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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Current Electricity
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The supply voltage to a room is $120\: V\: DC$. 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?

$\begin {array} {1 1} (A)\;0V & \quad (B)\;2.9V \\ (C)\;13.3V & \quad (D)\;10.04V \end {array}$

 

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1 Answer

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Resistance of bulb = $ \large\frac{120 \times 120}{60}$ $ = 240\Omega$
Resistance of heater = $ \large\frac{120 \times 120}{240} $ $ =60\Omega$
Voltage across bulb before heater is switched on, $V_1 =\large\frac{120}{246}$ $ \times 240$=117.07
Voltage across bulb after heater is switched on, $V_2 =\large\frac{120}{54}$ $ \times 48$=106.66
Decrease in the voltage =$ V_1 – V_2 = 10.04$ (approximately)
Ans : (D)
Note: Here supply voltage is taken as rated voltage.

 

answered Feb 8, 2014 by thanvigandhi_1
edited Sep 10, 2014 by thagee.vedartham
 

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