The power dissipated by a light bulb with 4 ohm resistance when connected in parallel to 12V battery is

36 W

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3.6 W

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0.36 W

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0.036 W

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Solution

The correct option is A 36 W
Given : $V = 12$ volts $R = 4 Ω$
Thus power dissipated $P = \frac{V^{2}}{R} = \frac{12^{2}}{4} = 36$ W

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The power dissipated by a light bulb with 4 ohm resistance when connected in parallel to 12V battery is

The correct option is A 36 WGiven : V=12 volts R=4ΩThus power dissipated P=V2R=1224=36 W

The correct option is A 36 W
Given : $V = 12$ volts $R = 4 Ω$
Thus power dissipated $P = \frac{V^{2}}{R} = \frac{12^{2}}{4} = 36$ W