The supply voltage to a room is 120 V. 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?

Options

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

Explanation

`P=V^2/R`
Resistance of the bulb
Initially with only bulb in circuit
`R=(120 times 120)/60=240 Omega`
`R_(eq)=240+6=246 Omega implies I_1=V/R_(eq)=120/246`
`V_1=I_1R_1=120/246 times 240=17.073 V`
Resistance of the heater `=V^2/P=(120 times 120)/246=60 Omega`
As bulb end heater are connected in parallel
Net resistance `=(240 times60)/300=48 Omega`
Total resistance `R_2=48+6=54 Omega`
Potential across heater = Potential across bulb
`V_2=120/54 times 48=106.66V, V_1-V_2=117.073-106.66=10.4V`