An electric bulb rated for $500 W$ at $100 V$ is used in a circuit having a $200 V$ supply. The resistance $R$ that must be put in series with the bulb, so that the bulb draws $500 W$ , is

5 Ω

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20 Ω

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10 Ω

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30 Ω

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Solution

The correct option is A $20 Ω$
Here, $P = V I \Rightarrow I = P / V = 500 / 100 = 5 A$
This is the current through bulb as well as the resistor $R$ .
Voltage across $R$ is $V_{R} = I R$
$\Rightarrow 100 = 5 R \Rightarrow R = 20 Ω$

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An electric bulb rated for 500 W at 100 V is used in a circuit having a 200 V supply. The resistance R that must be put in series with the bulb, so that the bulb draws 500 W, is

The correct option is A 20ΩHere, P=VI⇒I=P/V=500/100=5A This is the current through bulb as well as the resistor R.Voltage across R is VR=IR⇒100=5R⇒R=20Ω

An electric bulb rated for $500 W$ at $100 V$ is used in a circuit having a $200 V$ supply. The resistance $R$ that must be put in series with the bulb, so that the bulb draws $500 W$ , is

The correct option is A $20 Ω$
Here, $P = V I \Rightarrow I = P / V = 500 / 100 = 5 A$
This is the current through bulb as well as the resistor $R$ .
Voltage across $R$ is $V_{R} = I R$
$\Rightarrow 100 = 5 R \Rightarrow R = 20 Ω$