Power dissipated in an RL circuit

In summary, the conversation discusses a circuit with two unknown elements, and the current and phase angle when connected to a 2.16×102 V, 60.0 Hz source. The two elements are determined to be an LR circuit and the power dissipated in the circuit is calculated using the real component of the complex power and the given current's phase angle. The conversation also mentions using RMS values and considering the power factor of the circuit.
  • #1
pengumon
4
0

Homework Statement



a) A circuit contains two elements, but it is not known if they are L, R or C. The current in this circuit when connected to a 2.16×102 V, 60.0 Hz source is 6.10 A and lags the voltage by 3.60×10^1 degrees. What are the two elements?

c) What is the power dissipated in this circuit?

Homework Equations



P=IV

The Attempt at a Solution



I determined that it was an LR circuit and calculated the impedance Z for part b) of the question but I'm stuck on the power. I did a simple P=IR to get an answer of 1.32E3 W but that was wrong, and I also tried dividing the I and V given by √2 to get rms values, then multiplying, which gave me 659 W but it's still wrong. Any insights or nudges in the right direction would be much appreciated!

Edit: I just realized my angle is 36 degrees, not 360 like I originally thought. That means I need to have a cos36 in there somewhere right? I'm still unsure as to whether I'm supposed to be using rms values or not though.
 
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  • #2
pengumon said:

Homework Statement



a) A circuit contains two elements, but it is not known if they are L, R or C. The current in this circuit when connected to a 2.16×102 V, 60.0 Hz source is 6.10 A and lags the voltage by 3.60×10^1 degrees. What are the two elements?

c) What is the power dissipated in this circuit?

Homework Equations



P=IV

The Attempt at a Solution



I determined that it was an LR circuit and calculated the impedance Z for part b) of the question but I'm stuck on the power. I did a simple P=IR to get an answer of 1.32E3 W but that was wrong, and I also tried dividing the I and V given by √2 to get rms values, then multiplying, which gave me 659 W but it's still wrong. Any insights or nudges in the right direction would be much appreciated!

Edit: I just realized my angle is 36 degrees, not 360 like I originally thought. That means I need to have a cos36 in there somewhere right? I'm still unsure as to whether I'm supposed to be using rms values or not though.

You can assume that the voltage and current values given are RMS values unless otherwise indicated.

Note that the given current has a phase angle associated with it, which implies a power factor for the circuit. When they ask for power they want the real component of the complex power, which is the magnitude of the total power multiplied by the power factor.
 
  • #3
I got it, thank you!
 

1. What is power dissipation in an RL circuit?

Power dissipation in an RL circuit refers to the amount of energy that is converted into heat and lost as the current flows through the circuit. This heat is produced due to the resistance of the components in the circuit, and is measured in watts (W).

2. How is power dissipated in an RL circuit calculated?

The power dissipated in an RL circuit can be calculated using the formula P = I^2R, where P is power in watts, I is the current in amperes (A), and R is the resistance in ohms (Ω). Alternatively, you can also use the formula P = VI, where V is the voltage in volts (V) and I is the current.

3. What factors affect power dissipation in an RL circuit?

The main factor that affects power dissipation in an RL circuit is the resistance. The higher the resistance, the more power will be dissipated. Additionally, the type and size of the components used in the circuit can also affect power dissipation.

4. How does power dissipation impact the overall performance of an RL circuit?

Power dissipation can have a significant impact on the overall performance of an RL circuit. Too much power dissipation can lead to overheating and damage to components, potentially causing the circuit to fail. It can also result in a decrease in efficiency and accuracy of the circuit.

5. What can be done to reduce power dissipation in an RL circuit?

One way to reduce power dissipation in an RL circuit is to lower the resistance by using components with lower resistance values. Additionally, using larger components or implementing cooling measures can also help dissipate heat and reduce power dissipation. Proper circuit design and layout can also play a role in reducing power dissipation.

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