Question

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

Answer 1

We will use the concept of equivalent resistance in a parallel combination of resistors. We can then use the formula for power for parallel circuits. Finally, we will substitute the values of the corresponding values of the parameters in the formula and evaluate the value of the drawn current.

Formulae Used:
$P = \dfrac{{{V^2}}}{R}$
Where, $P$ is the power of the appliance, $V$ is the potential difference across the circuit and $R$ is the resistance of the appliance.
$I = \dfrac{V}{R}$
Where $I$ is the current through the appliance.
${R_{eq}} = \dfrac{{{R_1}{R_2}}}{{{R_1} + {R_2}}}$

Complete step by step answer:
We are given some values. We will firstly note down the values with respect to the corresponding parameters.
Given, Power of the first bulb, ${P_1} = 100W$
Voltage rating of the first bulb, ${V_1} = 220V$
Power of the second bulb, ${P_2} = {\text{ 6}}0W$
Voltage rating of the second bulb, ${V_2} = 220V$
Supply Voltage, ${V_o} = 220V$
Now, we know that the power of an appliance can be written as
${P_1} = \dfrac{{{V_1}^2}}{{{R_1}}}$
Putting in the values, we get
$100 = \dfrac{{{{\left( {220} \right)}^2}}}{{{R_1}}}$
Further, we get
${R_1} = \dfrac{{{{\left( {220} \right)}^2}}}{{100}}$
Calculating further, we get
${R_1} = \dfrac{{4840}}{{10}}$
Thus, resistance of the first bulb is
${R_1} = \dfrac{{4840}}{{10}}$

Again, we use the formula of power for the second bulb
${P_2} = \dfrac{{{V_2}^2}}{{{R_2}}}$
Putting in the values, we get
$60 = \dfrac{{{{\left( {220} \right)}^2}}}{{{R_2}}}$
Further, we get
${R_2} = \dfrac{{48400}}{{60}}$
Thus, we get
${R_2} = \dfrac{{4840}}{6}$
Thus, resistance of the first bulb is
${R_2} = \dfrac{{4840}}{6}$
Now according to the question, the bulbs are connected in parallel with each other. Thus, the equivalent resistance of the two bulbs can be calculated using the formula
${R_{eq}} = \dfrac{{{R_1}{R_2}}}{{{R_1} + {R_2}}}$
Substituting in the values, we get
${R_{eq}} = \dfrac{{\dfrac{{4840}}{{10}} \times \dfrac{{4840}}{6}}}{{\dfrac{{4840}}{{10}} + \dfrac{{4840}}{6}}}$

Further, we get
${R_{eq}} = \dfrac{{\dfrac{{{{\left( {4840} \right)}^2}}}{{60}}}}{{\dfrac{{16 \times 4840}}{{60}}}}$
After further calculations, we get
${R_{eq}}{\text{ = }}\dfrac{{4840}}{{16}}$
Thus, ${R_{eq}} = 302.5\Omega$
Now, we know that the supply voltage, ${V_o} = 220V$
Then, we can evaluate the total drawn current from the line is
$I = \dfrac{{{V_o}}}{{{R_{eq}}}}$
Putting in the values, we get
$I = \dfrac{{220}}{{302.5}}$
Finally, we get
$\therefore I = 0.73A$

Hence, $0.73A$ current is drawn from the supply line.

Note: Students should be very careful while evaluating the equivalent resistance. Precisely speaking, they have to be very clever while selecting the value of the parameters as sometimes selecting the values in fraction helps to reduce the chances of falling into the trap of clumsy calculations.