First slide

Electrical power

Question

Two electric lamps of 40 watt each are connected in parallel. The power consumed by the combination will be

Moderate

A

20 watt
B

60 watt
C

80 watt
D

100 watt

Solution

$P_{1} = \frac{V^{2}}{R_{1}} and P_{2} = \frac{V^{2}}{R_{2}}$
When bulbs are connected in parallel, then equivalent resistance R is given by
$\begin{array}{l} R_{p} = (\frac{R_{1} R_{2}}{R_{1} + R_{2}}) \\ ∴ P_{p} = \frac{V^{2}}{(\frac{R_{1} R_{2}}{R_{1} + R_{2}})} = \frac{V^{2} (R_{1} + R_{2})}{R_{1} R_{2}} \\ = \frac{V^{2} [(V^{2} / P_{1}) + (V^{2} / P_{2})]}{(V^{2} / P_{1}) \times (V^{2} / P_{2})} = \frac{V^{4} [\frac{P_{1} + P_{2}}{P_{1} P_{2}}]}{V^{4} (1 / P_{1} P_{2})} \\ = P_{1} + P_{2} = 40 + 40 = 80 watt \end{array}$
Note. For series combination
$\begin{array}{l} P_{s} = \frac{V^{2}}{(R_{1} + R_{2})} = \frac{V^{2}}{(V^{2} / P_{1}) + (V^{2} / P_{2})} \\ = \frac{P_{1} P_{2}}{P_{1} + P_{2}} \end{array}$

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