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Question 07: Find ya(t) in the circuit shown in Figure 7. Note that, since the voltage input is multiplied by u(t), the voltage source is a short for all t < 0 and iL(0) = 0. Copyright© The McGraw-Hill Companies, Inc. Permission required for reproduction or display 12 75e 2u(1) V 2 H 3 2Ω v,(1) Figure 7. ll

Question 07: Find ya(t) in the circuit shown in Figure 7. Note that, since the voltage input is multiplied by u(t), the voltage source is a short for all t < 0 and iL(0) = 0. Copyright© The McGraw-Hill Companies, Inc. Permission required for reproduction or display 12 75e 2u(1) V 2 H 3 2Ω v,(1) Figure 7. ll

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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### Question 07: **Problem Statement:** Find \( v_o(t) \) in the circuit shown in Figure 7. Note that, since the voltage input is multiplied by \( u(t) \), the voltage source is a short for all \( t < 0 \) and \( i_L(0) = 0 \). **Illustrative Diagram:** The circuit diagram is represented as follows: - The voltage source is given by \( 75e^{-2t}u(t) \) V. - The circuit includes: - A resistor of \( 1 \Omega \) in series with the voltage source. - A parallel connection of: - An inductor of \( 2 H \). - A resistor of \( 2 \Omega \). - The output voltage \( v_o(t) \) is measured across the \( 2 \Omega \) resistor in the parallel branch. **Figure 7 Explanation:** The circuit diagram consists of a voltage source \( 75e^{-2t}u(t) \) V, connected in series with a \( 1 \Omega \) resistor. Following this series combination, the connection splits into two parallel branches: 1. One branch contains a \( 2 H \) inductor. 2. The other branch contains a \( 2 \Omega \) resistor. The output voltage \( v_o(t) \) is taken across the \( 2 \Omega \) resistor. **Additional Notes:** - The unit step function \( u(t) \) implies that the voltage source is active for \( t \geq 0 \). - The initial current through the inductor \( i_L(0) \) is zero, indicating there was no prior energy stored in the inductor.
Transcribed Image Text:### Question 07: **Problem Statement:** Find \( v_o(t) \) in the circuit shown in Figure 7. Note that, since the voltage input is multiplied by \( u(t) \), the voltage source is a short for all \( t < 0 \) and \( i_L(0) = 0 \). **Illustrative Diagram:** The circuit diagram is represented as follows: - The voltage source is given by \( 75e^{-2t}u(t) \) V. - The circuit includes: - A resistor of \( 1 \Omega \) in series with the voltage source. - A parallel connection of: - An inductor of \( 2 H \). - A resistor of \( 2 \Omega \). - The output voltage \( v_o(t) \) is measured across the \( 2 \Omega \) resistor in the parallel branch. **Figure 7 Explanation:** The circuit diagram consists of a voltage source \( 75e^{-2t}u(t) \) V, connected in series with a \( 1 \Omega \) resistor. Following this series combination, the connection splits into two parallel branches: 1. One branch contains a \( 2 H \) inductor. 2. The other branch contains a \( 2 \Omega \) resistor. The output voltage \( v_o(t) \) is taken across the \( 2 \Omega \) resistor. **Additional Notes:** - The unit step function \( u(t) \) implies that the voltage source is active for \( t \geq 0 \). - The initial current through the inductor \( i_L(0) \) is zero, indicating there was no prior energy stored in the inductor.
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