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At time, t = 0 the current in an RL - series circuit is zero Amperes. Suppose that a variable resistor, R = N and a 4 - Henry inductor are connected in series with constant electromotive force of Evolts. Determine E so that i = 10A when t = 10sec.

At time, t = 0 the current in an RL - series circuit is zero Amperes. Suppose that a variable resistor, R = N and a 4 - Henry inductor are connected in series with constant electromotive force of Evolts. Determine E so that i = 10A when t = 10sec.

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
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The topic is about Applications of First Order Differential Equations. The problem given is about RL Circuits. The first image attached is how I want to answer the given problem which can be seen on the second image attached. 

Please provide a complete solution. (of how I want to answer it) Thanks! Will give you a thumbs up.

At time, t = 0 the current in an RL - series circuit is zero Amperes. Suppose that a variable resistor, R = 3t+34 N and a 4 - Henry inductor are connected in series with constant electromotive force of E volts. Determine E so that i = 10A when t = 10sec.
Transcribed Image Text:At time, t = 0 the current in an RL - series circuit is zero Amperes. Suppose that a variable resistor, R = 3t+34 N and a 4 - Henry inductor are connected in series with constant electromotive force of E volts. Determine E so that i = 10A when t = 10sec.
A battery giving a constant voltage of 40 V is connected in series to a resistor of resistance 20 Q and an inductor of inductance 1 H. If the i = Ke-20t + 2 initial current in the circuit is 3 amperes, find the current after t seconds. What is the steady state current? i(0) = 3D ЗА: K = 1.000 -20t .: i(t) = e + 2 di + Ri = E dt Given: Solution: E = 40 V -20t di + 20i = 40 dt B] lim i(t) = lim e + lim 2 R = 20 N 1 L = 1 H P(t) = 20; Q(t) = 40 lim i(t) + 2 -20(0) e i(0) = 3 A Ø = eS 20dt = e20t lim i(t) = 2 A Required: | e 20t dt 20t a) i(t) .. i· e = 40 Transient Term: e-20t b) lim i(t) i• e20t = 2e20t + K
Transcribed Image Text:A battery giving a constant voltage of 40 V is connected in series to a resistor of resistance 20 Q and an inductor of inductance 1 H. If the i = Ke-20t + 2 initial current in the circuit is 3 amperes, find the current after t seconds. What is the steady state current? i(0) = 3D ЗА: K = 1.000 -20t .: i(t) = e + 2 di + Ri = E dt Given: Solution: E = 40 V -20t di + 20i = 40 dt B] lim i(t) = lim e + lim 2 R = 20 N 1 L = 1 H P(t) = 20; Q(t) = 40 lim i(t) + 2 -20(0) e i(0) = 3 A Ø = eS 20dt = e20t lim i(t) = 2 A Required: | e 20t dt 20t a) i(t) .. i· e = 40 Transient Term: e-20t b) lim i(t) i• e20t = 2e20t + K
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