# Current in a inductor

1. Apr 15, 2010

### Angello90

1. The problem statement, all variables and given/known data
If the current in the inductor 10mH decreases exponentially from its
initial value of 14 A with a time constant of 3 seconds, what is the voltage
across the inductor at the instant when the current reaches 7 A?

2. Relevant equations
V(t)=L di/dt

3. The attempt at a solution
di = 7, dt=3
thus
V=(10e-3)(7/3)=0.0233v

Am I correct? Seems to be too simple for me

2. Apr 15, 2010

### xcvxcvvc

you have to use the equation for a discharging inductor in an RL circuit.
$$V=IR$$
$$V_{inductor} - V_{resistor} = 0$$
$$L\frac{di}{dt} = IR$$
$$\frac{R}{L}dt = \frac{di}{i}$$
$$\frac{Rt}{L}= ln(i) + c_1$$
$$i(t) = c_2e^{\frac{Rt}{L}}$$
$$c_2$$ is the initial current (which is given)

3. Apr 16, 2010

### Angello90

but there is no resistor in the circuit

4. Apr 16, 2010

### xcvxcvvc

We know that the time constant is equal to:
[tex]\frac{L}{R}[/tex[
Since the inductor has a time constant other than infinity, we know that there is a resistance. If resistance was zero, the time constant would be infinity, and the inductor would never lose current (i.e. it would never lose energy) since there would be no resistive losses.

Since we know L and Timeconsant, we can solve for resistance if you wanted to know it.

5. Apr 17, 2010

### Angello90

Ok now I get it. It's all about internal resistance right?

Thanks a lot xcvxcvc