Current Electricity 3 Question 4
4. Two electric bulbs rated at $25 W, 220 V$ and $100 W, 220 V$ are connected in series across a $220 V$ voltage source. If the $25 W$ and $100 W$ bulbs draw powers $P _1$ and $P _2$ respectively, then
(2019 Main, 12 Jan I)
(a) $P _1=16 W, P _2=4 W$
(b) $P _1=4 W, P _2=16 W$
(c) $P _1=9 W, P _2=16 W$
(d) $P _1=16 W, P _2=9 W$
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Solution:
- Resistance of a bulb of power $P$ and with a voltage source $V$ is given by
$$ R=\frac{V^{2}}{P} $$
Resistance of the given two bulbs are
and
$$ \begin{aligned} & R _1=\frac{V^{2}}{P _1}=\frac{(220)^{2}}{25} \\ & R _2=\frac{V^{2}}{P _2}=\frac{(220)^{2}}{100} \end{aligned} $$
Since, bulbs are connected in series. This means same amount of current flows through them.
$\therefore$ Current in circuit is
$$ i=\frac{V}{R _{\text {total }}}=\frac{220}{\frac{(220)^{2}}{25}+\frac{(220)^{2}}{100}}=\frac{1}{11} A $$
Power drawn by bulbs are respectively,
$$ \begin{aligned} & \qquad P _1=i^{2} R _1=\frac{1}{11}^{2} \times \frac{220 \times 220}{25}=16 W \\ & \text { and } P _2=i^{2} R _2=\frac{1}{11}^{2} \times \frac{220 \times 220}{100}=4 W \end{aligned} $$
