Explanation

(b) Resistance of bulb 1
[image:3_image166.png]
$R_{1} = \frac{V^{2}}{P} = \frac{(220)^{2}}{100}$
$R_{1} = 484\ \Omega$
Resistance of bulb 2
$R_{2} = \frac{(220\ V)^{2}}{60\ W} = \frac{4840}{6} = 806.6\ \Omega$
$R_{eq} = R_{1} + R_{2} = (220)^{2}\left( \frac{1}{100} + \frac{1}{60} \right)$
$\Rightarrow$ Current flowing$\ I = \frac{220\ V}{R_{eq}} = \frac{220}{(220)^{2}}\left( \frac{100 \times 60}{160} \right)$
$I = \frac{1}{220}\left( \frac{75}{2} \right),\ I = \frac{15}{88}A$
Power consumed by $100W$ bulb $= I^{2}R_{1}$
$= \left( \frac{15}{88} \right)^{2} \times \frac{(220)^{2}}{100} = \frac{225}{16} = 14W$