Three identical bulbs $B_{1}, B_{2} and B_{3}$ are connected in series parallel combination as shown. Rating of each bulb is 45 W, 100 $V o l t$ . If a 100V supply is connected across the combination, the power consumed by the system will be

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90 W

160 W

180 W

30 W

answer is D.

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Detailed Solution

If R be the resistance of each bulb, then $R = \frac{V^{2}}{P}$

Now, $R_{C} = \frac{R \times R}{R + R} + R = \frac{3 R}{2}$

$i =$ current flowing through $B_{3} = \frac{V}{3 R / 2} = \frac{2}{3} \frac{V}{R}$

$P^{1} = \{{(\frac{i}{2})}^{2} . R\} \times 2 + i^{2} R = \frac{3}{2} i^{2} R$

$= \frac{3}{2} {(\frac{2 V}{3 R})}^{2} R = \frac{2}{3} \frac{V^{2}}{R} = \frac{2}{3} P$

$∴ P^{1} = \frac{2}{3} \times 45 W = 30 W$

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