Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science : Circuits & Electronics Problem Set #1 Solution


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1 Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.2: Circuits & Electronics Problem Set # Solution Exercise. The three resistors form a series connection. Therefore, by directly applying equation (2.59) in the text, we obtain: eq 2 The three resistors form a parallel connection. This time, by applying (2.8) in the text, we obtain: eq 2 Note the pleasing symmetry in the solution. 2 eq 2 2 (iii) Split up the problem into two parts. First, apply the result from to 2,, and. Call this solution p. Second, apply the result from to, p, and 5 : 2 p 2 2 eq p 5 Exercise.2.6Ω is equivalent to /5Ω. Because a series connection cannot achieve this, we try to apply a parallel connection first: A B 5 A B By setting B =, we obtain A =.5Ω..5Ω can easily be achieved using additional resistors, as the following diagram shows: 2/Ω can be achieved using a and a 2/Ω resistor. Using the same method from,
2 we see immediately that th/ω resistor can be synthesized from a 2Ω and a resistor. Therefore, the solution looks like the following: Problem. Using KVL equations around loop L and L2, we obtain the following equations: V V V 2V L L2 V 2 Loop L: 2V V V = Loop L2: V 2 V 2V = Therefore, the unknown voltages are: V V V 2 V The bottom portion of the circuit is exactly the same as ; therefore, we can immediately obtain V and V. Using KVL around loop L, we obtain the following equation: V 2 V V L V 2V V Loop L: V 2 V V = Therefore, the unknown voltages are: V V V 2 2V V V We can verify the calculation by perform KVL on the exterior loop consisting of V 2, V, and the V drop. This gives V 2V V =, which confirms the solution. (iii) Applying KCL at node N, we obtain the following equation: 2
3 A N I Node N: A = I Therefore, the unknown current is: I A (iv) First we apply KCL at N since that is the only node with unknown variable. Upon solving I, I 2 can be found next by KCL at N2. Finally, I is obtained by KCL at N. N I I 2 N2 I A N Node N: I = I 2 Node N2: I 2 I = Node N: = I A Therefore, the unknown currents are: I A I 2 A I A Problem.2 In particular, notice that some of the currents as defined in the diagram are negative. There are a total of nodes in the circuit, as indicated by the diagram below. The KCL equations for each node are as follows: V V N I N2 I I I 2 2 I s V 2 V s V N Node N: I I 2 I = Node N2: I = I Node N: I I 2 I = Only 2 of the KCL equations are independent. This can be seen by substituting I in place of I, since they are the same as indicated by the N2 equation, into N and
4 observing that the N and N equations become identical. There are a total of loops in the circuit, as indicated by the diagram below. The KVL equations for each node are as follows: L2 V I I 2 I V 2 I s V 2 V s L L V Loop L: V V 2 = Loop L2: V V V = Loop L: V 2 V V = Again, only 2 of the KVL equations are independent. This can be seen by substituting V in place of V 2, since they are the same as indicated by the L equation, into L and observing that the L2 and L equations become identical. (iii) The key point to notice in this part is that the voltage across an ideal current source and the current across an ideal voltage source cannot be defined on their own; instead the rest of the circuit dictates their values. Voltage Source: V S = V Current Source: I S = I : = V 2 / I 2 2 : 2 = V / I (iv) The 8 equations include N2, either N or N, L, either L2 or L, and the equations from part (iii). From the voltage and current source equations, we immediately have: I A V V Next we substitute the and 2 constitutive laws into N; together with L, we now have to a system of 2 equations and 2 unknowns (we have already solved I ): Equation : Equation 2: V 2 I V 2 V 2 V V By substituting V 2 from equation 2 into equation and simplifying, we obtain: V I V 2
5 Therefore, substituting in numbers gives us V 2V V 2 5V Finally, by substituting V 2 and V into the and 2 constitutive laws, we obtain the remaining unknown values: I 2 2.5A I.5A The following table summarizes all the unknown currents and voltages: Number 2 Current (A) Voltage (V) (v) Using equation (.9) from the text and noting the defined direction of the currents and the polarity of the voltages in the circuit, we obtain the following: Problem. P current = I V = 5W (source) P = I 2 V 2 = 2.5W (sink) P 2 = I V = W (sink) P voltage = I V =.5W (sink) Since the branch powers add up to, we can be confident that the calculations were done correctly. Note that as stated on page 5 of the text, an element is considered sourcing power if the power is negative under the associated variables convention while it is considered sinking power if the power is positive. All the current and voltage definitions in the circuit conform to the convention, with the current entering on the positive terminal of the elements. The resistor network in this problem is also known as a bridge circuit. Often times, we use this circuit in the lab to determine unknown resistor value within to using the remaining. Pag6 of the text shows how this circuit can be used (note that because and are switched compared to the book, the bridge branch is not balanced). e = 2 = I s I 5 = I 2 I 5 I I = = All node voltages e through are defined positively with respect to the ground node. The current directions have been chosen arbitrarily; wrong direction will be indicated by a final current value that is negative. Now, applying KCL at th nodes, we have: 5
6 Node : Nod: Nod: e I s e e e In order to facilitate computation, we turn th equations with unknown into a matrix and perform row reduction on the result: 7 7 e I s 7 7 e I s 7 2 e I s e I s I s 9 5 e I s I s I s Now, the node voltages by inspection are: 5 e I 2 s I s Applying the constitutive law for resistors, it is now trivial to obtain the branch currents: e I I e 2 I 2 I 2 2 I I I I I 5 I 5 6
7 Problem. Using the first experiment, as diagramed below, we observe that because is floating, the full A from the current source at port 2 must flow through 2 and. This also means that there is no voltage drop across. By defining a convenient reference node, as shown in the diagram, we can solve for : V 2V A 2 A A 2V V A 2Ω Next we apply the result of the second experiment, as diagramed below. By the KCL equation at node N, we can find the current, and hence the voltage, of. The value of and 2 follows: 9V N 2V A V V = 2V A 2Ω 2 V A 2Ω V 2V V 2V A Ω 2V V 2 2 7
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