# Simple RL Circuit

1. Jan 28, 2010

### KillerZ

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

The switch has been closed for a long time.

a) Current flowing through inductor. Voltage across it. IS any current flowing through the resistor.

b) The switch opens at t = 0s. Find an equation for the inductor voltage as a function of time for t > 0. How long does it take for the inductor to reach 10% of its maximum value?

2. Relevant equations

Kirchhoff's Voltage Law (KVL)

3. The attempt at a solution

a) 1A is flowing throw through the inductor as the circuit is stable and the inductor is replaced with a short circuit this causes the resistor to have no current through it and the inductor to have no voltage across it.

b) This is where I am not sure about:

t > 0

I used the KVL equation around the circuit to get:

$$-(100)10^{-3}\frac{di_{L}}{dt} + 1i_{L} = 0$$

I am not sure if this is the right way to do this.

2. Jan 30, 2010

### ehild

-L *di/dl is the emf arising if the current change. So your equation is

$$-0.1\frac{di_{L}}{dt} = 1i_{L}$$

The starting condition is $$I_{L}(0)=1 A$$

The equation is easy to solve.

ehild

3. Jan 31, 2010

### KillerZ

Ok I attempted this and here is what I got:

KVL:

$$iR + v = 0$$

$$iR + L\frac{di}{dt} = 0$$

$$\frac{di}{dt} = -\frac{iR}{L}$$

$$-\frac{di}{i} = \frac{R}{L}dt$$

$$-lni + C = \frac{R}{L}t$$

$$I_{L}(0)=1 A$$

$$-ln1 + C = \frac{R}{L}0$$

$$C = 0$$

$$-lni + 0 = \frac{R}{L}t$$

$$0 = \frac{R}{L}t + lni$$

$$1 = e^{\frac{R}{L}t} + i$$

$$i = 1 - e^{\frac{R}{L}t}$$

$$v = L\frac{di}{dt}$$

$$v = L\frac{d(1 - e^{\frac{R}{L}t})}{dt}$$

$$v = -Re^{\frac{R}{L}t}$$

4. Jan 31, 2010

### ehild

Is is correct up to here, but wrong from here.

You have a "* " instead of "+"

$$i = e^{-\frac{R}{L}t}$$

$$v = \frac{di}{dt}$$

$$v = L\frac{d(e^{-\frac{R}{L}t})}{dt}$$

$$v = -Re^{-\frac{R}{L}t}$$

ehild

5. Jan 31, 2010

### KillerZ

Ok I got it thanks for the help.