Current Electricity 3 Question 7

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:

  1. 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,

$$ \begin{aligned} & \qquad P _1=i^{2} R _1=\frac{1}{11}^{2} \times \frac{220 \times 220}{25}=16 W \\ & \text { and } P _2=i^{2} R _2=\frac{1}{11}^{2} \times \frac{220 \times 220}{100}=4 W \end{aligned} $$



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