**Problem:**
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 bulb B_{1}, B_{2} and B_{3} respectively. Then,
(IIT JEE 2002)

- $W_1 > W_2 = W_3$
- $W_1 > W_2 > W_3$
- $W_1 < W_2 = W_3$
- $W_1 < W_2 < W_3$

**Solution:**
Let the given power ratings be defined at an operating voltage V. The resistances of the three bulbs are given by
\begin{align}
R_1&={V^2}/{100},\\
R_2&={V^2}/{60},\\
R_3&={V^2}/{60}.\nonumber
\end{align}
In the given configuration, the current through $B_1$ and $B_2$ is
\begin{alignat}{2}
i&=\frac{250}{R_1+R_2} \\
&=\frac{250}{V^2}\left(\frac{100\times 60}{100+60}\right) \\
&=\frac{250}{V^2}\left(\frac{75}{2}\right).\nonumber
\end{alignat}
Substitute the values of $i$, $R_1$, and $R_2$ to get the output powers
\begin{align}
&W_1=i^2R_1 \approx 14\, \left({250}/{V}\right)^{2}, \nonumber\\
&W_2=i^2R_2 \approx 23\, \left({250}/{V}\right)^{2}, \nonumber
\end{align}
and
\begin{align}
&W_3={(250)^2}/{R_3}=60\, \left({250}/{V}\right)^{2}.\nonumber
\end{align}
It is interesting to note that $B_1$ (100 W) is dimmer than $B_2$ (60 W) which in turn is dimmer than $B_3$ (60 W).

- Series and Parallel Arrangements of Resistances
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