Question

The supply voltage to the 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) zero volt
B) $2.9$ volt
C) $13.3$ volt
D) $10.4$ volt

Answer 1

Hint:
The power, $P = VI = \dfrac{{{V^2}}}{R}$. Calculate each resistance across the heater and bulb. Calculate the equivalent resistance from the parallel connection. Then obtain the voltage drop, one across the bulb, another one across the bulb, and heater.

Formula used:
Power, $P = VI = \dfrac{{{V^2}}}{R}$ .
Equivalent resistance $R = \dfrac{1}{{{R_b}}} + \dfrac{1}{{{R_h}}}$ .

Complete step by step solution:
Let, the supply voltage is $V$, and current is $I$.
So, the power, $P = VI$
Now, according to Ohm’s law, let, temperature and other physical conditions remain constant, the current flowing through a conductor is directly proportional to the potential difference between its two ends.
So, suppose $V$ is the potential difference between the two ends of a current carrying conductor and $I$ is the current through the conductor. Then, according to the law, $V = IR$ [Where, $R$ is the constant of proportionality, known as resistance].
So, $P = VI = \dfrac{{{V^2}}}{R}$
Let, the resistance of the bulb is ${R_b}$ .
It is given that $P = 60W$ bulb is switched on and supply voltage is $V = 120V$ .
So, $P = \dfrac{{{V^2}}}{{{R_b}}}$ or, ${R_b} = \dfrac{{{{\left( {120} \right)}^2}}}{{60}} = \dfrac{{120 \times 120}}{{60}} = 240\Omega $
Similarly, let the resistance of the heater is ${R_h}$ .
It is given that the $P = 240W$ heater is switched on and supply voltage is $V = 120V$ .
So, $P = \dfrac{{{V^2}}}{{{R_h}}}$ or, ${R_h} = \dfrac{{{{\left( {120} \right)}^2}}}{{240}} = \dfrac{{120 \times 120}}{{240}} = 60W$
Now, the resistance of the heater and the bulb are in parallel connection.
So, equivalent resistance between the ${R_b}$ and ${R_h}$ is $R = \dfrac{1}{{{R_b}}} + \dfrac{1}{{{R_h}}} = \dfrac{{{R_b}{R_h}}}{{{R_b} + {R_h}}} = \dfrac{{240 \times 60}}{{\left( {240 + 60} \right)}} = 48\Omega $
Now, there are two conditions. One is before connecting the heater another is after connecting the heater. It is given that the resistance of the lead wires is $6\Omega $ .
Before, connecting the heater, the voltage drop across the bulb is
${V_1} = \dfrac{{V \times {R_b}}}{{{R_b} + 6}} = \dfrac{{120 \times 240}}{{240 + 6}} = 117.07V$
After connecting the heater, the voltage drop is
${V_2} = \dfrac{{V \times R}}{{R + 6}} = \dfrac{{120 \times 48}}{{48 + 6}} = 106.66V$
So, the decrease of the voltage across the bulb is $\left( {{V_1} - {V_2}} \right) = \left( {117.07 - 106.66} \right)V = 10.41V \approx 10.4V$.
Option (D) is correct.

Note:
The equivalent resistance of the combination is defined as the single resistance, which is used instead of a combination of resistances, to keep the current unchanged in the circuit. In the equivalent resistance, the same current flows through each resistance. Also, the total potential difference across the combination is equal to the sum of the individual potential difference across each resistance.