How Do You Calculate Different Parameters in an RL Circuit?

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Homework Help Overview

The discussion revolves around calculating various parameters in an RL circuit, specifically focusing on energy stored in the inductor, the rate of energy storage, and the power delivered by the battery. The circuit parameters include an inductance of 9.00 H, resistance of 5.00 ohms, and a voltage source of 24.0 V.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the energy stored in the inductor using the formula U=1/2 LI^2 and explore the rate of energy storage by attempting to relate current, voltage, and resistance through various equations. There are questions about the application of formulas for power calculations and the interpretation of results.

Discussion Status

Some participants have found specific formulas and attempted calculations, with one noting a result for the rate of energy storage. However, there is uncertainty regarding the application of these formulas to different parts of the problem, and suggestions for alternative approaches are being explored.

Contextual Notes

Participants are navigating potential complexities in the relationships between current, voltage, and power in the circuit, with some expressing confusion over the application of certain equations and the implications of their results.

GreenLantern674
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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?
 
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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?
 
GreenLantern674 said:
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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