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A generator has armature resistance of $0.1 \Omega$ and develops an induced emf of $120 \mathrm{~V}$ when driven at its rated speed. Its terminal voltage when a current of $50 \mathrm{~A}$ is being drawn is
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The correct answer is:
$115 \mathrm{~V}$
$115 \mathrm{~V}$
Armature resistance $R=0.1 \Omega$
Induced emf, $e=120 \mathrm{~V}$
Current drown, $I=50 \mathrm{~A}$
We know that,
$e=V+I R$
or, $V=e-I R=120-50 \times 0.1=115 \mathrm{~V}$
Induced emf, $e=120 \mathrm{~V}$
Current drown, $I=50 \mathrm{~A}$
We know that,
$e=V+I R$
or, $V=e-I R=120-50 \times 0.1=115 \mathrm{~V}$
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