Question

The supply voltage to 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. $10.04\text{Volt}$
• B. $13.3\text{Volt}$
• C. $2.9\text{Volt}$
• D. zero $\text{Volt}$

Answer 1

The correct option is A $10.04\text{Volt}$

Let resistance of the bulb is $R.$
Power, $P=\frac{V^2}{R}\Rightarrow R=\frac{V^2}{P}$

Resistance of bulb:

$R=\frac{120\times 120}{60}=240\Omega$

$R_{eq}=240+6=246\Omega$

$V_1=\frac{240}{246}\times 120=117.073\text{Volt}$

Resistance of heater = $\frac{V^2}{P}$

$\frac{120\times 120}{240}=60\Omega$

As bulb and heater are connected in parallel.
Net resistance = $\frac{240\times 60}{300}=48\Omega$

Total resistance, $R_2=48+6=54\Omega$

and total current, $I_2=\frac{V}{R_2}=\frac{120}{54}$

Potential across heater = Potential across bulb

$V_2=\frac{120}{54}\times 48=106.66\text{Volt}$

So, $V_1-V_2=117.073-106.66=10.04\text{Volt}$

Hence, option $\left(D\right)$ is correct.