When is it safe to use the V^2/R equation for Power

[September24]

I just watched coursesaver's circuits video and he brought up a great point. There are two equations for power, I^2R and V^2/R. The MCAT may ask what how the overall power changes when a resistor is added/removed. Before, this was easy. If the circuit has a parallel circuit...Use V^2/R since voltage stays constant. If the resistors are in series, use I^2R.

However, in the video, the instructor shows a problem in which we have to calculate the power. The battery was provided 12V which was going into a parallel branch with resistors. Also in "series" with this "parallel branch" was another resistor. We were asked what the power was through one resistor in the parallel branch and I immediately thought of using V^2/R (due to parallel nature). However, it wasn't as simple as using 12^2/R (with R being the resistor in question). Since there was ANOTHER resistor in series, we simply cant use 12.

My question is, when is it safe to use V^2/R. Is there a quick way to determine what voltage TO USE in the example I provided. I believe Kirchoff's laws would be useful, but I'm not sure.

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[DrknoSDN]

They are just rearranged versions of the same equation for convenience.
I^2*R = V^2/R
because V=IR ... so (IR)(IR)/R = IIR = I^2R

Interchangeable for all intensive purposes. In the example you probably have to look at the voltage drop across each resistor before calculating power. If total circuit power is 12V then you should determine what the parallel resistors single resistor equivalent would be (convert the parallels to 1 resistor). Then you can find the voltage drop across the parallel set and then calculate power for the individual resistor of the parallel set.

It's safe to use V^2/R if you actually have the correct voltage drop across the resistor. Here 12 is not going to be correct.

To determine voltage look at the two resistors in parallel and calculate the combined resistance using 1/Req = 1/R1 + 1/R2 or the better shortcut:
Req = (R1*R2)/(R1+R2) <--(quick way)
Hypothetical 12 and 6 ohm resistors in parallel would give:
R = (12*6)/(12+6) = 72/18 = 4 ohm equivalent

You then add that equivalent resistance to the other resistor in the circuit to get a total resistance. Assuming the other series resistor was also 4 ohm would make it easy. You know that the parallel setup has a 6 volt drop (50% of the total 12 volts). You then determine current through each resistor by using V=IR or I =V/R
I = 6V/12 ohm for R1 = 0.5 Amp
I = 6V/6 ohm for R2 = 1.0 Amp

Power at R1 = IIR = (.5)(.5)(12) = 3 watts.
Power at R1 = V^2/R = (6)(6)/(12) = 3 watts.

Power at R2 = IIR = (1)(1)(6) = 6 watts.
Power at R2 = V^2/R = (6)(6)/(6) = 6 watts.

Both equations work. Sorry if I butchered something, very possible when making up numbers. Hope it helps. GL =D
*Also, caution, the 'shortcut' only works for 2 parallel resistors. More than that requires different methods.

 

[NextStepTutor_1]

Also, remember that for any general circuit component, the energy dissipated (or produced) can be found by the equation P = I*V. For a resistor V = IR, so if we plug that in for the first equation we can see that it does give us the two equations above (I^2 *R and V^2/R), however, the strength of using this equation is that you can simply find the voltage and current on each resistor to find power, and not have to worry about when to use each equation. Just remember that circuits in parallel with the voltage source share the same voltage, and the ones in series share the same current. By using these laws with Kirchoff's laws you can get the power dissipated.