Question

A string of holiday lights can be wired in series, but all the bulbs go out if one burns out because that breaks the circuit. Most lights today are wired in series, but each bulb has a special fuse that short-circuits the bulb—making a connection around it—if it burns out, thus keeping the other lights on. Suppose a string of $50$ lights is connected in this way and plugged into a $120\mathrm{V}$ outlet. By what factor does the power dissipated by each remaining bulb increase when the first bulb burns out?

Short answer

The power dissipated by each bulb increases by factor of$1.04.$

Step-by-step solution

Given information

We have given,

Number of bulb in series = $50$

Voltage supply =$120\mathrm{V}$

We have to find the power dissipated by the single bulb when one bulb will fused.

Simplify 

Let us consider every bulb has resistance R. Then

voltage across each bulb is =$\frac{125}{50}=2.4V$

power dissipated by one bulb = $\frac{{\mathrm{V}}^2}{\mathrm{R}}=\frac{5.76}{R}$

After one bulb is fused,

voltage across the each bulb = $\frac{120}{49}=2.45V$

Power dissipated by each bulb = $\frac{5.998}{\mathrm{R}}$

Divided the above two values

factor=$\frac{5.998}{5.76}=1.04$