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Does resistor power rating mean 'real' power?
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Given a mixed capacitive-resistive circuit, I have a 36 ohm resistor through which a current of 9.22mA @ 86º passes.
My AC theory is rusty, but wouldn't I do:
P = I^2 R = 3.06 mW @ 172º
Preal = Pcos172º = -3.03 mW
This is a negative value, which is probably wrong. Which figure should I be comparing against a resistor specsheet's power rating? (I realize nearly all resistors will accommodate 3mW; this is a question to straighten out my theory.)
Edit
I'd totally forgotten about the conjugate rule. With the adjusted calculation, I get
(9.22mA)^2 * 36 = 3.06 mW (peak)
(6.52mA)^2 * 36 = 1.53 mW (RMS)
power ac
asked 1 hour ago
Reinderien
689312
add a comment |Â
up vote
2
down vote
favorite
Given a mixed capacitive-resistive circuit, I have a 36 ohm resistor through which a current of 9.22mA @ 86º passes.
My AC theory is rusty, but wouldn't I do:
P = I^2 R = 3.06 mW @ 172º
Preal = Pcos172º = -3.03 mW
This is a negative value, which is probably wrong. Which figure should I be comparing against a resistor specsheet's power rating? (I realize nearly all resistors will accommodate 3mW; this is a question to straighten out my theory.)
Edit
I'd totally forgotten about the conjugate rule. With the adjusted calculation, I get
(9.22mA)^2 * 36 = 3.06 mW (peak)
(6.52mA)^2 * 36 = 1.53 mW (RMS)
power ac
asked 1 hour ago
Reinderien
689312
Resistors don't create reactive power, so what else would it mean?
– immibis
1 min ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Given a mixed capacitive-resistive circuit, I have a 36 ohm resistor through which a current of 9.22mA @ 86º passes.
My AC theory is rusty, but wouldn't I do:
P = I^2 R = 3.06 mW @ 172º
Preal = Pcos172º = -3.03 mW
This is a negative value, which is probably wrong. Which figure should I be comparing against a resistor specsheet's power rating? (I realize nearly all resistors will accommodate 3mW; this is a question to straighten out my theory.)
Edit
I'd totally forgotten about the conjugate rule. With the adjusted calculation, I get
(9.22mA)^2 * 36 = 3.06 mW (peak)
(6.52mA)^2 * 36 = 1.53 mW (RMS)
power ac
asked 1 hour ago
Reinderien
689312
Given a mixed capacitive-resistive circuit, I have a 36 ohm resistor through which a current of 9.22mA @ 86º passes.
My AC theory is rusty, but wouldn't I do:
P = I^2 R = 3.06 mW @ 172º
Preal = Pcos172º = -3.03 mW
This is a negative value, which is probably wrong. Which figure should I be comparing against a resistor specsheet's power rating? (I realize nearly all resistors will accommodate 3mW; this is a question to straighten out my theory.)
Edit
I'd totally forgotten about the conjugate rule. With the adjusted calculation, I get
(9.22mA)^2 * 36 = 3.06 mW (peak)
(6.52mA)^2 * 36 = 1.53 mW (RMS)
power ac
power ac
asked 1 hour ago
Reinderien
689312
asked 1 hour ago
Reinderien
689312
edited 49 mins ago
asked 1 hour ago
Reinderien
689312
asked 1 hour ago
Reinderien
689312
asked 1 hour ago
Reinderien
689312
689312
Resistors don't create reactive power, so what else would it mean?
– immibis
1 min ago
add a comment |Â
Resistors don't create reactive power, so what else would it mean?
– immibis
1 min ago
Resistors don't create reactive power, so what else would it mean?
– immibis
1 min ago
Resistors don't create reactive power, so what else would it mean?
– immibis
1 min ago
add a comment |Â
1 Answer
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I have a 36 ohm resistor through which a current of 9.22mA @ 86º passes.
If the frequencies involved are not too high, the voltage associated with this current will also have a phase angle of 86º.
As a result, the power of the resistor will be 100% real power. There will be no reactive power. In order to have reactive power, you must have reactive elements: inductors or capacitors.
P = I^2 R = 3.06 mW @ 172º
The formula $P=I^2R$ isn't meant for phasor cases. For phasors, you should use
$$P=I^star IR$$
where $I^star$ is the complex conjugate of $I$.
In general, the power consumed by a component with phasor voltage $V$ and current $I$ is
$$S=VI^star$$
giving complex power $S$, with $S=P+jQ$ where $P$ is real power and $Q$ is reactive power.
With this correction you will get a real result, as expected.
Which figure should I be comparing against a resistor specsheet's power rating?
Since the power used by a resistor is entirely real power, you need to calculate the real power and compare it to the rating.
If the frequency applied is very low, you may need to compare the peak power rather than the average power, since it could be possible for the resistor to burn out at the peak of the waveform.
answered 1 hour ago
The Photon
79.8k394189
Awesome! I've fixed my calculation in the question. Hopefully what I've done for RMS conversion also makes sense.
– Reinderien
43 mins ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
I have a 36 ohm resistor through which a current of 9.22mA @ 86º passes.
If the frequencies involved are not too high, the voltage associated with this current will also have a phase angle of 86º.
As a result, the power of the resistor will be 100% real power. There will be no reactive power. In order to have reactive power, you must have reactive elements: inductors or capacitors.
P = I^2 R = 3.06 mW @ 172º
The formula $P=I^2R$ isn't meant for phasor cases. For phasors, you should use
$$P=I^star IR$$
where $I^star$ is the complex conjugate of $I$.
In general, the power consumed by a component with phasor voltage $V$ and current $I$ is
$$S=VI^star$$
giving complex power $S$, with $S=P+jQ$ where $P$ is real power and $Q$ is reactive power.
With this correction you will get a real result, as expected.
Which figure should I be comparing against a resistor specsheet's power rating?
Since the power used by a resistor is entirely real power, you need to calculate the real power and compare it to the rating.
If the frequency applied is very low, you may need to compare the peak power rather than the average power, since it could be possible for the resistor to burn out at the peak of the waveform.
answered 1 hour ago
The Photon
79.8k394189
Awesome! I've fixed my calculation in the question. Hopefully what I've done for RMS conversion also makes sense.
– Reinderien
43 mins ago
add a comment |Â
up vote
2
down vote
I have a 36 ohm resistor through which a current of 9.22mA @ 86º passes.
If the frequencies involved are not too high, the voltage associated with this current will also have a phase angle of 86º.
As a result, the power of the resistor will be 100% real power. There will be no reactive power. In order to have reactive power, you must have reactive elements: inductors or capacitors.
P = I^2 R = 3.06 mW @ 172º
The formula $P=I^2R$ isn't meant for phasor cases. For phasors, you should use
$$P=I^star IR$$
where $I^star$ is the complex conjugate of $I$.
In general, the power consumed by a component with phasor voltage $V$ and current $I$ is
$$S=VI^star$$
giving complex power $S$, with $S=P+jQ$ where $P$ is real power and $Q$ is reactive power.
With this correction you will get a real result, as expected.
Which figure should I be comparing against a resistor specsheet's power rating?
Since the power used by a resistor is entirely real power, you need to calculate the real power and compare it to the rating.
If the frequency applied is very low, you may need to compare the peak power rather than the average power, since it could be possible for the resistor to burn out at the peak of the waveform.
answered 1 hour ago
The Photon
79.8k394189
Awesome! I've fixed my calculation in the question. Hopefully what I've done for RMS conversion also makes sense.
– Reinderien
43 mins ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
I have a 36 ohm resistor through which a current of 9.22mA @ 86º passes.
If the frequencies involved are not too high, the voltage associated with this current will also have a phase angle of 86º.
As a result, the power of the resistor will be 100% real power. There will be no reactive power. In order to have reactive power, you must have reactive elements: inductors or capacitors.
P = I^2 R = 3.06 mW @ 172º
The formula $P=I^2R$ isn't meant for phasor cases. For phasors, you should use
$$P=I^star IR$$
where $I^star$ is the complex conjugate of $I$.
In general, the power consumed by a component with phasor voltage $V$ and current $I$ is
$$S=VI^star$$
giving complex power $S$, with $S=P+jQ$ where $P$ is real power and $Q$ is reactive power.
With this correction you will get a real result, as expected.
Which figure should I be comparing against a resistor specsheet's power rating?
Since the power used by a resistor is entirely real power, you need to calculate the real power and compare it to the rating.
If the frequency applied is very low, you may need to compare the peak power rather than the average power, since it could be possible for the resistor to burn out at the peak of the waveform.
answered 1 hour ago
The Photon
79.8k394189
I have a 36 ohm resistor through which a current of 9.22mA @ 86º passes.
If the frequencies involved are not too high, the voltage associated with this current will also have a phase angle of 86º.
As a result, the power of the resistor will be 100% real power. There will be no reactive power. In order to have reactive power, you must have reactive elements: inductors or capacitors.
P = I^2 R = 3.06 mW @ 172º
The formula $P=I^2R$ isn't meant for phasor cases. For phasors, you should use
$$P=I^star IR$$
where $I^star$ is the complex conjugate of $I$.
In general, the power consumed by a component with phasor voltage $V$ and current $I$ is
$$S=VI^star$$
giving complex power $S$, with $S=P+jQ$ where $P$ is real power and $Q$ is reactive power.
With this correction you will get a real result, as expected.
Which figure should I be comparing against a resistor specsheet's power rating?
Since the power used by a resistor is entirely real power, you need to calculate the real power and compare it to the rating.
If the frequency applied is very low, you may need to compare the peak power rather than the average power, since it could be possible for the resistor to burn out at the peak of the waveform.
answered 1 hour ago
The Photon
79.8k394189
edited 1 hour ago
answered 1 hour ago
The Photon
79.8k394189
answered 1 hour ago
The Photon
79.8k394189
answered 1 hour ago
The Photon
79.8k394189
79.8k394189
Awesome! I've fixed my calculation in the question. Hopefully what I've done for RMS conversion also makes sense.
– Reinderien
43 mins ago
add a comment |Â
Awesome! I've fixed my calculation in the question. Hopefully what I've done for RMS conversion also makes sense.
– Reinderien
43 mins ago
Awesome! I've fixed my calculation in the question. Hopefully what I've done for RMS conversion also makes sense.
– Reinderien
43 mins ago
Awesome! I've fixed my calculation in the question. Hopefully what I've done for RMS conversion also makes sense.
– Reinderien
43 mins ago
add a comment |Â
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Resistors don't create reactive power, so what else would it mean?
– immibis
1 min ago