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Need clarity, kindly explain! - Current Electricity - BITSAT

The supply voltage to a 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?

 

  • Option 1)

    10.04 Volt

  • Option 2)

    Zero Volt

  • Option 3)

    2.9 Volt

  • Option 4)

    13.3 Volt

 
Answers (3)
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Need details on this question

N neha

As discussed in

Power dissipiated in external resistance -

P=(\frac{E}{R+r})^{2}R

-

 

 

  P=\frac{v^{2}}{R}\:=R\:=\frac{v^{2}}{P}=\frac{120\times 120}{60}\:=240\Omega

Power of heater = 240\Omega

R=\frac{120\times 120}{240}=60\Omega

Voltage across bulb before heater switched on

v_{1}=\frac{240}{246}\times 120 = 117.73v

When Heater and bulb in parallel then R_{eq}=\frac{240*60}{300}=\frac{240}{5}

Voltage across bulb after heater switched on:

v_{2}=I_{net}R_{eq}=(\frac{120}{6+(\frac{240}{5})})\frac{240}{5}=(\frac{120*5}{270})\frac{240}{5}=\frac{320}{3}=106.66

 

v_{2}-v_{1}=117.073-106.66=10.04


Option 1)

10.04 Volt

Option 2)

Zero Volt

Option 3)

2.9 Volt

Option 4)

13.3 Volt

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