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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. - Physics

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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.

बेरीज

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

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पाठ 8: Current Electricity - Exercise 8 (C) 3 [पृष्ठ २१३]

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सेलिना Physics [English] Class 10 ICSE
पाठ 8 Current Electricity
Exercise 8 (C) 3 | Q 15.3 | पृष्ठ २१३
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