# Figure Circuit for Question 1. Figure Circuit for Question 2

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1 Exercises 10.7 Exercises Multiple Choice 1. For the circuit of Figure the time constant is A. 0.5 ms µs 2, 000 s D. 0.2 ms 4 Ω 2 Ω 12 Ω 1 mh 12u 0 () t V Figure Circuit for Question 1 2. For the circuit of Figure the time constant is A. 50 ms 100 ms 190 ms D ms 4 KΩ 5 KΩ 5u 0 () t A 6 KΩ 10 KΩ 10 µf Figure Circuit for Question 2 3. The forced response component of the inductor current for the circuit of Figure is i Lf A. 16 A 10 A Circuit Analysis I with MATLAB Applications 10-33

2 Chapter 10 Forced and Total Response in RL and RC Circuits 6 A D. 2 A 16u 0 () t A 4 Ω 12 Ω 5 Ω 1 mh Figure Circuit for Question 3 4. The forced response component of the capacitor voltage for the circuit of Figure is v Cf A. 10 V 2 V 32 3 V D. 8 V 16u 0 () t V 4 KΩ 12 KΩ 2 KΩ 1 µf Figure Circuit for Question 4 5. For the circuit of Figure i L ( 0 ) 2 A. For t > 0 the total response of i L () t is A. 6 A 6e 5000t A 6 6e 5000t A D. 6 4e 5000t A Circuit Analysis I with MATLAB Applications

3 Exercises 16u 0 () t A 4 Ω 12 Ω 5 Ω 1 mh i L () t Figure Circuit for Question 5 6. For the circuit of Figure v C ( 0 ) 5 V. For t > 0 the total response of v C () t is A. 12 V 10 5e 500t V 12 7e 200t V D. 12 7e 200t V 16u 0 () t V 4 KΩ 12 KΩ 2 KΩ 1 µf v C () t Figure Circuit for Question 6 7. For the circuit of Figure i L ( 0 ) 2 A. For t > 0 the total response of v L () t is A. 20e 5000t V 20e 5000t V 32e 8000t V D. 32e 8000t V Circuit Analysis I with MATLAB Applications 10-35

4 Chapter 10 Forced and Total Response in RL and RC Circuits 16u 0 () t A 4 Ω 12 Ω 5 Ω 1 mh v L () t Figure Circuit for Question 7 8. For the circuit of Figure v C ( 0 ) 5 V. For t > 0 the total response i C () t is A. 1400e 200t A 1.4e 200t A 3500e 500t A D. 3.5e 500t A 16u 0 () t V 4 KΩ 12 KΩ 2 KΩ 1 µf i C () t Figure Circuit for Question 8 9. The waveform of Figure can be expressed as A. D. 3tu 0 () t A 3u [ 0 () t ] 3u [ 0 ( t 3) ] A 3t[ u 0 () t u 0 ( t 1) ] ( 1.5t 4.5) [ u 0 ( t 1) u 0 ( t 3) ] A 3t[ u 0 () t u 0 ( t 3) ] ( 1.5t 4.5) [ u 0 () t u 0 ( t 3) ] A Circuit Analysis I with MATLAB Applications

5 Exercises 3 i L () t A t ( s) Figure Waveform for Question The waveform of Figure can be expressed as A. D. 21 ( e αt e βt )u 0 () t V ( 2 2e αt )[ u 0 () t u 0 ( t 2) ] ( 2e βt )[ u 0 ( t 2) u 0 ( t 3) ] V ( 2 2e αt )[ u 0 () t u 0 ( t 2) ] ( 2e βt )[ u 0 ( t 2) u 0 ( t 3) ] V ( 2 2e αt )[ u 0 () t ] ( 2e βt )[ u 0 ( t 3) ] V 2 1 v C () t V 2 2e αt 1 2 2e βt 3 t ( s) Figure Waveform for Question 10 Problems 1. In the circuit of Figure 10.54, the voltage source () t varies with time as shown by the waveform of Figure Compute, sketch, and express v LOAD () t as a sum of unit step functions for 0 t 5 s. Circuit Analysis I with MATLAB Applications 10-37

6 Chapter 10 Forced and Total Response in RL and RC Circuits 6 Ω 6 Ω 12 Ω () t 10 Ω v LOAD () t Figure Circuit for Problem 1 () t (V) Figure Waveform for Problem 1 t(s) 2. In the circuit of Figure 10.56, () t 15u 0 () t 30u 0 ( t 2) V. Compute i L () t for t > 0. R () t 3 KΩ L i L () t 1 mh Figure Circuit for Problem 2 3. In the circuit of Figure (a), the excitation () t is a pulse as shown in Figure (b). a. Compute i L () t for 0< t< 0.3 ms b. Compute and sketch i L () t for all t > Circuit Analysis I with MATLAB Applications

7 Exercises i L () t () t 3 Ω 1 mh 24 V 6 Ω 6 Ω () t (a) 0.3 (b) t (ms) Figure Circuit and waveform for Problem 3 4. In the circuit of Figure 10.58, switch S has been open for a very long time and closes at t 0. Compute and sketch i L () t and i SW () t for t > 0. 4 Ω t 0 i SW () t 6 Ω S 20 V 8 Ω 1 H i L () t Figure Circuit for Problem 4 5. For the circuit of Figure 10.59, compute v C () t for t > 0. 10u 0 () t A 38 Ω 24 V 2 Ω 50 µf vc () t Figure Circuit for Problem 5 6. For the circuit of Figure 10.60, compute v out () t for t > 0 in terms of R, C, and v in u 0 () t given that v C ( 0 ) 0 Circuit Analysis I with MATLAB Applications 10-39

8 Chapter 10 Forced and Total Response in RL and RC Circuits v in u 0 () t R C v out () t Figure Circuit for Problem 6 7. In the circuit of Figure 10.61, switch S has been open for a very long time and closes at t 0. Compute and sketch v C () t and v R3 () t for t> 0. R 1 2 R KΩ t 0 100u 0 ( t) V S 50 V 25 KΩ C 1 µf R KΩ v C () t v R3 () t Figure Circuit for Problem 7 8. For the circuit of Figure 10.62, it is given that v C ( 0 ) 5 V. Compute i C () t for t > 0. Hint: Be careful in deriving the time constant for this circuit. i C () t C 1 F v C () t R 1 18 Ω R 2 12 Ω 10i C () t Figure Circuit for Problem Circuit Analysis I with MATLAB Applications

9 Answers to Exercises 10.8 Answers to Exercises Multiple Choice 1. D 2. B 3. C 4. E 5. E 6. C 7. D 8. A 9. C 10. B 12 V 6 4e 8000t A Problems 1. We replace the given circuit shown below with its Thevenin equivalent. () t x 6 Ω 6 Ω 12 Ω 10 Ω y v LOAD () t v TH 10 Ω 10 Ω v LOAD () t () t () t (V) t(s) For the Thevenin equivalent voltage at different time intervals is as shown below. Circuit Analysis I with MATLAB Applications 10-41

10 Chapter 10 Forced and Total Response in RL and RC Circuits and v LOAD () t v TH () t v 18 s () t v 18 s () t v 18 s () t v 18 s () t v 20 TH () t 0.5v TH () t V 0< t< 1 s V 1< t< 3 s ( 60) 40 V 3< t< 4 s V t > 4 s V 0 < t < 1 s V 1 < t < 3 s 0.5 ( 40) 20 V 3< t< 4 s V t > 4 s The waveform of the voltage across the load is as shown below. v LOAD () t ( V) t ( s) The waveform above can now be expressed as a sum of unit step functions as follows: v LOAD () t 20u 0 () t 20u 0 ( t 1) 40u 0 ( t 1) 40u 0 ( t 3) 20u 0 ( t 4) 20u 0 ( t 3) 20u 0 ( t 4) 20u 0 ( t 4) 20u 0 () t 20u 0 ( t 1) 60u 0 ( t 3) 40u 0 ( t 4) 2. The circuit at t is as shown below and since we are not told otherwise, we will assume that i L ( 0 ) 0 0 For t > 0 we let i L1 () t be the inductor current when the 15u 0 () t voltage source acts alone and i L2 () t when the 30u 0 ( t 2) voltage source acts alone. Then, i L TOTAL () t i L1 () t i L2 () t Circuit Analysis I with MATLAB Applications

11 Answers to Exercises 3 kω 1 mh il 0 ( 0 ) 0 ( ) 0 For 0 < t < 2 s the circuit is as shown below. 3 KΩ 15 V i L1() t 1 mh Then, i L1 () t i L1f i L1n where i L1f ma and i 3 KΩ L1n A 1 e ( R L)t Ae t Thus, i L1 () t 5 A 1 e t ma and using the initial condition i L ( 0 ) i L ( 0 ) 0, we get i L1 ( 0) 5 A 1 e 0 ma or A 1 5. Therefore, i L1 () t 5 5e t (1) Next, with the 30u 0 ( t 2) voltage source acts alone the circuit is as shown below. 30 V 3 KΩ i L () t 1 mh Then, i L2 () t i L2f i L2n, i L2f ma and i 3 KΩ L2n A 2 e ( R L) ( t 2) Be ( t 2) Thus, i L2 () t 10 A 2 e ( t 2) ma and the initial condition at t 2 is found from (1) above as i L1 t 6 t 5 5e ma. Therefore, 2 s 5 10 A or 2 e ( 2 2) ma A 2 15 i L2 i t 2 s L1 t 2 s and i L2 () t 10 15e t 2 ( ) ma (2) Circuit Analysis I with MATLAB Applications 10-43

12 Chapter 10 Forced and Total Response in RL and RC Circuits Therefore, the total current when both voltage sources are present is the summation of (1) and (2), that is, i L TOTAL () t i L1 () t i L2 () t 5 5e t 10 15e t 2 ( ) ma e t 15e ( t 2) ma a. For this circuit () t 24[ u 0 () t u 0 ( t 0.3) ] and since we are not told otherwise, we will assume that i L ( 0 ) 0. For 0< t< 0.3 ms the circuit and its Thevenin equivalent are as shown below. 3 Ω 6 Ω 6 Ω 1 mh i L () t () t () t 24[ u 0 () t u 0 ( t 0.3) ] v TH 8 Ω 1 mh i L () t v TH () t () t 16[ u 0 () t u 0 ( t 0.3) ] Then, and at t 0 i L () t i Lf i Ln 16 8 A 1 e ( R L)t 2 Ae 8000t i L ( 0) i L ( 0 ) 0 2 A 1 e 0 or A 1 2 and thus for 0 < t < 0.3 ms i L () t 2 2e 8000t (1) b. For t > 0.3 ms the circuit is as shown below. For this circuit i L () t A 2 e ( R L) ( t 0.3) A 2 e 8000 ( t 0.3 ) (2) and A 2 is found from the initial condition at t 0.3 ms, that is, with (1) above we get 3 Ω () t 0 6 Ω 6 Ω 1 mh i L () t 8 Ω 1 mh i L () t Circuit Analysis I with MATLAB Applications

13 i L 2 2e t 0.3 ms and by substitution into (2) above or A Therefore for t > 0.3 ms 2 2e A i L 1.82 A t 0.3 ms 2 e ( ) i L () t 1.82e ( t 0.3 ms ) The waveform for the inductor current i L () t for all t > 0 is shown below i L ( A) Answers to Exercises 0.3 t ( ms) 4. At t 0 the circuit is as shown below where i L ( 0 ) 20 ( 4 6) 2 A and thus the initial condition has been established. 4 Ω 6 Ω 20 V i L 0 ( ) For and t> 0 the circuit and its Thevenin equivalent are as shown below where 8 v TH V 4 8 R 8 4 TH Ω 8 4 Circuit Analysis I with MATLAB Applications 10-45

14 Chapter 10 Forced and Total Response in RL and RC Circuits 4 Ω 6 Ω 20 V 8 Ω Closed Switch 1 H i L () t R TH v TH 1 H 40 3 V 26 3 Ω i L () t Then, and A i L () t i Lf i 40 3 ( Ln Ae R L )t Ae is evaluated from the initial condition, i.e., ( )t i L ( 0 ) i L ( 0 ) Ae 0 from which A 6 13 and thus for t > 0 i L () t e ( 26 3)t e 8.67t (1) Next, to find () t we observe that this current flows also through the 8 Ω resistor and this can be found from i SW v 8 Ω shown on the circuit below. 4 Ω i SW () t 6 Ω 20 V 8 Ω v 8 Ω 1 H i L () t Now, and v 8 Ω v 6 Ω v L () t 6i L () t L di L dt 6( e 8.67t d ) ( e 8.67t ) dt e 8.67t e 8.67t e 8.67t Circuit Analysis I with MATLAB Applications

15 Answers to Exercises i SW v () t i 8 Ω e 8.67t 8 Ω e 8.67t 8 8 Therefore, from the initial condition, (1) and (2) above we have (2) i L ( 0 ) 2 i L ( ) 1.54 i SW ( 0 ) i SW ( ) 1.16 and with these values we sketch i L () t and i SW () t as shown below. i L () t A i SW () t A t t 5. At t 0 the circuit is as shown below where v C ( 0 ) 24 V and thus the initial condition has been established. 38 Ω 24 V 2 Ω 50 µf vc () t The circuit for source. t > 0 is shown below where the current source has been replaced with a voltage 38 Ω 24 V 2 Ω 20 V 50 µf vc () t 4 V 40 Ω 50 µf v C () t Now, v C () t v Cf v Cn 4 Ae 1 ( RC) t 4 Ae 500t Circuit Analysis I with MATLAB Applications 10-47

16 Chapter 10 Forced and Total Response in RL and RC Circuits and with the initial condition v C ( 0 ) v C ( 0 ) 24 V 4 Ae 0 from which A 20 we get v C () t 4 20e 500t 6. For t > 0 the op amp circuit is as shown below. v in () t R v C v out () t Application of KCL at the minus () input yields and since or v 0 v v in C dv C R dt C dv C dt dv C dt v in R v in RC Integrating both sides and observing that v out () t v C () t we get v out () t v in t k RC where k is the constant of integration of both sides and it is evaluated from the given initial condition. Then, or k 0. Therefore, v C ( 0 ) v C ( 0 ) 0 0 k v out () t v in t u RC 0 () t and v in RC is the slope as shown below Circuit Analysis I with MATLAB Applications

17 Answers to Exercises slope v in RC 7. At t 0 the circuit is as shown below where v C ( 0 ) 150 V and thus the initial condition has been established. 175 KΩ 150 V 1 µf v C 0 ( ) The circuit for t> 0 is shown below where the voltage source 1 is absent for all positive time and the 50 KΩ is shorted out by the closed switch. 125 KΩ 50 V 1 µf v C () t For the circuit above v C () t v Cf v Cn 50 Ae t RC) 50 Ae 8t and with the initial condition v C ( 0 ) v C ( 0 ) Ae 0 from which A 100 and thus for t> 0 V v C () t e 8t To find v R3 () t we will first find i C () t from the circuit below where i C () t C dv C ( e 8t ) dt Circuit Analysis I with MATLAB Applications 10-49

18 Chapter 10 Forced and Total Response in RL and RC Circuits 50 V 25 KΩ 1 µf 100 KΩ v C () t v R3 () t Then, v R3 () t ( 100 KΩ)i C 10 5 ( e 8t ) 80e 8t V The sketches below show v C () t and v R3 () t as they approach their final values. 150 v C () t V v R3 () t V t t 8. For this circuit we cannot short the dependent source and therefore we cannot find by combining the resistances R 1 and R 2 in parallel combination in order to find the time constant τ R Instead, we will derive the time constant from the differential equation of (9.9) of the previous chapter, that is, R eq From the given circuit shown below dv C dt v C RC v C i C () t C 1 F v C () t R 1 18 Ω R 2 12 Ω 10i C () t or v i C 10i C v C C R 1 R Circuit Analysis I with MATLAB Applications

19 Answers to Exercises or or C dv C dt vc -----C dv C dt R 1 R 2 R dv C C R 1 R vc 0 dt R 1 R 2 R 1 dv C dt R 1 R R 1 R v C 0 C R 1 and from this differential equation we see that the coefficient of v C is and thus R eq C τ v C () t Ae t and with the given initial condition v C ( 0 ) V 0 A 5 V we get Then, using v C () t 5e t we find that for t > 0 and the minus () sign indicates that the i C C dv C dt i C () t ( 1) ( e t ) e t i C () t direction is opposite to that shown. Circuit Analysis I with MATLAB Applications 10-51

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