A car battery of $e.m.f$. $12\,V$ and internal resistance $5 \times {10^{ - 2}}\,\Omega $, receives a current of $60\; A$ from external source, then terminal voltage of battery is
AIPMT 2000, Easy
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(c) $V = E + ir$= $12 + 60 \times 5 \times {10^{ - 2}}$= $15\,V$.
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