Reference Manual, 6 th edition, Professional Publications, Inc, 2002.

FE/EIT eview Circuits Instructor: uss Tatro eferences John A. Camara, Electrical Engineering eference Manual, 6 th edition, Professional Publications, Inc, 00. John A. Camara, Practice Problems for the

### Chapter 10 Instructor Notes

G. izzoni, Principles and Applications of lectrical ngineering Problem solutions, hapter 10 hapter 10 nstructor Notes hapter 10 introduces bipolar junction transistors. The material on transistors has

### ECE137B Final Exam. There are 5 problems on this exam and you have 3 hours There are pages 1-19 in the exam: please make sure all are there.

ECE37B Final Exam There are 5 problems on this exam and you have 3 hours There are pages -9 in the exam: please make sure all are there. Do not open this exam until told to do so Show all work: Credit

### Lecture 6: Impedance (frequency dependent. resistance in the s- world), Admittance (frequency. dependent conductance in the s- world), and

Lecture 6: Impedance (frequency dependent resistance in the s- world), Admittance (frequency dependent conductance in the s- world), and Consequences Thereof. Professor Ray, what s an impedance? Answers:

### Transient response of RC and RL circuits ENGR 40M lecture notes July 26, 2017 Chuan-Zheng Lee, Stanford University

Transient response of C and L circuits ENG 40M lecture notes July 26, 2017 Chuan-Zheng Lee, Stanford University esistor capacitor (C) and resistor inductor (L) circuits are the two types of first-order

### Unit 2: Modeling in the Frequency Domain. Unit 2, Part 4: Modeling Electrical Systems. First Example: Via DE. Resistors, Inductors, and Capacitors

Unit 2: Modeling in the Frequency Domain Part 4: Modeling Electrical Systems Engineering 582: Control Systems I Faculty of Engineering & Applied Science Memorial University of Newfoundland January 20,

### Review of DC Electric Circuit. DC Electric Circuits Examples (source:

Review of DC Electric Circuit DC Electric Circuits Examples (source: http://hyperphysics.phyastr.gsu.edu/hbase/electric/dcex.html) 1 Review - DC Electric Circuit Multisim Circuit Simulation DC Circuit

### Energy Storage Elements: Capacitors and Inductors

CHAPTER 6 Energy Storage Elements: Capacitors and Inductors To this point in our study of electronic circuits, time has not been important. The analysis and designs we have performed so far have been static,

### Operational Amplifiers

Operational Amplifiers A Linear IC circuit Operational Amplifier (op-amp) An op-amp is a high-gain amplifier that has high input impedance and low output impedance. An ideal op-amp has infinite gain and

### Electrical Engineering Fundamentals for Non-Electrical Engineers

Electrical Engineering Fundamentals for Non-Electrical Engineers by Brad Meyer, PE Contents Introduction... 3 Definitions... 3 Power Sources... 4 Series vs. Parallel... 9 Current Behavior at a Node...

### Lecture 9 Time Domain vs. Frequency Domain

. Topics covered Lecture 9 Time Domain vs. Frequency Domain (a) AC power in the time domain (b) AC power in the frequency domain (c) Reactive power (d) Maximum power transfer in AC circuits (e) Frequency

### In Chapter 15, you learned how to analyze a few simple ac circuits in the time

In Chapter 15, you learned how to analyze a few simple ac circuits in the time domain using voltages and currents expressed as functions of time. However, this is not a very practical approach. A more

### EECE 2150 Circuits and Signals Final Exam Fall 2016 Dec 9

EECE 2150 Circuits and Signals Final Exam Fall 2016 Dec 9 Name: Instructions: Write your name and section number on all pages Closed book, closed notes; Computers and cell phones are not allowed You can

### a + b Time Domain i(τ)dτ.

R, C, and L Elements and their v and i relationships We deal with three essential elements in circuit analysis: Resistance R Capacitance C Inductance L Their v and i relationships are summarized below.

### R-L-C Circuits and Resonant Circuits

P517/617 Lec4, P1 R-L-C Circuits and Resonant Circuits Consider the following RLC series circuit What's R? Simplest way to solve for is to use voltage divider equation in complex notation. X L X C in 0

### QUESTION BANK SUBJECT: NETWORK ANALYSIS (10ES34)

QUESTION BANK SUBJECT: NETWORK ANALYSIS (10ES34) NOTE: FOR NUMERICAL PROBLEMS FOR ALL UNITS EXCEPT UNIT 5 REFER THE E-BOOK ENGINEERING CIRCUIT ANALYSIS, 7 th EDITION HAYT AND KIMMERLY. PAGE NUMBERS OF

### University of Alabama Department of Physics and Astronomy. Problem Set 4

University of Alabama Department of Physics and Astronomy PH 26 LeClair Fall 20 Problem Set 4. A battery has an ideal voltage V and an internal resistance r. A variable load resistance R is connected to

### SCHOOL OF COMPUTING, ENGINEERING AND MATHEMATICS SEMESTER 1 EXAMINATIONS 2012/2013 XE121. ENGINEERING CONCEPTS (Test)

s SCHOOL OF COMPUTING, ENGINEERING AND MATHEMATICS SEMESTER EXAMINATIONS 202/203 XE2 ENGINEERING CONCEPTS (Test) Time allowed: TWO hours Answer: Attempt FOUR questions only, a maximum of TWO questions

### One-Port Networks. One-Port. Network

TwoPort s Definitions Impedance Parameters dmittance Parameters Hybrid Parameters Transmission Parameters Cascaded TwoPort s Examples pplications OnePort s v i' 1 OnePort pair of terminals at which a signal

### Learnabout Electronics - AC Theory

Learnabout Electronics - AC Theory Facts & Formulae for AC Theory www.learnabout-electronics.org Contents AC Wave Values... 2 Capacitance... 2 Charge on a Capacitor... 2 Total Capacitance... 2 Inductance...

### Electrochemical methods : Fundamentals and Applications

Electrochemical methods : Fundamentals and Applications Lecture Note 7 May 19, 2014 Kwang Kim Yonsei University kbkim@yonsei.ac.kr 39 8 7 34 53 Y O N Se I 88.91 16.00 14.01 78.96 126.9 Electrochemical

### Electric Circuit Theory

Electric Circuit Theory Nam Ki Min nkmin@korea.ac.kr 010-9419-2320 Chapter 18 Two-Port Circuits Nam Ki Min nkmin@korea.ac.kr 010-9419-2320 Contents and Objectives 3 Chapter Contents 18.1 The Terminal Equations

### EXPERIMENT 07 TO STUDY DC RC CIRCUIT AND TRANSIENT PHENOMENA

EXPERIMENT 07 TO STUDY DC RC CIRCUIT AND TRANSIENT PHENOMENA DISCUSSION The capacitor is a element which stores electric energy by charging the charge on it. Bear in mind that the charge on a capacitor

### Switched-Capacitor Circuits David Johns and Ken Martin University of Toronto

Switched-Capacitor Circuits David Johns and Ken Martin University of Toronto (johns@eecg.toronto.edu) (martin@eecg.toronto.edu) University of Toronto 1 of 60 Basic Building Blocks Opamps Ideal opamps usually

### Review of Linear Time-Invariant Network Analysis

D1 APPENDIX D Review of Linear Time-Invariant Network Analysis Consider a network with input x(t) and output y(t) as shown in Figure D-1. If an input x 1 (t) produces an output y 1 (t), and an input x

### Circuits. Fawwaz T. Ulaby, Michel M. Maharbiz, Cynthia M. Furse. Solutions to the Exercises

Circuits by Fawwaz T. Ulaby, Michel M. Maharbiz, Cynthia M. Furse Solutions to the Exercises Chapter 1: Circuit Terminology Chapter 2: Resisitive Circuits Chapter 3: Analysis Techniques Chapter 4: Operational

### Mod. Sim. Dyn. Sys. Amplifiers page 1

AMPLIFIERS A circuit containing only capacitors, amplifiers (transistors) and resistors may resonate. A circuit containing only capacitors and resistors may not. Why does amplification permit resonance

### Poles, Zeros, and Frequency Response

Complex Poles Poles, Zeros, and Frequency esponse With only resistors and capacitors, you're stuck with real poles. If you want complex poles, you need either an op-amp or an inductor as well. Complex

### BASIC NETWORK ANALYSIS

SECTION 1 BASIC NETWORK ANALYSIS A. Wayne Galli, Ph.D. Project Engineer Newport News Shipbuilding Series-Parallel dc Network Analysis......................... 1.1 Branch-Current Analysis of a dc Network......................

### Q. 1 Q. 25 carry one mark each.

Q. Q. 5 carry one mark each. Q. Consider a system of linear equations: x y 3z =, x 3y 4z =, and x 4y 6 z = k. The value of k for which the system has infinitely many solutions is. Q. A function 3 = is

### Steady State Frequency Response Using Bode Plots

School of Engineering Department of Electrical and Computer Engineering 332:224 Principles of Electrical Engineering II Laboratory Experiment 3 Steady State Frequency Response Using Bode Plots 1 Introduction

### Lecture 24. Impedance of AC Circuits.

Lecture 4. Impedance of AC Circuits. Don t forget to complete course evaluations: https://sakai.rutgers.edu/portal/site/sirs Post-test. You are required to attend one of the lectures on Thursday, Dec.

### Basic Electrical Circuits Analysis ECE 221

Basic Electrical Circuits Analysis ECE 221 PhD. Khodr Saaifan http://trsys.faculty.jacobs-university.de k.saaifan@jacobs-university.de 1 2 Reference: Electric Circuits, 8th Edition James W. Nilsson, and

### Driven RLC Circuits Challenge Problem Solutions

Driven LC Circuits Challenge Problem Solutions Problem : Using the same circuit as in problem 6, only this time leaving the function generator on and driving below resonance, which in the following pairs

### 11. The Series RLC Resonance Circuit

Electronicsab.nb. The Series RC Resonance Circuit Introduction Thus far we have studied a circuit involving a () series resistor R and capacitor C circuit as well as a () series resistor R and inductor

### Electrical Circuit & Network

Electrical Circuit & Network January 1 2017 Website: www.electricaledu.com Electrical Engg.(MCQ) Question and Answer for the students of SSC(JE), PSC(JE), BSNL(JE), WBSEDCL, WBSETCL, WBPDCL, CPWD and State

### 12. Introduction and Chapter Objectives

Real Analog - Circuits 1 Chapter 1: Steady-State Sinusoidal Power 1. Introduction and Chapter Objectives In this chapter we will address the issue of power transmission via sinusoidal or AC) signals. This

### ESE319 Introduction to Microelectronics. Output Stages

Output Stages Power amplifier classification Class A amplifier circuits Class A Power conversion efficiency Class B amplifier circuits Class B Power conversion efficiency Class AB amplifier circuits Class

### Overview of Bode Plots Transfer function review Piece-wise linear approximations First-order terms Second-order terms (complex poles & zeros)

Overview of Bode Plots Transfer function review Piece-wise linear approximations First-order terms Second-order terms (complex poles & zeros) J. McNames Portland State University ECE 222 Bode Plots Ver.

### E2.2 Analogue Electronics

E2.2 Analogue Electronics Instructor : Christos Papavassiliou Office, email : EE 915, c.papavas@imperial.ac.uk Lectures : Monday 2pm, room 408 (weeks 2-11) Thursday 3pm, room 509 (weeks 4-11) Problem,

### RLC Circuit (3) We can then write the differential equation for charge on the capacitor. The solution of this differential equation is

RLC Circuit (3) We can then write the differential equation for charge on the capacitor The solution of this differential equation is (damped harmonic oscillation!), where 25 RLC Circuit (4) If we charge

### Capacitance, Resistance, DC Circuits

This test covers capacitance, electrical current, resistance, emf, electrical power, Ohm s Law, Kirchhoff s Rules, and RC Circuits, with some problems requiring a knowledge of basic calculus. Part I. Multiple

### 12 Chapter Driven RLC Circuits

hapter Driven ircuits. A Sources... -. A ircuits with a Source and One ircuit Element... -3.. Purely esistive oad... -3.. Purely Inductive oad... -6..3 Purely apacitive oad... -8.3 The Series ircuit...

### CHAPTER.6 :TRANSISTOR FREQUENCY RESPONSE

CHAPTER.6 :TRANSISTOR FREQUENCY RESPONSE To understand Decibels, log scale, general frequency considerations of an amplifier. low frequency analysis - Bode plot low frequency response BJT amplifier Miller

### ENGR 2405 Chapter 8. Second Order Circuits

ENGR 2405 Chapter 8 Second Order Circuits Overview The previous chapter introduced the concept of first order circuits. This chapter will expand on that with second order circuits: those that need a second

### Department of Electrical Engineering and Computer Sciences University of California, Berkeley. Final Exam: Friday, August 10, 2012

Electrical Engineering 42/00 Summer 202 Instructor: Tony Dear Department of Electrical Engineering and Computer Sciences University of California, Berkeley Final Exam: Friday, August 0, 202 Last Name:

### AC Circuits. The Capacitor

The Capacitor Two conductors in close proximity (and electrically isolated from one another) form a capacitor. An electric field is produced by charge differences between the conductors. The capacitance

### VI. Transistor amplifiers: Biasing and Small Signal Model

VI. Transistor amplifiers: iasing and Small Signal Model 6.1 Introduction Transistor amplifiers utilizing JT or FET are similar in design and analysis. Accordingly we will discuss JT amplifiers thoroughly.

### Physics 4B Chapter 31: Electromagnetic Oscillations and Alternating Current

Physics 4B Chapter 31: Electromagnetic Oscillations and Alternating Current People of mediocre ability sometimes achieve outstanding success because they don't know when to quit. Most men succeed because

### 0-2 Operations with Complex Numbers

Simplify. 1. i 10 2. i 2 + i 8 3. i 3 + i 20 4. i 100 5. i 77 esolutions Manual - Powered by Cognero Page 1 6. i 4 + i 12 7. i 5 + i 9 8. i 18 Simplify. 9. (3 + 2i) + ( 4 + 6i) 10. (7 4i) + (2 3i) 11.

### Analogue Filters Design and Simulation by Carsten Kristiansen Napier University. November 2004

Analogue Filters Design and Simulation by Carsten Kristiansen Napier University November 2004 Title page Author: Carsten Kristiansen. Napier No: 04007712. Assignment title: Analogue Filters Design and

### 4.2 Graphs of Rational Functions

4.2. Graphs of Rational Functions www.ck12.org 4.2 Graphs of Rational Functions Learning Objectives Compare graphs of inverse variation equations. Graph rational functions. Solve real-world problems using

### 8 sin 3 V. For the circuit given, determine the voltage v for all time t. Assume that no energy is stored in the circuit before t = 0.

For the circuit given, determine the voltage v for all time t. Assume that no energy is stored in the circuit before t = 0. Spring 2015, Exam #5, Problem #1 4t Answer: e tut 8 sin 3 V 1 For the circuit

### Tactics Box 23.1 Using Kirchhoff's Loop Law

PH203 Chapter 23 solutions Tactics Box 231 Using Kirchhoff's Loop Law Description: Knight/Jones/Field Tactics Box 231 Using Kirchhoff s loop law is illustrated Learning Goal: To practice Tactics Box 231

### BRIDGE CIRCUITS EXPERIMENT 5: DC AND AC BRIDGE CIRCUITS 10/2/13

EXPERIMENT 5: DC AND AC BRIDGE CIRCUITS 0//3 This experiment demonstrates the use of the Wheatstone Bridge for precise resistance measurements and the use of error propagation to determine the uncertainty

### Adjoint networks and other elements of circuit theory. E416 4.Adjoint networks

djoint networks and other elements of circuit theory One-port reciprocal networks one-port network is reciprocal if: V I I V = Where and are two different tests on the element Example: a linear impedance

### Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science Electronic Circuits Fall 2000.

Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.002 Electronic Circuits Fall 2000 Final Exam Please write your name in the space provided below, and circle

### Chapter 9. Estimating circuit speed. 9.1 Counting gate delays

Chapter 9 Estimating circuit speed 9.1 Counting gate delays The simplest method for estimating the speed of a VLSI circuit is to count the number of VLSI logic gates that the input signals must propagate

### INSTRUMENTAL ENGINEERING

INSTRUMENTAL ENGINEERING Subject Code: IN Course Structure Sections/Units Section A Unit 1 Unit 2 Unit 3 Unit 4 Unit 5 Unit 6 Section B Section C Section D Section E Section F Section G Section H Section