# Average power dissipated as the heat in the resistor

I’m trying to calculate the average power dissipated as the heat in the resistor.

A 12 V amplitude sinusoidal signal is applied across a 500 ohm resistor.

I have determined the current amplitude to be 0.024 A and the root mean square to be 16.97 mA.

I use P = I x V = 0.024 x 12 = 0.288 W

Where P is the power, I is the current and V is the voltage.

Is this correct ?

• No. You need to multiply RMS values of both current and voltage. – Eugene Sh. Dec 21 '18 at 18:12
• So it should be 0.144 W ? – Dan Khan Dec 21 '18 at 18:13
• "12V amplitude" - is that 12Vrms, peak, or peak-peak? – Reinderien Dec 21 '18 at 18:27
• @Reinderien - or average (mean )? – WhatRoughBeast Dec 21 '18 at 18:41
• @WhatRoughBeast unlikely. Since there is no mention of a DC rail, the mean would be 0. – Reinderien Dec 21 '18 at 18:44

With AC the real power is Cosθ is 1 because you have resisitive load.

Thus you have to calculate with RMS values.

Note that power can also be calculated as

and RMS voltage is Thus your power is 