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Transient analysis: 1) Use Thevenin's theorem to convert the circuit of Figure 1 to the circuit of Figure 2 by finding out the values of V1 and R1. Now using the circuit of Figure 2, find out the expression of i_c(t) as a function of time. Figure 1: 1 k? 1.25 k t = 0 8 V 3 kl 2 mA 1 µF- i.(t) Figure 2: R1 =? kN Vị =? V 1 µF vi(t) wps.com 2) Now we want to check if the superposition principle works in this case or not. First we turn the voltage source off and keep the current source on in the following Figure 1. Then we convert it to the circuit of Figure 2 using Thevenin's theorem. Now using the circuit of Figure 2, find out the expression of ic_1(t) as a function of time. Figure 1: 1 kN 1.25 kN t = 0 ov 3 kn 2 mA ( 1 µF Figure 2: R =? kN t= 0 V =? V 1 µF vic_1(t) 3) Now we turn the current source off and keep the voltage source on in the following Figure 1. Then we convert it to the circuit of Figure 2 using Thevenin's theorem. Now using the circuit of Figure 2, find out the expression of ic_2(t) as a function of time. Figure 1: 1 kN 1.25 k 8 V 3 kN, O mA 1 µF Figure R1 =? kf? t= 0 Vị =? V 1 µF v ic_2(t) 4) Is ic¡(1) + ic,(t) = ic(t) ? ( ic(t) being the expression that we found in step 1) Does the superposition principle work in this case for time varying current also?

Transient analysis: 1) Use Thevenin's theorem to convert the circuit of Figure 1 to the circuit of Figure 2 by finding out the values of V1 and R1. Now using the circuit of Figure 2, find out the expression of i_c(t) as a function of time. Figure 1: 1 k? 1.25 k t = 0 8 V 3 kl 2 mA 1 µF- i.(t) Figure 2: R1 =? kN Vị =? V 1 µF vi(t) wps.com 2) Now we want to check if the superposition principle works in this case or not. First we turn the voltage source off and keep the current source on in the following Figure 1. Then we convert it to the circuit of Figure 2 using Thevenin's theorem. Now using the circuit of Figure 2, find out the expression of ic_1(t) as a function of time. Figure 1: 1 kN 1.25 kN t = 0 ov 3 kn 2 mA ( 1 µF Figure 2: R =? kN t= 0 V =? V 1 µF vic_1(t) 3) Now we turn the current source off and keep the voltage source on in the following Figure 1. Then we convert it to the circuit of Figure 2 using Thevenin's theorem. Now using the circuit of Figure 2, find out the expression of ic_2(t) as a function of time. Figure 1: 1 kN 1.25 k 8 V 3 kN, O mA 1 µF Figure R1 =? kf? t= 0 Vị =? V 1 µF v ic_2(t) 4) Is ic¡(1) + ic,(t) = ic(t) ? ( ic(t) being the expression that we found in step 1) Does the superposition principle work in this case for time varying current also?

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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I know you cant solve all due to the rules ,But the question is interrelated, Kindly solve all, its a request
Transient analysis: 1) Use Thevenin's theorem to convert the circuit of Figure 1 to the circuit of Figure 2 by finding out the values of V1 and R1. Now using the circuit of Figure 2, find out the expression of i_c(t) as a function of time. Figure 1: 1 k! 1.25 k? t = 0 3 kN. 2 mA 1 µF- i(t) Figure 2: R1 =? kN t=0 Vị =? V 1 µF vi(t) wps.com 2) Now we want to check if the superposition principle works in this case or not. First we turn the voltage source off and keep the current source on in the following Figure 1. Then we convert it to the circuit of Figure 2 using Thevenin's theorem. Now using the circuit of Figure 2, find out the expression of ic_1(t) as a function of time. Figure 1: 1 k 1.25 ko t= 0 ov 3 kN 2 ma () 1 µF Figure 2: R1 =? k£! t =0 V =? V 1 µF v ic_1(t) 3) Now we turn the current source off and keep the voltage source on in the following Figure 1. Then we convert it to the circuit of Figure 2 using Thevenin's theorem. Now using the circuit of Figure 2, find out the expression of ic_2(t) as a function of time. Figure 1: 1 kN 1.25 k? 8 V 3 kN. O mA 1 µF Figure R1 =? k? t= 0 Vị =? V 1 µF v ic_2(t) 4) Is ic¡(t) + ic,(t) = ic(t)? ( ic(t) being the expression that we found in step 1) Does the superposition principle work in this case for time varying current also?
Transcribed Image Text:Transient analysis: 1) Use Thevenin's theorem to convert the circuit of Figure 1 to the circuit of Figure 2 by finding out the values of V1 and R1. Now using the circuit of Figure 2, find out the expression of i_c(t) as a function of time. Figure 1: 1 k! 1.25 k? t = 0 3 kN. 2 mA 1 µF- i(t) Figure 2: R1 =? kN t=0 Vị =? V 1 µF vi(t) wps.com 2) Now we want to check if the superposition principle works in this case or not. First we turn the voltage source off and keep the current source on in the following Figure 1. Then we convert it to the circuit of Figure 2 using Thevenin's theorem. Now using the circuit of Figure 2, find out the expression of ic_1(t) as a function of time. Figure 1: 1 k 1.25 ko t= 0 ov 3 kN 2 ma () 1 µF Figure 2: R1 =? k£! t =0 V =? V 1 µF v ic_1(t) 3) Now we turn the current source off and keep the voltage source on in the following Figure 1. Then we convert it to the circuit of Figure 2 using Thevenin's theorem. Now using the circuit of Figure 2, find out the expression of ic_2(t) as a function of time. Figure 1: 1 kN 1.25 k? 8 V 3 kN. O mA 1 µF Figure R1 =? k? t= 0 Vị =? V 1 µF v ic_2(t) 4) Is ic¡(t) + ic,(t) = ic(t)? ( ic(t) being the expression that we found in step 1) Does the superposition principle work in this case for time varying current also?
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