Solve ysing step by step method t3DEz3H§32i,SAG,(6) for toVsing step bey Stepmethad

Answered: t=0 E, 3H @SA §32 TH

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

Solve ysing step by step method

t3D
Ez
3H
§32
i,
SA
G,(6) for to
Vsing step bey Step
methad

Expert Solution

Step 1

Before opening the switch at t=0, the switch was closed for long time. Hence all inductors are replaced by short circuit.

The circuit before opening the switch i.e. at t=0- is

Equivalent resistance of parallel combination of $3 Ω a n d 2 Ω i s \frac{3 \times 2}{3 + 2} = \frac{6}{5} = 1.2 Ω$

Hence 5 A current is divided between $8 Ω a n d 1.2 Ω$ .

Applying current division

$i = \frac{8}{8 + 1.2} \times 5 = \frac{40}{9.2} = 4.35 A$

Current i is further divided into i₁ and i₂.

Applying current division

$i_{1} (t = 0 -) = \frac{2}{3 + 2} i = \frac{2}{5} \times 4.35 = 1.74 A a n d i_{2} (t = 0 -) = \frac{3}{3 + 2} i = \frac{3}{5} \times 4.35 = 2.61 A$

Step 2

The circuit just after opening the switch i.e. at t=0+ is

1 H and 2 H inductors can be replaced by their equivalent i.e. $\frac{1 \times 2}{1 + 2} = \frac{2}{3} H$ inductor.

Hence the circuit is modified to

The currents i₁(t=0+) and i₂(t=0+) are same as that of i₁(t=0-) and i₂(t=0-) as inductor oppose sudden change in current.

Gradullay i₁ will increase and i₂will decrease. And at steady state both will be equal in magnitude but opposite in direction as they are the currents in the ssame loop.

Applying Kirchoff's loop law

$3 i_{1} (t) + \frac{2}{3} \frac{d i_{1} (t)}{d t} + 2 i_{1} (t) + 3 \frac{d i_{1} (t)}{d t} = 0 \Rightarrow 5 i_{1} (t) + \frac{11}{3} \frac{d i_{1} (t)}{d t} = 0 \Rightarrow \frac{d i_{1} (t)}{d t} + \frac{15}{11} i_{1} (t) = 0 \Rightarrow \frac{d i_{1} (t)}{i_{1} (t)} = - \frac{15}{11} d t I n t e g r a t i n g b o t h s i d e s \ln (i_{1} (t)) - \ln (i_{1} (0 +)) = - \frac{15}{11} t \Rightarrow i_{1} (t) = i_{1} (0 +) e^{- \frac{15}{11} t} \Rightarrow i_{1} (t) = 1.74 e^{- \frac{15}{11} t}$

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