First slide

Electrical power

Question

The resistance of the filament of an electric bulb changes with temperature. If an electric bulb rated 220 volt and 100 watt is connected to a (220 x .8) volt source, then the actual power would be

Moderate

A

$100 \times 0 .8 watt$
B

$100 \times {(0 .8)}^{2} watt$
C

between 100 x 0.8 watt and 100 watt
D

between 100 x (0.8)² watt and 100 x 0.8 watt

Solution

$P_{1} = \frac{{(220)}^{2}}{R_{1}} and P_{1} = \frac{{(220 \times 0 .8)}^{2}}{R_{2}}$
$\frac{P_{2}}{P_{1}} = \frac{{(220 \times 0 .8)}^{2}}{{(220)}^{2}} \times \frac{R_{1}}{R_{2}} \Rightarrow \frac{P_{2}}{P_{1}} = {(0 .8)}^{2} \times \frac{R_{1}}{R_{2}}$
Here R₂ < R₁
(because voltage decreases from 220 V $\to$ 220 x 0.8 V. It means heat produced $\to$ decreases)
$So \frac{R_{1}}{R_{2}} > 1 \Rightarrow P_{2} {(0 .8)}^{2} P_{1} \Rightarrow P_{2} > {(0 .8)}^{2} \times 100 W$
$Also \frac{P_{2}}{P_{1}} = \frac{(220 \times 0 .8) i_{2}}{220 i_{1}}, Since i_{2} < i_{1} (we expect)$
$So \frac{P_{2}}{P_{1}} < 0 .8 \Rightarrow P_{2} < (100 \times 0 .8)$
Hence, the actual power would be between 100 x (0.8)² W and (100 x 0.8) W

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