A 100 W Bulb B1 and two 60 W bulbs B2 and B3...


Problem: A 100 W bulb B1 and two 60 W bulbs B2 and B3, are connected to a 250 V source, as shown in the figure. Now W1, W2 and W3 are the output powers of the bulb B1, B2 and B3 respectively. Then,  (IIT JEE 2002)

A 100 W Bulb B1 and two 60 W bulbs B2 and B3
  1. $W_1 > W_2 = W_3$
  2. $W_1 > W_2 > W_3$
  3. $W_1 < W_2 = W_3$
  4. $W_1 < W_2 < W_3$

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).

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