JEE Main 2013 — Current Electricity Question with Solution

Resistance of bulb, $R_b=\frac{V^2}{P_b}=\frac{120\times 120}{60}=240 \mathrm{\Omega }$
Resistance of heater, $R_h=\frac{V^2}{P_h}=\frac{120\times 120}{240}=60 \mathrm{\Omega }$

Voltage across bulb before the heater is not connected, $V_1=\frac{V\times R_b}{R_b+6}=\frac{120\times 240 \mathrm{V}}{246}=117.07 \mathrm{V}$

Now the bulb and heater are connected in parallel.

Resistance of the combination is $R_{net}=\frac{240\times 60}{240+60}=48 \mathrm{\Omega }$
Voltage across bulb after the heater is switched on,
$V_2=\frac{V\times R_{net}}{R_{net}+6}=\frac{120\times 48}{48+6}=106.66 \mathrm{V}$

The decrease in the voltage is $V_1-V_2=117.07-106.66\simeq 10.4 \mathrm{V}$

Note: Here supply voltage is taken as rated voltage.