Current Electricity Ques 65
- 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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Answer:
Correct Answer: 65.(a)
Solution:
Formula:
- 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=\left(\frac{1}{11}\right)^{2} \times \frac{220 \times 220}{25}=16 W \\ & \text { and } P _2=i^{2} R _2=\left(\frac{1}{11}\right)^{2} \times \frac{220 \times 220}{100}=4 W \end{aligned} $
BETA
