Current Electricity 5 Question 9

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

(2013 Main)

(a) zero

(b) 2.9V

(c) 13.3V

(d) 10.4V

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Solution:

  1. As, P=V2R

where, P= power dissipates in the circuit,

V= applied voltage,

R= net resistance of the circuit

R=120×12060=240Ω [resistance of bulb]

Req=240+6=246Ωi1=VReq=120246 [before connecting heater] R=V2R=120×120240R=60Ω [resistance of heater] 

So, from figure,

V1=240246×120=117.073V[V=IR]i2=VR2=12054V2=4854×120=106.66VV1V2=10.41V



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