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Question
A battery of emf E and internal resistance r is connected to a rheostat. When a current of 2 A is drawn from the battery, the potential difference across the rheostat is 5 V. The potential difference becomes 4 V when a current of 4 A is drawn from the battery. Calculate the value of E and r.
Numerical
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Solution
When a current of 2 A is drawn from the battery, the potential difference across the rheostat is 5 V.
∴ 5 = E − 2r ...(i)
When a current of 4 A is drawn from the battery, the potential difference across the rheostat is 4 V.
∴ 4 = E − 4r ...(ii)
Subtracting Eq. (i) from (ii);
r = `1/2` ohm
Putting the value of r in Eq. (i),
5 = `E - 2 xx 1/2`
5 = E − 1
E = 5 + 1
E = 6 V
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