## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

The power dissipated by each bulb is $~18.75~W$
We can write an expression for the power: $P = I^2~R$ We know that the current is $I = \frac{V}{R}$. If two bulbs are connected in series, then the total resistance is doubled, so the current is halved. Then the power dissipated by one bulb is reduced by a factor of $\frac{1}{4}$ The power dissipated by each bulb is $\frac{75~W}{4} = 18.75~W$