Current in Inductor: Calculate Voltage at 7A

[Angello90]

Homework Statement

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?

Homework Equations

V(t)=L di/dt

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

 

[xcvxcvvc]

Homework Statement

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?

Homework Equations

V(t)=L di/dt

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

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)

 

[Angello90]

but there is no resistor in the circuit

 

[xcvxcvvc]

but there is no resistor in the circuit

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

 

[Angello90]

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

Thanks a lot xcvxcvc