The brightness of the bulbs arranged in the above circuits is determined by the power dissipated, $P = V^2 R_{\text{eq}}$. Thus, the bulb brightness is inversely proportional to the equivalent resistance, $R_{\text{eq}}$. The supplied voltage, $V$, is Light Bulbs the same for both circuits. $R_{\text{eq}}$ is the equivalent resistance of four 60 W arranged either in series, or parallel, each bulb with resistance $R$.

For the series arrangement of four bulbs,

$$R_{\text{eq}} = R_{\text{s}} = R + R + R + R = 4R$$

For the parallel arrangement of four bulbs,

$$R_{\text{eq}} = R_{\text{p}} = \frac{1}{ \frac{1}{R} + \frac{1}{R} + \frac{1}{R} + \frac{1}{R}} = \frac{R}{4}$$

Since the brightness of each circuit of bulbs is inversely proportional to $R_{\text{eq}}$, we see that the parallel circuit of bulbs should be brighter than the series circuit by a factor of 16. In fact the parallel circuit is less than 16 times brighter because of the effect a hotter filament has on bulb resistance. We have neglected this non-ohmic effect, and the measured value of power ratio is closer to a factor of 8 instead of 16.