Current Electricity 3 Question 10

9. A $100 W$ bulb $B _1$, and two $60 W$ bulbs $B _2$ and $B _3$, are connected to a $250 V$ source, as shown in the figure. Now $W _1, W _2$ and $W _3$ are the output powers of the bulbs $B _1, B _2$ and $B _3$ respectively. Then,

(2002, 2M)

(a) $W _1>W _2=W _3$

(b) $W _1>W _2>W _3$

(c) $W _1<W _2=W _3$

(d) $W _1<W _2<W _3$

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Solution:

  1. $\quad P=\frac{V^{2}}{R}$ so, $R=\frac{V^{2}}{P}$

$$ \therefore \quad R _1=\frac{V^{2}}{100} \quad \text { and } \quad R _2=R _3=\frac{V^{2}}{60} $$

$$ \text { Now, } \quad W _1=\frac{(250)^{2}}{\left(R _1+R _2\right)^{2}} \cdot R _1 $$

$$ \begin{gathered} W _2=\frac{(250)^{2}}{\left(R _1+R _2\right)^{2}} \cdot R _2 \quad \text { and } \quad W _3=\frac{(250)^{2}}{R _3} \\ W _1: W _2: W _3=15: 25: 64 \text { or } W _1<W _2<W _3 \end{gathered} $$



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