Power
Recall that power is a measure of how quickly energy is transformed, measured in [J/s = Watts = W]. The power dissipate by a resistor is
where / is the current through the resistor and is the potential difference across it.

Because of Ohm’s law, we may also write
P_{res} = I^{2}R = (ΔV)^{2}/R
(It is better to remember how to derive this equation than to memorize it.) The power provided by a DC cell to a circuit is
where / is the current through the cell and is the potential difference across it.

Example
Light bulbs that you use around the house are designed to have 120 V across their terminals. For instance, a 120W bulb uses 120 Watts of power when placed in a socket with a 120V potential difference. (Actually it is a little more complicated. See Section F.) With that in mind,
 what is the resistance of a “120Watt” bulb?
 What is the resistance of a “30Watt” bulb?
 If these two bulbs are connected in series, and plugged into a wall outlet, which bulb would be brighter (Figure 1520)?
Figure 1520
Solution
 Figure 1521 shows the circuit diagram for a single 120W bulb plugged into a potential source. The equation for power is
P = IΔV
120W = I(120V)
I = 1A
By Ohm’s law, we have
ΔV = IR
120V = (1A)RR = 120 Ω.
 A similar calculation gives a resistance 480 Ω for the “30W” bulb.
 If we place the bulbs in series (Figure 1522), then the current through the bulbs is the same. Since
P = I^{2}R
the power is proportional to the resistance. So the 480Ω bulb (that is, the “30W” bulb) is brighter because the bulbs are in series. (Usually the “120W” bulb is brighter because the bulbs are connected in parallel.)
Figure 1521

Figure 1522
