Q

# Need clarity, kindly explain! - Current Electricity - BITSAT

The supply voltage to a room is 120 V. The resistance of the lead wires is 6 . 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

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N

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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