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Question

A bulb is connected to a battery of potential difference 4 V and internal resistance 2.5 Ω. A steady current of 0.5 A flows through the circuit. The energy dissipated in the bulb in 10 minutes is :

A
125J
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B
825J
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C
120J
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D
240J
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Solution

The correct option is B 825J
Energy (power×time) is measured in Joules and by including time (t) in the power formulae, the energy dissipated by a component or circuit can be calculated.

Energy dissipated = Pt or VI×tasP=VI.

In the question, it is given that the current I is 0.5 A and voltage V is 4 V. Thus the power dissipated is calculated as P=VI=4V×0.5A=2W.

Substituting V=IR (Ohm's law) in the above formula, we get, P=I2×R.

Therefore, the total resistance in the circuit is calculated using the relation R=PI2=20.52=8ohms.

The internal resistance of the cell is 2.5 ohms, hence, the resistance of the bulb is 82.5=5.5 ohms.

The power dissipated by the bulb in 10 minutes is calculated from the formula P=I2×R×t, where I=0.5A,R=5.5 ohms and t=10 minutes, that is 600 seconds.

Hence, P=0.52×5.5×600=825J.

The power dissipated by the bulb in 10 minutes is 825 Joules.

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