Question 335: Current Electricity

Question

The supply voltage to room is $$120V.$$ 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?

Accepted Answer

Power of bulb $$=60W$$ $$\left( {given} ight)$$
Resistance of bulb $$ = {{120 	imes 120} \over {60}} = 240\Omega $$
$$\left[ {\,\,} ight.$$ $$\left. {\,P = {{{V^2}} \over R}\,} ight]$$
Power of heater $$=240W$$ (given)
Resistance of heater $$ = {{120 	imes 120} \over {240}} = 60\Omega $$
Voltage across bulb before heater is switched  on,
$${V_1} = {{240} \over {246}} 	imes 120 = 117.73\,\,$$ volt
Voltage across bulb after heater is switched on,
$${V_2} = {{48} \over {54}} 	imes 120 = 106.66$$ volt
Hence decrease in voltage
$${V_1} - {V_2} = 117.073 - 106.66 = 10.04$$ Volt (approximately)

Answer

zero

Answer

$$2.9$$ Volt

Answer

$$13.3$$  Volt

Answer

$$10.04$$ Volt

Current Electricity

The supply voltage to room is $120V. The resistance of the lead wires is 6\Omega . A 60 W bulb is already switched on. What is the decrease of voltage across the bulb, when a 240 W$ heater is switched on in parallel to the bulb?

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