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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?

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

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.

edited Sep 10, 2014