### Power Calculations A circuit for working Ohm’s Law problems

You can calculate the power P, in watts, in a dc circuit such as that shown in above figure, by using the formula P = EI. This formula tells us that the power in watts is the product of the voltage in volts and the current in amperes. If you are not given the voltage directly, you can calculate it if you know the current and the resistance.

Recall the Ohm’s Law formula for obtaining voltage: E = IR. If you know I and R but you don’t know E, you can get the power P this way:
P = EI = (IR)I = I 2R
Suppose you’re given only the voltage and the resistance. Remember the Ohm’s Law formula for obtaining current: I = E/R. Therefore:
P = EI = E(E/R) = E2/R

#### Problem 1

Suppose that the voltmeter in above figure reads 12 V and the ammeter shows 50 mA. What is the power dissipated by the potentiometer?
Use the formula P = EI. First, convert the current to amperes, getting I = 0.050 A. (Note that the last 0 counts as a significant digit.) Then multiply by 12 V, getting P = EI = 12 × 0.050 = 0.60 W.

#### Problem 2

If the resistance in the circuit of above figure is 999 Ω and the voltage source delivers 3 V, what is the power dissipated by the potentiometer?
Use the formula P = E2/R = 3 × 3/999 = 9/999 = 0.009 W = 9 mW. You are justified in going to only one significant figure here.

#### Problem 3

Suppose the resistance in above figure is 47 kΩ and the current is 680 mA. What is the power dissipated by the potentiometer?
Use the formula P = I 2R, after converting to ohms and amperes. Then P = 0.680 × 0.680 × 47,000 = 22,000 W = 22 kW. (This is an unrealistic state of affairs: an ordinary potentiometer, such as the type you would use as the volume control in a radio, dissipating 22 kW, several times more than a typical household!)

#### Problem 4

How much voltage would be necessary to drive 680 mA through a resistance of 47 kΩ, as is described in the previous problem?
Use Ohm’s Law to find the voltage: E = IR = 0.680 × 47,000 = 32,000 V = 32 kV. That’s the level of voltage you’d expect to find on a major utility power line, or in a high-power tube-type radio broadcast transmitter.