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

3.6 W

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36 W

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0.36 W

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None of these

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Solution

The correct option is B
36 W

Step 1: Given
Bulb resistance given $R = 4 Ω$
The voltage of the battery that is connected $V = 12 v o l t s$
Step 2: Calculate power-
The power dissipated can be given by the formula
$P = \frac{V^{2}}{R}$
By substituting values we get
$P = \frac{12^{2}}{4} = 36 W$
Hence, the power dissipated is (B) i.e. 36 W.

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

The correct option is B 36 WStep 1: GivenBulb resistance given R=4ΩThe voltage of the battery that is connected V=12voltsStep 2: Calculate power-The power dissipated can be given by the formulaP=V2RBy substituting values we getP=1224=36WHence, the power dissipated is (B) i.e. 36 W.