# Electric power question

1. Feb 15, 2012

### zak8000

Hi

if i have a voltage supply of 100V connected across a 66 ohm resistor which has a power rating of 155W. then P=V^2/R=10000/66=151W but say if i had the same situation except i add another resistor in series to increase the resistance of the overall circuit by adding a 100 ohm resistor which has a power ratting of 1W. then P=10000/166=60W.

but the question i have is will the 100 ohm resistor blow up because there is 60W across both resistors as it has a power rating of 1W?

Last edited: Feb 16, 2012
2. Feb 16, 2012

### _maxim_

Hi,

when one sets resistors in series, the current (rather than the voltage) is the same over the whole circuit. Thus the power dissipation should be calculated by using the current.

So you have: R1=66 ohm and R2=100 ohm, Rtot=166 ohm
Itot = Vtot / Rtot = 100/166 = 0.602 A
P1 = Itot2 * R1 = 23.95 W and P2 = Itot2 * R2 = 36.29 W

So, R1 dissipates 23W as Joule effect, and if its power rating is 155 W you're in the safe side, while R2 needs to be > 36W in order to avoid cooking.

This makes sense because by adding a second resistor of the same order of magnitude, the total power will be roughly distributed between the two (23W + 36W).

Last edited by a moderator: Feb 16, 2012
3. Feb 16, 2012

### zak8000

so if i wanted to measure the total power dissipation for this circuit would be 59W ?

4. Feb 16, 2012

### _maxim_

You're right: when you increase a load resistance while keeping the same voltage, the current decreases then the power decreases accordingly (Joule effect).