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The voltage and power of an electric bulb are 220V and 100W respectively. What will be its power when the voltage becomes 110 V?
Given:
Voltage, $V$ = 220V
Power, $P$ = 100W
To find: Power, $(P)$ when the bulb is operated on 110 V.
Solution:
We know that power is given as-
$P=\frac {V^2}{R}$
Substituting the given values we get-
$100=\frac {220^2}{R}$
$R=\frac {220\times {220}}{100}$
$R=22\times {22}$
$R=484\Omega$
Thus, the resistance of the bulb is $484\Omega$
Now,
When, the voltage, $V$ becomes 110 V, then the power consumed is given as-
$P=\frac {V^2}{R}\Rightarrow\frac {110^2}{484}\Rightarrow\frac {12100}{484}\Rightarrow25W$.
So, 25 W power is consumed when the bulb is operated on 110 V.
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