Question

you are given two electric bulbs operating on 220 v ,one is of 60 w and the other of 100 w .if they are connected in series and then the combination is connected to 440 v which of the two bulbs will fuse?

Answer 1

We know the resistance of the bulb is given by $R=\frac{V^2}{W}$, where V is the voltage difference and W is the power. Now for the 60 watt bulb the resistance is $R_{60}=\frac{{220}^2}{60}=806.67\text{Ω}$. And the resistance of the 100 watt bulb is $R_{100}=\frac{{220}^2}{100}=484\text{Ω}$. So the total resistance is $806.67+484=1290.67$. So the current through the circuit is $I=\frac{440}{1290.67}=0.341\text{A}$. So the voltage drop across the 60 watt bulb is $V_{60}=I{R}_{60}=0.341\times 806.67=275.07\text{V}$ and voltage drop across the 100 watt bulb is $V_{100}=I{R}_{100}=0.341\times 484=165.04\text{A}$. Now since the voltage drop across the 60 watt bulb is more than what it should be the 60 watt bulb will be fused.