- #1

Truman I

- 16

- 4

- Homework Statement
- 1) For the circuit shown in the figure, the switch has been open for a very long time.

(a) What is the potential drop across the 15.0-mH inductor just after closing the switch?

- Relevant Equations
- ε = L*(di/dt)

V_R = iR

At t=0, I believe that the current is

*instantaneously*0 Amps. If that is correct, then technically at that instant there is no voltage drop across any of the resistors due to Ohm's Law. So I replaced the resistors with wire. Next, I tried replacing all of the capacitors with open circuits to simplify everything further. When I did that, there was only one closed loop for the current to travel through, and that's the loop with my 15-mH inductor. But that's where I'm getting lost.

My only other thought is that I need to use the equation ε = L*(di/dt) but there are two unknowns. I want to find ε for the 15 mH inductor, but I don't know di/dt.