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Power in an RL Circuit

An RL circuit in which L = 9.00 H and R = 5.00 is connected to a 24.0 V battery at t = 0.
(a) What energy is stored in the inductor when the current is 0.500 A?
(b) At what rate is energy being stored in the inductor when I = 1.00 A?
(c) What power is being delivered to the circuit by the battery when I = 0.500 A?

I got the energy; that was pretty straightforward. Just U=1/2 LI^2.

For the rate of energy, I tried solving for energy at 1.00 A, then solving for t when inductor is at 1A. To do that I used I=V/R(1-e^-Rt/L), then I divided energy by the time to find power but that didn't work. Any tips?
 

Answers and Replies

454
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Any power delivered by the battery not used up in the resistor is stored in the inductor
 
Okay, I found the formula IE=RI^2+IL(dI/dt) and I used that to solve (b), which turned out to be 19 W, but that formula didn't work when I tried it on (c). I think what I have to do is find the total power in the system and subtract my answer from (b) from that. But I'm not sure. Any suggestions?
 
454
0
Okay, I found the formula IE=RI^2+IL(dI/dt) and I used that to solve (b), which turned out to be 19 W, but that formula didn't work when I tried it on (c). I think what I have to do is find the total power in the system and subtract my answer from (b) from that. But I'm not sure. Any suggestions?
c) seems suspiciously easy. The power that a battery delivers is just V*I. Using
I = V/R(1-e^-Rt/L) and Power = RI^2+IL(dI/dt) works as well but seems rather complicated compared with just computing I*V
 

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