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प्रश्न
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.
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उत्तर
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
संबंधित प्रश्न
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