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) $W _1<W _2<W _3$
Show Answer
Solution:
- $\quad P=\frac{V^{2}}{R}$ so, $R=\frac{V^{2}}{P}$
$$ \therefore \quad R _1=\frac{V^{2}}{100} \quad \text { and } \quad R _2=R _3=\frac{V^{2}}{60} $$
$$ \text { Now, } \quad W _1=\frac{(250)^{2}}{\left(R _1+R _2\right)^{2}} \cdot R _1 $$
$$ \begin{gathered} W _2=\frac{(250)^{2}}{\left(R _1+R _2\right)^{2}} \cdot R _2 \quad \text { and } \quad W _3=\frac{(250)^{2}}{R _3} \\ W _1: W _2: W _3=15: 25: 64 \text { or } W _1<W _2<W _3 \end{gathered} $$
