Question

The supply voltage to 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?

• A. zero Volt
• B. 2.9 Volt
• C. 13.3 Volt
• D. 10.4 Volt

Answer 1

The correct option is D 10.4 Volt

The resistance of the bulb is given as,

$R_1=\frac{V^2}{P_1}$

$=\frac{{\left(120\right)}^2}{60}$

$=240\Omega$

The resistance of the heater is given as,

$R_2=\frac{V^2}{P_2}$

$=\frac{{\left(120\right)}^2}{240}$

$=60\Omega$

The equivalent resistor of the bulb and heater together is given as,

$R=\frac{R_1{R}_2}{R_1+R_2}$

$=\frac{240\times 60}{240+60}$

$=48\Omega$

The voltage drop across the bulb before the heater was connected is given as,

$V_1=\frac{120}{240+6}\times 240$

$=117.07V$

The voltage drop across the bulb after the heater was connected is given as,

$V_2=\frac{120}{48+6}\times 48$

$=106.66V$

The potential difference is given as,

$V_1-V_2=117.07-106.66$

$=10.41V$

The decrease of voltage across the bulb is $10.41V$.