Two identical light bulbs are connected to identical batteries in two different ways. First they are connected in series and then in parallel. In which circuit will the bulbs shine brightly? What is the ratio of the power supplied in series to the power supplied in parallel?

###### Solution.

The power dissipated by a bulb is given by

P = V^{2}/R (1),

where V is the voltage on the light bulb and R is the electrical resistance.

First, we simplify the circuit where the light bulbs are in series.

The total resistance of the series circuit is

R_{series} = R + R = 2R.

Secondly, we simplify the circuit where the light bulbs are in parallel.

The total resistance of the parallel circuit is

1/R_{parallel} = 1/R + 1/R

or

R_{parallel} = R/2.

Thereby, the power for the light bulbs in series is

P_{series} = V^{2}/2R (2)

and the power for the light bulbs in parallel is

P_{parallel} = 2V^{2}/R (3)

Now we can divide equation (2) by equation (3) and we get

P_{series}/P_{parallel} = 1/4

Finally, we can see that in parallel circuit the bulbs will shine brightly.