How Do You Calculate Different Parameters in an RL Circuit?

Click For Summary
In an RL circuit with an inductor of 9.00 H and a resistor of 5.00 ohms connected to a 24.0 V battery, the energy stored in the inductor at 0.500 A is calculated using the formula U=1/2 LI^2. For the rate of energy storage at 1.00 A, the formula IE=RI^2+IL(dI/dt) was used, yielding a result of 19 W. The power delivered by the battery at 0.500 A can be computed simply as P=VI, which is more straightforward than using complex formulas. Suggestions for finding the total power in the system involve subtracting the power used by the resistor from the total power delivered. The discussion emphasizes the importance of using the simplest calculations for efficiency.
GreenLantern674
Messages
27
Reaction score
0
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?
 
Physics news on Phys.org
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
 
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

Similar threads

  • · Replies 1 ·
Replies
1
Views
868
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 15 ·
Replies
15
Views
832
Replies
6
Views
1K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K