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The p.d. across the terminals of a battery is 50 V when 11A are drawn and 60 V when 1A is drawn. The emf and internal resistance of the battery are

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$64 V, 2 Ω$

$63 V, 1 Ω$

$62 V, 2 Ω$

$61 V, 1 Ω$

answer is C.

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Detailed Solution

$V = E - ir ∴ 50 = E - 11 r \to (1) 60 = E - 1 r \to (2) Solving (1) \& (2), r = 1 Ω ∴ E = ir + V E = 11 \times 1 + 50 = 61 V$

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