Question

Two electric bulbs, rated at $\left( {25W,220V} \right)$ and $\left( {100W,220V} \right)$ are connected in series across a $220V$ voltage source. If the $25W$ and $100W$ bulbs draw powers ${P_1}$ and ${P_2}$ respectively, then,
A) ${P_1} = 9W,{P_2} = 16W$
B) ${P_1} = 4W,{P_2} = 16W$
C) ${P_1} = 16W,{P_2} = 4W$
D) ${P_1} = 16W,{P_2} = 9W$

Answer 1

Hint: In this question the power of each bulb is given and also the voltage is given for each bulb. To solve the question, we can use this value to find the resistance of each bulb and then the total current used by the bulbs. From the currents we can find the values of power used by each bulb.

Formula used:
$R = \dfrac{{{V^2}}}{P}$
where $R$ is the resistance, $V$ is the voltage reading and $P$ is the power reading.
${R_{series}} = {R_1} + {R_2}$
Where ${R_{series}}$ is the net resistance in series connection, ${R_1}$ is the resistance in 1^st substance and
${R_2}$ is the resistance in 2^nd substance
$I = \dfrac{V}{R}$
Here $I$ is the total current, $V$ is the potential and $R$ is the net resistance.
$P = {I^2}R$
Where $P$ is the power supplied, $I$ is the total current and $R$ is the resistance.

Complete step by step answer:
To find the power for each bulb we need to find the resistance and the current through each bulb.
For bulb 1, the resistance will be
$ \Rightarrow {R_1} = \dfrac{{{V_1}^2}}{{{P_1}}}$
where ${R_1}$ is the resistance of the bulb 1 , ${V_1}$ is the voltage reading of the bulb 1 and ${P_1}$ is the power reading the bulb 1.
 $ \Rightarrow {R_1} = \dfrac{{{{220}^2}}}{{25}} = 1936\Omega $
For bulb 2, the resistance will be
$ \Rightarrow {R_2} = \dfrac{{{V_2}^2}}{{{P_2}}}$
where ${R_2}$ is the resistance of the bulb 2 , ${V_2}$ is the voltage reading of the bulb 2 and ${P_2}$ is the power reading the bulb 2.
$ \Rightarrow {R_2} = \dfrac{{{{220}^2}}}{{100}} = 484\Omega $
As the bulbs are connected in series, the net resistance will be,
$ \Rightarrow {R_{series}} = {R_1} + {R_2}$
Where ${R_{series}}$ is the net resistance in series connection, ${R_1}$ is the resistance of bulb 1 and
${R_2}$ is the resistance in bulb 2.
$ \Rightarrow {R_{series}} = 1936 + 484 = 2420\Omega $
The total current in the series will be
$ \Rightarrow I = \dfrac{V}{R}$
where $I$is the total current, $V$ is the potential and $R$ is the net resistance in series.
$ \Rightarrow I = \dfrac{{220}}{{2420}} = 0.090A$
So the power in each bulb will be,
$P = {I^2}R$
For bulb 1,
$ \Rightarrow {P_1} = {I^2}{R_1}$
$ \Rightarrow {P_1} = {(.090)^2}1936 = 16W$
For bulb 2,
$ \Rightarrow {P_2} = {I^2}{R_2}$
$ \Rightarrow {P_2} = {(.090)^2}484 = 4W$

Hence, the correct option is option (C).

Note: While solving this question we have to be careful about the formulas and when and where they are used. Here the power formula is used differently, one with resistance and voltage and the other with resistance and current. Also the connection of the bulbs are important. To get correct value, the net resistance should be properly checked.