Chapter 5 Solution P5.22, 3, 6 P5.33, 5, 8, 15 P5.43, 6, 8, 16 P5.52, 4, 6, 11 P5.62, 4, 9


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1 Chapter 5 Solution P5.22, 3, 6 P5.33, 5, 8, 15 P5.43, 6, 8, 16 P5.52, 4, 6, 11 P5.62, 4, 9 P Consider the circuit of Figure P Find i a by simplifying the circuit (using source transformations) to a singleloop circuit so that you need to write only one KVL equation to find i a. Figure P Finally, apply KVL: ia + 4 ia = 0 ia = 2.19 A 3
2 (checked using LNAP 8/15/02) P Find o using source transformations if i = 5/2 A in the circuit shown in Figure P Hint: Reduce the circuit to a single mesh that contains the oltage source labeled o. Answer: o = 28 V Figure P Source transformation at left; equialent resistor for parallel 6 and 3 Ω resistors: Equialents for series resistors, series oltage source at left; series resistors, then source transformation at top: Source transformation at left; series resistors at right:
3 Parallel resistors, then source transformation at left: Finally, apply KVL to loop 6 + i (9 + 19) 36 = 0 i = 5 / 2 = (5 / 2) = 28 V o o (checked using LNAP 8/15/02) P Use source transformations to find the alue of the oltage a in Figure P Answer: a = 7 V Figure P A source transformation on the right side of the circuit, followed by replacing series resistors with an equialent resistor: Source transformations on both the right side and the left side of the circuit: Replacing parallel resistors with an equialent resistor and also replacing parallel current sources with an equialent current source:
4 Finally, ( ) ( ) ( ) a = 0.21 = 0.21 = 7 V (checked using LNAP 8/15/02) P5.33 The circuit shown in Figure P5.33 has two inputs, s and i s, and one output i o. The output is related to the inputs by the equation io = ais + bs Gien the following two facts: and The output is i o = 0.45 A when the inputs are i s = 0.25 A and s = 15 V. The output is i o = 0.30 A when the inputs are i s = 0.50 A and s = 0 V. Determine the alues of the constants a and b and the alues of the resistances are R 1 and R 2. Answers: a = 0.6 A/A, b = 0.02 A/V, R 1 = 30 Ω and R 2 = 20 Ω. Figure P5.33 From the 1 st fact: 0.45 = a b 15 ( ) ( ) From the 2nd fact: = a( 0.50) + b( 0) a= = ( 0.60)( 0.25) Substituting gies 0.45 = ( 0.60)( 0.25) + b( 15) b= = Next, consider the circuit: R 1 ai s = io1 = i o = i s s = 0 R1 R so 0.60 = R 2 R1 3 R2 R + R = 1 2
5 and bs = io2 = io = is = 0 R so R1 R2 1 2 s + R = 50 R + R + = 0.02 = Ω Soling these equations gies R 1 = 30 Ω and R 2 = 20 Ω. Figure P5.35 P5.35 Determine (t), the oltage across the ertical resistor in the circuit in Figure P Solution; We ll use superposition. Let 1 (t) the be the part of (t) due to the oltage source acting alone. Similarly, let 2 (t) the be the part of (t) due to the oltage source acting alone. We can use these circuits to calculate 1 (t) and 2 (t). Notice that 1 (t) is the oltage across parallel resistors. Using equialent resistance, we calculate = 8 Ω. Next, using oltage diision we calculate 8 1 ( t ) = ( 12) = 2 V Similarly 2 (t) is the oltage across parallel resistors Using equialent resistance we first determine = 20 Ω and then calculate ( t 20 2 ) = ( 12cos( 5 )) 8cos( 5 ) V t = + t t = 1 t + 2 t = 2 + 8cos 5 t V Using superposition ( ) ( ) ( ) ( )
6 P Use superposition to find the alue of the current i x in Figure P Answer: i = 3.5 ma Figure P5.38 Consider 8 V source only (open the 2 A source) Let i 1 be the part of i x due to the 8 V oltage source. Apply KVL to the supermesh: ( i1) ( i1) ( i1) = 0 Consider 2 A source only (short the 8 V source) 8 2 i 1 = = A 12 3 Let i 2 be the part of i x due to the 2 A current source. Apply KVL to the supermesh: ( ) ( ) 6 i + 3 i i = i = = A Finally, ix = i1+ i2 = = A P The circuit shown in Figure P has three inputs: 1, i 2, and 3. The output of the circuit is the current i o. The output of the circuit is related to the inputs by i 1 = a o + b 2 + ci 3 where a, b, and c are constants. Determine the alues of a, b, and c. Figure P
7 i ( 40 10) ( 40 10) 1 o = + i ( ) ( ) So i = + i o a = 0.05, b= 0.1 and c= (checked: LNAP 6/19/04) P The circuit shown in Figure P 5.43b is the Théenin equialent circuit of the circuit shown in Figure P 5.43a. Find the alue of the opencircuit oltage, oc, and Théenin resistance, R t. Answer: oc = 2 V and R t = 4 Ω Figure P The circuit from Figure P5.43a can be reduced to its Theenin equialent circuit in fie steps: (a) (b) (c)
8 (d) (e) Comparing (e) to Figure P5.43b shows that the Theenin resistance is R t = 4 Ω and the open circuit oltage, oc = 2 V. (checked using LNAP 8/15/02) P Find the Théenin equialent circuit for the circuit shown in Figure P Find oc : Figure P a a a Apply KCL at the top, middle node: = a = 18 V 3 6 The oltage across the righthand 3 Ω resistor is zero so: a = oc = 18 V Find i sc : Apply KCL at the top, middle node: 2 a a a a = + 3+ a = 18 V 3 6 3
9 Apply Ohm s law to the righthand 3 Ω resistor : Finally: R t oc 18 = = = 3 Ω i 6 sc i sc a 18 = = = 6 V 3 3 (checked using LNAP 8/15/02) P A resistor, R, was connected to a circuit box as shown in Figure P The oltage,, was measured. The resistance was changed, and the oltage was measured again. The results are shown in the table. Determine the Théenin equialent of the circuit within the box and predict the oltage,, when R = 8 kω. Figure P From the gien data: = oc Rt oc = 1.2 V 4000 Rt = 1600 Ω 2 = oc Rt = R R + R t oc When R = 8000 Ω, 8000 = ( 1.2) = 1.5 V P An ideal oltmeter is modeled as an open circuit. A more realistic model of a oltmeter is a large resistance. Figure P a shows a circuit with a oltmeter that measures the oltage m. In Figure P b the oltmeter is replaced by the model of an ideal oltmeter, an open circuit. The oltmeter measures mi, the ideal alue of m. As R m, the oltmeter becomes an ideal oltmeter and m mi. When R m <, the oltmeter is not ideal and m > mi. The difference between m and mi is a measurement error caused by the fact that the oltmeter is not ideal. (a) Determine the alue of mi. (b) Express the measurement error that
10 (c) occurs when R m = 1000 Ω as a percentage of mi. Determine the minimum alue of R m required to ensure that the measurement error is smaller than 2 percent of mi. Figure P Replace the circuit by its Theenin equialent circuit: R m m = 5 Rm + 50 = lim = 5 V (a) mi m R m (b) When Rm = 1000 Ω, m = V so % error = = 4.76% 5 R m 5 5 Rm + 50 Rm (c) Rm 2450 Ω 5 R + 50 m (checked: LNAP 6/16/04)
11 P Two black boxes are shown in Figure P Box A contains the Théenin equialent of some linear circuit, and box B contains the Norton equialent of the same circuit. With access to just the outsides of the boxes and their terminals, how can you determine which is which, using only one shorting wire? Figure P When the terminals of the boxes are opencircuited, no current flows in Box A, but the resistor in Box B dissipates 1 watt. Box B is therefore warmer than Box A. If you short the terminals of each box, the resistor in Box A will draw 1 amp and dissipate 1 watt. The resistor in Box B will be shorted, draw no current, and dissipate no power. Then Box A will warm up and Box B will cool off. P Find the Norton equialent circuit for the circuit shown in Figure P Figure P 5.54
12 P The circuit shown in Figure P 5.56b is the Norton equialent circuit of the circuit shown in Figure P 5.56a. Find the alue of the shortcircuit current, i sc, and Théenin resistance, R t. Answer: i sc = 24 A and R t = 3 Ω Figure P To determine the alue of the short circuit current, I sc, we connect a short circuit across the terminals of the circuit and then calculate the alue of the current in that short circuit. Figure (a) shows the circuit from Figure 4.65a after adding the short circuit and labeling the short circuit current. Also, the nodes hae been identified and labeled in anticipation of writing node equations. Let 1, 2 and 3 denote the node oltages at nodes 1, 2 and 3, respectiely. In Figure (a), node oltage 1 is equal to the negatie of the oltage source oltage. Consequently, 1 = 24 V. The oltage at node 3 is equal to the oltage across a short, 3 = 0. The controlling oltage of the VCCS, a, is equal to the node oltage at node 2, i.e. a = 2. The oltage at node 3 is equal to the oltage across a short, i.e. 3 = 0. Apply KCL at node 2 to get Apply KCL at node 3 to get = = = 3 a a = 16 V = isc a = isc isc = ( 16) = 24 A Figure (a) Calculating the short circuit current, I sc, using mesh equations. To determine the alue of the Theenin resistance, R th, first replace the 24 V oltage source by a 0 V oltage source, i.e. a short circuit. Next, connect a current source circuit across the terminals of the circuit and then label the oltage across that current source as shown in Figure (b). The Theenin resistance will be calculated from the current and oltage of the current source as
13 R th = i T T Also, the nodes hae been identified and labeled in anticipation of writing node equations. Let 1, 2 and 3 denote the node oltages at nodes 1, 2 and 3, respectiely. In Figure (b), node oltage 1 is equal to the across a short circuit, i.e. 1 = 0. The controlling oltage of the VCCS, a, is equal to the node oltage at node 2, i.e. a = 2. The oltage at node 3 is equal to the oltage across the current source, i.e. 3 = T. Apply KCL at node 2 to get Apply KCL at node 3 to get = = 32 T = a Finally, it = iT = i = a T T 3 + 6i = 0 2 = 6i R t T T T T T T = = 3 Ω i T Figure (b) Calculating the Theenin resistance, R T th =, using mesh equations. it To determine the alue of the open circuit oltage, oc, we connect an open circuit across the terminals of the circuit and then calculate the alue of the oltage across that open circuit. Figure (c) shows the circuit from Figure P 4.65a after adding the open circuit and labeling the open circuit oltage. Also, the nodes hae been identified and labeled in anticipation of writing node equations. Let 1, 2 and 3 denote the node oltages at nodes 1, 2 and 3, respectiely. In Figure (c), node oltage 1 is equal to the negatie of the oltage source oltage. Consequently, 1 = 24 V. The controlling oltage of the VCCS, a, is equal to the node oltage at node 2, i.e. a = 2. The oltage at node 3 is equal to the open circuit oltage, i.e. 3 = oc. Apply KCL at node 2 to get
14 = = oc = Apply KCL at node 3 to get = = 0 9a = 6 3 Combining these equations gies ( ) = 9 = = 72 V oc a oc oc oc a Figure (c) Calculating the open circuit oltage, oc, using node equations. As a check, notice that ( )( ) R I = 3 24 = 72 = V th sc oc (checked using LNAP 8/16/02) P Determine alues of R t and i sc that cause the circuit shown in Figure P b to be the Norton equialent circuit of the circuit in Figure P a. Answer: R t = 3 Ω and i sc = 2 A Figure P i a 12 ia = ia = 3 A 6 = 2i = 6 V oc a
15 12 + 6i = 2i i = 3 A a a a 2 3isc = 2ia isc = ( 3) = 2 A 3 R t 6 = = 3 Ω 2 P For the circuit in Figure P 5.62, (a) find R such that maximum power is dissipated in R and (b) calculate the alue of maximum power. Answer: R = 60 Ω and P max = 54 mw Figure P a) For maximum power transfer, set R L equal to the Theenin resistance: R L = R = = 101 Ω t b) To calculate the maximum power, first replace the circuit connected to R L be its Theenin equialent circuit: 101 L = 100 = 50 V L 50 pmax = = = W R 101 The oltage across R L is ( ) Then L
16 P Find the maximum power to the load R L if the maximum power transfer condition is met for the circuit of Figure P Answer: max p L = 0.75 W Figure P L R L = S RS + RL p = = R 2 2 L S L L 2 RL ( RS + RL) By inspection, p L is max when you reduce R S to get the smallest denominator. set R S = 0 P A resistie circuit was connected to a ariable resistor, and the power deliered to the resistor was measured as shown in Figure P Determine the Théenin equialent circuit. Answer: R t = 20 Ω and oc = 20 V Figure P From the plot, the maximum power is 5 W when R = 20 Ω. Therefore: and 2 oc R t = 20 Ω pmax = oc = pmax 4 Rt = 5( 4) 20 = 20 V 4 R t
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