How Do You Calculate Different Parameters in an RL Circuit?

In summary, when an RL circuit with L = 9.00 H and R = 5.00 is connected to a 24.0 V battery at t = 0, the energy stored in the inductor is found by U=1/2 LI^2. The rate of energy being stored in the inductor at 1.00 A is 19 W, and the power being delivered to the circuit by the battery at 0.500 A is 12 W. The power delivered by the battery can also be found by multiplying the voltage and current.
  • #1
GreenLantern674
27
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?
 
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  • #2
Any power delivered by the battery not used up in the resistor is stored in the inductor
 
  • #3
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?
 
  • #4
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
 

Related to How Do You Calculate Different Parameters in an RL Circuit?

1. What is power in an RL circuit?

Power in an RL (resistor-inductor) circuit refers to the rate at which energy is transferred from the power source to the circuit. It is measured in watts (W) and represents the amount of work that can be done per unit of time.

2. How is power calculated in an RL circuit?

Power in an RL circuit can be calculated using the formula P = I^2R, where P is power in watts, I is current in amperes, and R is resistance in ohms. It can also be calculated using the formula P = VI, where V is voltage in volts and I is current in amperes.

3. What is the relationship between power and current in an RL circuit?

In an RL circuit, power is directly proportional to current. This means that as current increases, power also increases, and vice versa. This relationship is described by the formula P = I^2R, where P is power, I is current, and R is resistance.

4. How does the presence of an inductor affect power in an RL circuit?

The presence of an inductor in an RL circuit affects power by causing a delay in the flow of current. This causes a phase shift between the voltage and current, resulting in a decrease in power. However, once the inductor reaches its steady state, power will be equal to the value calculated using the formula P = I^2R.

5. What is the significance of power factor in an RL circuit?

Power factor in an RL circuit is a measure of the efficiency of power transfer from the source to the circuit. It is calculated by dividing the real power (P) by the apparent power (S), and is represented by the symbol cos(Φ). A higher power factor indicates a more efficient use of power, while a lower power factor means that some of the power is being wasted.

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