Power Dissipated By Resistor is LRC Circuit

In summary: So, if you want to use Vrms, you need to use the Vrms across the resistor. If you use the maximum voltage, your answer will be incorrect.In summary, to find the average power dissipated by a 25-Ω resistor in an LRC series ac circuit with a power factor of 0.25 and a maximum voltage of 8.0 V, you need to use the formula Pav = (1/2)(Vrms)(Irms)cos(phi). This is because Ohm's Law for an AC circuit is Vrms=Irms*R, and the Vrms across the resistor is not the same as the given maximum voltage. Using the maximum voltage will result in an incorrect answer.
  • #1
mj23
2
0
[Note by mentor: this post does not use the homework template because it was originally posted in a non-homework forum.]

---------------------------------------------

Problem:
What is the average power dissipated by a 25-Ω resistor in an LRC series ac circuit for which the power factor is equal to 0.25 and the maximum voltage of the ac source is 8.0 V?

Answer: .08W

Attempt at Solution:
We know that Pav = (1/2)(V)(I)cos(phi) where cos(phi) is the power factor. So if we substitute I for V/I (ohms law) we get Pav = (1.2)((V^2)/R)cos(phi). Plugging in all the values I get that Pav = .32W which is clearly incorrect...

What did I do wrong?
Thanks for the advice!
 
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  • #2
mj23 said:
[Note by mentor: this post does not use the homework template because it was originally posted in a non-homework forum.]

---------------------------------------------

Problem:
What is the average power dissipated by a 25-Ω resistor in an LRC series ac circuit for which the power factor is equal to 0.25 and the maximum voltage of the ac source is 8.0 V?

Answer: .08W

Attempt at Solution:
We know that Pav = (1/2)(V)(I)cos(phi) where cos(phi) is the power factor. So if we substitute I for V/I (ohms law) we get Pav = (1.2)((V^2)/R)cos(phi). Plugging in all the values I get that Pav = .32W which is clearly incorrect...

What did I do wrong?
Thanks for the advice!
Check your formulas. What is Ohm's Law for an AC circuit? How do you get the average power in terms of Vrms and R?
 
  • #3
ok i understand i made a typo by saying substitute I in for V/I meant V/R as you can see in the next step, but why do we need to use the Vrms in this situation rather than V?
 
  • #4
mj23 said:
ok i understand i made a typo by saying substitute I in for V/I meant V/R as you can see in the next step, but why do we need to use the Vrms in this situation rather than V?
Well, you can use the maximum voltage and current if you like, but I=V/R is valid for the voltage across the resistor which is not the same as the given 8.0 V.
 

1. What is Power Dissipation?

Power dissipation is the process of converting electrical energy into heat energy. In other words, it is the rate at which energy is lost or used up in a circuit.

2. How is Power Dissipation related to Resistors?

Power dissipation in a circuit is directly proportional to the resistance of the circuit. This means that the higher the resistance of a circuit, the more power will be dissipated as heat.

3. What is an LRC Circuit?

An LRC circuit is a type of electrical circuit that contains inductance (L), resistance (R), and capacitance (C) components. These circuits are commonly used in electronic devices such as radios and televisions.

4. How does Power Dissipation occur in an LRC Circuit?

In an LRC circuit, power dissipation occurs due to the flow of current through the resistive component. As the current passes through the resistor, some of its electrical energy is converted into heat energy, resulting in power dissipation.

5. How can Power Dissipation be calculated in an LRC Circuit?

The power dissipated in an LRC circuit can be calculated using the formula P = I^2 * R, where P is power dissipation in watts, I is the current in amperes, and R is the resistance in ohms. Alternatively, it can also be calculated using the formula P = V^2 / R, where V is the voltage in volts.

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