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

Options

A. zero volt
B. 2.9 volt
C. 13.3 volt
D. 10.04 volt

Answer

d

Explanation

Here, `V = 120 V, P = 60 W`
`R_("bulb") = (V^(2))/(P) = ((120)^(2))/(60)= 240 Omega`

Refer to figure (a), `R_(eq) = 240 + 6 = 246 Omega`
Voltage across bulb `V_(B) = (120)/(246) xx 240 = 117.07V`
Refer to figure (b),
Resistance of heater of circuit is
`R'_(eq) = 6+(240 xx 60)/(240 + 60) = 6 ++ 48 = 54 Omega`
Voltage across bulb, `V'_(B) = (120)/(54) xx 48 = 106.66 V`