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

The power dissipated by each bulb is increased by a factor of $1.04$
Let $I_0$ be the original current when 50 bulbs are functioning. We can write an expression for the power: $P = I_0^2~R$ We know that the original current is $I_0 = \frac{V}{R_{eq}} = \frac{V}{50~R}$. If one of the bulbs connected in series is removed, then the total resistance is reduced. We can find the new current: $I = \frac{V}{R_{eq}}$ $I = \frac{V}{49~R}$ $I = \frac{50}{49} \times \frac{V}{50~R}$ $I = \frac{50}{49} \times I_0$ We can find an expression for the new power dissipated by each bulb: $P = I^2~R$ $P = (\frac{50}{49}~I_0)^2~R$ $P = 1.04\times I_0^2~R$ The power dissipated by each bulb is increased by a factor of $1.04$