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In Fig. 30-77,R1=8.0Ω,R2=10Ω,L1=0.30H,L2=0.20Hand the ideal battery hasε=6.0V. (a) Just after switch S is closed, at what rate is the current in inductor 1 changing? (b) When the circuit is in the steady state, what is the current in inductor 1?

Short Answer

Expert verified

a) The rate of change of current in inductor 1 is,

dibatdt=20A/s

b) The current in the inductor 1 is i=0.75A

Step by step solution

01

Given

R1=8.0ΩR2=10ΩL2=0.20Hε=6.0V

02

Understanding the concept

By using the equation 30-35, we can find the rate of current. After that by using the KVL

We can find the current in the inductor 1.

Formula:

ibat=εR1-eRtLdibatdt=εL1

03

(a) Calculate the rate of change of current in inductor 1.

We have, equation for self-induced emf as,

L1dibatdt=ε

So that the rate of current is,

dibatdt=εL1

By substituting the value

dibatdt=6.00.30dibatdt=20A/s

04

(b) Calculate the current in the inductor 1 at steady state.

Now we have to find the current in inductor 1 so that by Appling the KVL for outer loop we can writer as

ε-iR1-εL1-εL2=0

But as t→∞emfin the coil vanishes

ε-iR1=0ε-iR1i=εR1i=6.08.0i=0.75A

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