Solution:
Using \( E = I(R + r) \), we set up equations: \( E = 0.5(2 + r) \) and \( E = 0.25(5 + r) \). Equating them gives \( r = 1\ \Omega \), which yields \( E = 0.5(2 + 1) = 1.5\text{ V} \).
When a resistance of 2 ohm is connected across the terminals of a cell, the current is 0.5 amp. When the resistance is increased to 5 ohm, the current is 0.25 amp. The emf of the cell is:
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