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QJEEM13Phe0e3: JEE Main 2013 Physics - Class 12 JEEM13Phe0e3 Solution: The supply voltage to room is The resistance of the lead wires…

The supply voltage to room is 120V.120V. The resistance of the lead wires is 6Ω6\Omega . A 6060 WW bulb is already switched on. What is the decrease of voltage across the bulb, when a 240240 WW heater is switched on in parallel to the bulb?

Multiple Choice Options

  • A zero
  • B 2.92.9 Volt
  • C 13.313.3 Volt
  • D 10.0410.04 Volt

Step-by-Step Verified Solution

JEE Main 2013 (Offline) Physics - Current Electricity Question 306 English Explanation

Power of bulb =60W=60W (given)\left( {given} \right)

Resistance of bulb =120×12060=240Ω= {{120 \times 120} \over {60}} = 240\Omega

[  \left[ {\,\,} \right.  P=V2R ]\left. {\,P = {{{V^2}} \over R}\,} \right]

Power of heater =240W=240W (given)

Resistance of heater =120×120240=60Ω= {{120 \times 120} \over {240}} = 60\Omega

Voltage across bulb before heater is switched on,

V1=240246×120=117.73  {V_1} = {{240} \over {246}} \times 120 = 117.73\,\, volt

Voltage across bulb after heater is switched on,

V2=4854×120=106.66{V_2} = {{48} \over {54}} \times 120 = 106.66 volt

Hence decrease in voltage

V1−V2=117.073−106.66=10.04{V_1} - {V_2} = 117.073 - 106.66 = 10.04 Volt (approximately)

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