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A current of 2A flows through a 2Ω resistor when connected across a battery. The same battery supplies a current of 0.5A when connected across a 9Ω resistor. The internal resistance of the battery is
Options
(a) 0.5Ω
(b) 1/3Ω
(c) 1/4Ω
(d) 1Ω
Correct Answer:
1/3Ω
Explanation:
Let the internal resistance of the battery be r. Then the current flowing through the circuit is given by
i = E / R + r
In first case, 2 = E / 2 + r …(i)
In second case, 0.5 = E / 9 + r … (ii)
From (i) & (ii), 4 + 2r = 4.5 + 0.5r
1.5r = 0.5
r = 1/3Ω
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