# Change of flux in an L-R circuit

1. Feb 23, 2012

### arnab321

1. The problem statement, all variables and given/known data

2. Relevant equations

afaik, flux in inductor = Li.
at t=0, current through L is 0. so change of flux = LΔi = L(i-0)=Li

3. The attempt at a solution

at t=∞, i through cap. is 0.

so,

i get the following equations:
5(i1+i2)=20
5(i1+i2)+ 5i2=10

i get i2=-2 and i1 = 6
i through indctor = i1+i2 = 4

so, flux=Li= 0.5*4=2

book's ans. is different.

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2. Feb 23, 2012

### kushan

t-o current is zero in the inductor lol

3. Feb 23, 2012

### kushan

but capacitor was charged at t-0

4. Feb 23, 2012

### vela

Staff Emeritus
You did everything right except for assuming the current through the inductor is 0 right before the switch is closed.

5. Feb 23, 2012

### arnab321

Well, um.... Current through inductor is i=i0(1-e^(-t/tau)). Putting t=0 in the equation, I get i=0.

6. Feb 23, 2012

### vela

Staff Emeritus
That equation only applies to a simple LR circuit with the initial condition i(0-)=0. That's not what you have here.

7. Feb 23, 2012

### arnab321

edit: oh I'm sorry I didn't see the battery there before the switch is closed.

I got the ans. 1.5.

8. Feb 23, 2012

### vela

Staff Emeritus
That's not entirely accurate. At steady state, a capacitor acts like an open circuit, and an inductor, like a short circuit. I wouldn't say, however, the opposite is true at the beginning of a transient.

So suppose the switch has been open a long time before t=0. You replace the inductor with a short. What's the current flowing through the upper-left resistor? That will be the initial current through the inductor.