# Transient response of RL Circuit

1. Jan 4, 2014

### Cetullah

1. The problem statement, all variables and given/known data
Find the expression of the current "i" by terms of time.

2. Relevant equations

i(t)=I_final+(I_initial-I_final)*e^(-t/τ)

3. The attempt at a solution
I have found the initial current as 3 amps. However, after the switch had been changed for too long, I m not sure about the way how the inductor behaves. Will it be a short circuit? If so, there is some problem about the loop on the right, like 8ix-6ix=0
Can the dependant source behave like that? Also I have problem about finding the time constant, I find 1/τ as 60.

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2. Jan 4, 2014

### Staff: Mentor

The controlled source is going to throw a monkey wrench into a simplistic solution involving time constants based upon the passive components alone. The controlled source will alter the energy available in the circuit over time, so it won't behave like a typical passive RL circuit.

I suggest writing the differential equation for the loop based on KVL. You know the initial value of the current. (A very easy way is to us the Laplace domain/transform method).

I can confirm that the given answer is correct.

3. Jan 4, 2014

### rude man

Hard to read your image, but it looks like the controlled current source is 8 i_x and it's in series with i_x. That unfortunately is impossible.

4. Jan 4, 2014

### Staff: Mentor

Not desirable perhaps, but not impossible to solve. It means that there will be no steady state for as t → ∞. Sort of analogous to an amplifier with positive feedback. Take look at the proposed solution, and in particular, the sign of the exponent.

5. Jan 4, 2014

### rude man

Are you confirming that the 8i_x source is in series with i_x? 'Cause if you are, that is impossible.

6. Jan 4, 2014

### Staff: Mentor

I am confirming that the given solution can be derived from the circuit as shown.

EDIT: It just occurred to me that perhaps there is some confusion about the 8ix controlled source being in series with the ix current. That source is a controlled VOLTAGE source, so there is no contradiction in the math.

Last edited: Jan 4, 2014