Question 199 of 380

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

A$13.3V$B$10.4V$C$zeroV$D$2.9V$

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Answer: (B) $10.4V$

Resistance of bulb, ${R}_{b}=\dfrac{{V}^{2}}{{P}_{b}}=\dfrac{120\times 120}{60}=240\Omega$
Resistance of heater, ${R}_{h}=\dfrac{{V}^{2}}{{P}_{h}}=\dfrac{120\times 120}{240}=60\Omega$

Voltage across bulb before the heater is not connected, ${V}_{1}=\dfrac{V\times {R}_{b}}{{R}_{b}+6}=\dfrac{120\times 240V}{246}=117.07V$
Now the bulb and heater are connected in parallel.
Resistance of the combination is ${R}_{net}=\dfrac{240\times 60}{240+60}=48\Omega$
Voltage across bulb after the heater is switched on,
${V}_{2}=\dfrac{V\times {R}_{net}}{{R}_{net}+6}=\dfrac{120\times 48}{48+6}=106.66V$
The decrease in the voltage is ${V}_{1}-{V}_{2}=117.07-106.66≃10.4V$
Note: Here supply voltage is taken as rated voltage.