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A Bulb is Connected to a Battery of P.D. 4 V and Internal Resistanceselina Solutions Icse Class 10 Physics Chapter - Current Electricity.

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Question

A bulb is connected to a battery of p.d. 4 V and internal resistance 2.5 Ω . A steady current of 0.5 A flows through the circuit. Calculate:

The energy dissipated in the bulb in 10 minutes.

Sum
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Solution

Given , 

Voltage , V = 4 V

Resistance of the battery , RB = 2.5  Ω

Current , I = 0.5 A

Energy dissipated in the bulb in 10 min , E = I2Rt

E = (0.5)2 × 5.5 × 600 = 825 J

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