Power in Circuits

[G1SG2]

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So for resistors in series, the current is the same, and so the larger resistor will dissipate more power, as P=I^2*R.

For resistors in parallel, the voltage is the same, and so the larger resistor will dissipate less power, as P=V^2/R.

So, generally, for resistors in series, we use P=I^2*R and for resistors in parallel, we use P=V^2/R, right? Otherwise, it would be confusing (not numerically-I'm talking about those qualitative problems that ask what happens to the power in series/parallel circuits-I'm fine with manipulating the equations when it comes to answering a quantitative problem).

 

[G1SG2]

bump! (come on matth87, lol)

 

[amine2086]

This a good way to think about power in a circuit. But for power just know that P = VI and also ohm's law V = IR and you can derive either of the two equations you mentioned using these two formulas.

 

[matth87]

So for resistors in series, the current is the same, and so the larger resistor will dissipate more power, as P=I^2*R.

For resistors in parallel, the voltage is the same, and so the larger resistor will dissipate less power, as P=V^2/R.

So, generally, for resistors in series, we use P=I^2*R and for resistors in parallel, we use P=V^2/R, right? Otherwise, it would be confusing (not numerically-I'm talking about those qualitative problems that ask what happens to the power in series/parallel circuits-I'm fine with manipulating the equations when it comes to answering a quantitative problem).

Yup that works!
Really when it comes to circuits there are multiple ways to go about finding the answers, and any of the equations will work.

If you know the voltage or current or resistance of a resistor you can find the power dissipated through it by any of the formulas

Like amine2086 said, just remeber P = IV and V = IR and you can derive all the equations from those two equations.

The way I like to deal with circuits is condense it into the most simple 1 resistor circuit, start calculating your values, and then start expanding it back out into the circuit they want you to answers questions about. Kind of a fool proof method.

P.S. I was at the gym when you wrote your question haha, that all i do is eat, sleep, mcat, SND, and gym.

brb cooking food.

 

[G1SG2]

Yup that works!
Really when it comes to circuits there are multiple ways to go about finding the answers, and any of the equations will work.

If you know the voltage or current or resistance of a resistor you can find the power dissipated through it by any of the formulas

Like amine2086 said, just remeber P = IV and V = IR and you can derive all the equations from those two equations.

The way I like to deal with circuits is condense it into the most simple 1 resistor circuit, start calculating your values, and then start expanding it back out into the circuit they want you to answers questions about. Kind of a fool proof method.

P.S. I was at the gym when you wrote your question haha, that all i do is eat, sleep, mcat, SND, and gym.

brb cooking food.

Thanks guys. Haha, me too, except I don't sleep as much and have not been working out lately either. Bad, I know.

 

[BerlinDude]

So for resistors in series, the current is the same, and so the larger resistor will dissipate more power, as P=I^2*R.

For resistors in parallel, the voltage is the same, and so the larger resistor will dissipate less power, as P=V^2/R.

So, generally, for resistors in series, we use P=I^2*R and for resistors in parallel, we use P=V^2/R, right? Otherwise, it would be confusing (not numerically-I'm talking about those qualitative problems that ask what happens to the power in series/parallel circuits-I'm fine with manipulating the equations when it comes to answering a quantitative problem).

There's no difference whatsoever between those two equations - they are identical.
P=R*I^2=V^2/R

Just remember that you are dealing with voltage across that particular resistor and current through that particular resistor.

 

[thebillsfan]

uhh...im not sure if that's right. since the power is directly proportional to the SQUARE of I P=I^2R and since current is inversely proportional to resistance in V=IR, then a larger resistor will have more resistance (obv) but less current, and thus less power (since current is more of a factor than resistance thanks to I^2R=P). Does anyone agree?

 

[wanderer]

uhh...im not sure if that's right. since the power is directly proportional to the SQUARE of I P=I^2R and since current is inversely proportional to resistance in V=IR, then a larger resistor will have more resistance (obv) but less current, and thus less power (since current is more of a factor than resistance thanks to I^2R=P). Does anyone agree?

That's only if the resistors are parallel. If they're in series the current must be the same.