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प्रश्न
An auto lamp is joined to a battery of e.m.f. 4 V and internal resistance 2.5Ω. A steady current of 0.5 A flows through the circuit. Calculate the
(a) Total energy provided by battery in 10 minutes,
(b) Heat dissipated in the bulb in 10 minutes.
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उत्तर
Given, emf, e = 4 V, internal resistance r = 2.5Ω , current I= 0.5A
(a) Energy provided by battery in 10 mins =Power x time= (VIt)= 4 x 0.5 x 20= 40 watt-hour
(b) Heat dissipated in the bulb in 10 minutes= I2Rt
Let R be the resistance of the bulb, then:
I = `"e"/("R" + "r")`
or, 0.5 = `4/("R" + 2.5)`
or, R + 2.5 = 8
or, R = 5.5 Ω
Heat dissipated in the bulb in 10 minutes= I2Rt = ( 0.5)2 ( 5.5) (10 x 60) = 825 joules
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