ઓરડામાં સપ્લાય વોલ્ટેજ 120 V છે.લેડ વાયરનો અવ | vedclass

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Question

Power of bulb $=60\, \mathrm{W}$ (given)

Resistance of bulb $=\frac{120 	imes 120}{60}=240\, \Omega\left[\because \mathrm{P}=\frac{\mathrm{V}^{2}}

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Vedclass · Answer

Power of bulb $=60\, \mathrm{W}$ (given)

Resistance of bulb $=\frac{120 	imes 120}{60}=240\, \Omega\left[\because \mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}ight]$

Power of heater $=240\, \mathrm{W}$ (given)

Resistance of heater $=\frac{120 	imes 120}{240}=60\, \Omega$

Voltage across bulb before heater is switched on,

$\mathrm{V}_{1}=\frac{240}{246} 	imes 120=117.73\, 	ext { volt }$

Voltage across bulb after heater is switched on,

$\mathrm{V}_{2}=\frac{48}{54} 	imes 120=106.66\, \mathrm{volt}$

Hence decrease in voltage

$\mathrm{V}_{1}-\mathrm{V}_{2}=117.073-106.66=10.04\, \mathrm{Volt}$ (approximately)

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