# RL Circuit

1. Mar 7, 2014

### freezer

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

For the circuit shown below, the current through the inductor is defined as follows:

$$i(t)=\begin{cases} 0 & \text{if } -\infty <t<0 \\ 1-e^{-2t}& \text{if } 0\leq t < \infty \end{cases}$$

L = 1H, R2 = 2 ohms

For t=> 0 find:

a) VL(t)
b) Vr(t)
c) V1(t)
d) Power absorbed by the inductor
2. Relevant equations

$$i(t)=\frac{v}{R}(1-e^{-Rt/L})$$

3. The attempt at a solution

As t goes to infinity, current goes to 1A and the inductor becomes a wire. So V1 = 2V

But i don't think this is what it is asking for.
From the relevant equation section, solving for V would leave only 2

W = 0.5Li^2

W = 0.5(1)(1-e^{-2t})^2

Vr = (1-exp(-2t))/2

Vl = 0V
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Mar 8, 2014

### Simon Bridge

Looks like it wants t→0 ... for very small times, the exponential term is approximately linear.