Question

Two lamps, one rated $100{\rm{ W}}$ at $220{\rm{ V}}$, and the other $60{\rm{ W}}$ at $220{\rm{ V}}$, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is $220{\rm{ V}}$?

Answer 1

From the concept of Joule’s law of heating, we know that the power of a lamp is equal to the ratio of the square of potential difference and resistance of that lamp. We will also use the concept of resistors in parallel to find the equivalent resistance of the circuit.

Complete step by step answer:
Given:
The power of the first lamp is ${P_1} = 100{\rm{ W}}$.
The power of the second lamp is ${P_2} = 60{\rm{ W}}$.
The potential electric mains supply is $V = 220{\rm{ V}}$.
It is given that both the lamps are connected in parallel to electric supply, and we have to find the value of current that can be drawn from the line if the supply voltage is $220{\rm{ V}}$.
From the concept of Joule’s law, we can write:
$P = \dfrac{{{V^2}}}{R}$
Here, P is power, V is the potential difference, and R is the resistance.
Let us write the expression for power of the first lamp in terms of its voltage and resistance.
${P_1} = \dfrac{{{V^2}}}{{{R_1}}}$
Here ${P_1}$ is power and ${R_1}$ is the resistance of the first lamp.
On rearranging the above expression to get the value of resistance, we get:
${R_1} = \dfrac{{{V^2}}}{{{P_1}}}$
On substituting $220{\rm{ V}}$ for V and $100{\rm{ W}}$ for ${P_1}$ in the above expression, we get:

$${R_1} = \dfrac{{{{\left( {220{\rm{ V}}} \right)}^2}}}{{100{\rm{ W}}}}\\\ = 484{\rm{ }}\Omega$$

Let us write the expression for power of the second lamp in terms of its voltage and resistance.
${P_2} = \dfrac{{{V^2}}}{{{R_2}}}$
Here ${P_2}$ is power and ${R_2}$ is the resistance of the second lamp.
On rearranging the above expression to get the value of resistance, we get:
${R_2} = \dfrac{{{V^2}}}{{{P_2}}}$
On substituting $220{\rm{ V}}$ for V and $60{\rm{ W}}$ for ${P_2}$ in the above expression, we get:

$${R_2} = \dfrac{{{{\left( {220{\rm{ V}}} \right)}^2}}}{{60{\rm{ W}}}}\\\ = 806.67{\rm{ }}\Omega$$

We know that the expression for equivalent resistance when two resistors are connected in parallel can be written as:
${R_{eq}} = \dfrac{{{R_1}{R_2}}}{{{R_1} + {R_2}}}$
On substituting $484{\rm{ }}\Omega$ for ${R_1}$ and $806.67{\rm{ }}\Omega$ for ${R_2}$ in the above expression, we get:

$${R_{eq}} = \dfrac{{\left( {484{\rm{ }}\Omega } \right)\left( {806.67{\rm{ }}\Omega } \right)}}{{484{\rm{ }}\Omega + 806.67{\rm{ }}\Omega }}\\\ = 302.5{\rm{ }}\Omega$$

Using Ohm’s law for the given circuit, we can write:
$I = \dfrac{V}{{{R_{eq}}}}$
Here is the current drawn from the line.
On substituting $220{\rm{ V}}$ and $302.5{\rm{ }}\Omega$ for ${R_{eq}}$ in the above expression, we get:

$$I = \dfrac{{220{\rm{ V}}}}{{302.5{\rm{ }}\Omega }}\\\ = 0.727{\rm{ A}}$$

Therefore, $0.727{\rm{ A}}$ is the amount of current that can be drawn from the line if the supply voltage is $220{\rm{ V}}$.

Note: If the given two lamps were connected in series combination, then their equivalent resistance is equal to the summation of the individual resistances of each lamp. For a series combination of two resistors, mathematically we can write:
${R_{eq}} = {R_1} + {R_2}$