The supply voltage to room is $120\ V$. The resistance of the lead wires is $6\,\Omega$ . A $60\ W$ bulb is already switched on. What is the decrease of voltage across the bulb, when a $240\ W$ heater is switched on in parallel to the bulb? ............. $V$
JEE MAIN 2013, Medium
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Power of bulb $=60\, \mathrm{W}$ (given)
Resistance of bulb $=\frac{120 \times 120}{60}=240\, \Omega\left[\because \mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}\right]$
Power of heater $=240\, \mathrm{W}$ (given)
Resistance of heater $=\frac{120 \times 120}{240}=60\, \Omega$
Voltage across bulb before heater is switched on,
$\mathrm{V}_{1}=\frac{240}{246} \times 120=117.73\, \text { volt }$
Voltage across bulb after heater is switched on,
$\mathrm{V}_{2}=\frac{48}{54} \times 120=106.66\, \mathrm{volt}$
Hence decrease in voltage
$\mathrm{V}_{1}-\mathrm{V}_{2}=117.073-106.66=10.04\, \mathrm{Volt}$ (approximately)
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