Question

Three light bulbs of $40W$, $60W$ and $100W$ are connected in series with 220 volt source. Which of the bulbs will glow brightest?
(A) $60W$
(B) $40W$
(C) $100W$
(D) All with the same brightness.

Answer 1

Hint
We first need to find the resistance in each of the bulbs from the given power rating and the source potential. By calculating the equivalent resistance in the circuit, we can find the current in the circuit. The current will be the same in the three bulbs since they are connected in series. Therefore, using the value of the current and the resistance, we can find the actual power consumed by each of the bulbs. The one with the most power consumption will glow the brightest.

Formula Used: In the solution we will be using the following formula,
$\Rightarrow P = \dfrac{{{V^2}}}{R} = {I^2}R$ and $V = IR$
where $P$ is the power, $V$ is the voltage, $I$ is the current and $R$ is the resistance
$\Rightarrow {R_{eq}} = {R_1} + {R_2} + {R_3} +. ...$ where ${R_{eq}}$ is the equivalent resistance of a number of resistances placed in series connection.

Complete step by step answer
In the question we are given 3 bulbs with power rating $40W$, $60W$ and $100W$. Now these bulbs are connected across a $220V$ source. Therefore, the resistances of each of these bulbs will be given by the formula, $P = \dfrac{{{V^2}}}{R}$
We can rearrange the formula to get the resistance as, $R = \dfrac{{{V^2}}}{P}$
For the first bulb, substituting the values we get,
$\Rightarrow {R_1} = \dfrac{{{{\left( {220} \right)}^2}}}{{40}}$
On calculating we get,
$\Rightarrow {R_1} = 1210\Omega$
Similarly for the second bulb, we get the resistance by substituting the values as,
$\Rightarrow {R_2} = \dfrac{{{{\left( {220} \right)}^2}}}{{60}}$
On calculating we get,
$\Rightarrow {R_2} = 806.67\Omega$
For the third bulb on substituting we get,
$\Rightarrow {R_3} = \dfrac{{{{\left( {220} \right)}^2}}}{{100}}$
On calculating we get,
$\Rightarrow {R_3} = 484\Omega$
These three bulbs are in series. So the equivalent resistance in the circuit can be calculated from the formula, ${R_{eq}} = {R_1} + {R_2} + {R_3} +. ...$
So substituting the values we get,
$\Rightarrow {R_{eq}} = \left( {1210 + 806.67 + 484} \right)\Omega$
On adding we get,
$\Rightarrow {R_{eq}} = 2500.67\Omega$
Therefore the current flowing in the circuit due to the 220 volt source will be given by $V = IR$. Substituting we get,
$\Rightarrow 220 = I \times 2500.67$
Hence we get the current as,
$\Rightarrow I = 0.088A$
So this is the current that is flowing through all the three bulbs. So the power consumed by the bulbs in the circuit is calculated by the formula, $P = {I^2}R$
For the $40W$ bulb we get,
$\Rightarrow {P_1} = {\left( {0.088} \right)^2} \times 1210$
So we get the power as,
$\Rightarrow {P_1} = 9.37W$
Similarly for the $60W$ we get,
$\Rightarrow {P_2} = {\left( {0.088} \right)^2} \times 806.67$ which gives,
$\Rightarrow {P_2} = 6.24W$
And for the $100W$ bulb we have,
$\Rightarrow {P_3} = {\left( {0.088} \right)^2} \times 484$
Hence we get on calculating,
$\Rightarrow {P_3} = 3.74W$
So we can see the power consumed by the $40W$ bulb is the highest.
Thus, the correct option is (B).

Note
In the question, though we are given the power rating of each of the bulbs, still the power consumption of the bulbs is different from that. This is because the bulbs are connected in series. So the potential drop across each bulb is different. And the power consumption depends on the potential across the bulb. So the bulb with the maximum resistance glows the brightest.