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An electric bulb is rated 220 volt and 100 watt, power consumed by it when operated on 110 volt is
a) 50W
b) 85W
c) 90W
d) 25W

Last updated date: 20th Jun 2024
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Hint: The power dissipated across the bulb is given when operated at a particular potential difference. But when the bulb is operated at a different potential difference the power dissipated across it will change. The property of the bulb that does not change is its resistance. Hence from the given data lets calculate the resistance from the power expression and further use this resistance to calculate the power dissipated across it.

Complete step-by-step answer:
Let us first write the power dissipated across the resistance in terms of the potential difference across it and its resistance. The power dissipated in general is given by,
$\text{P=VI}....\text{(1)}$ where V is the potential generated across the bulb and I is the current flowing through the bulb. By Ohms law we can write the current through a resistor as, $I=\dfrac{V}{R}$ where V is the potential generated across the bulb and r is the resistance of the bulb. After substituting for current in equation 1 we get,
$\text{P=V}\dfrac{\text{V}}{\text{R}}=\dfrac{{{\text{V}}^{2}}}{\text{R}}\text{J/sec }....\text{(2)}$
It is given to us that the bulb is rated as 220 volt power dissipated is 100watt. Hence from equation 2 resistance of the bulb is
  & \text{P}=\dfrac{{{\text{V}}^{2}}}{\text{R}} \\
 & 100=\dfrac{{{220}^{2}}}{\text{R}} \\
 & R=\dfrac{{{220}^{2}}}{100}=\dfrac{48400}{100}=484\Omega \\
Now we wish to find the power consumed by it when the potential difference across it is 110 Volts. Hence using equation 2 we get,
  & \text{P}=\dfrac{{{\text{V}}^{2}}}{\text{R}} \\
 & P=\dfrac{{{110}^{2}}}{484}=\dfrac{12100}{484}=25\text{W} \\
The power dissipated across the bulb is 25 watt. Hence the correct answer is option d.

Note:The resistance theoretically we say does not change. But in reality it changes due to the heat generated across a resistor when high potential difference is generated across it. As a result the resistor gets thermally expanded. Since the resistance of a material is directly proportional to its length, on expansion the resistance increases.