A generator has an armature resistance of 0.1 | vedclass

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(C) Given:Armature resistance, $R = 0.1 \, \Omega$Induced emf, $e = 120 \, V$Current drawn, $I = 50 \, A$For a generator, the relationship between the

Vedclass · Accepted Answer

$115$

Vedclass · Answer

(C) Given:Armature resistance, $R = 0.1 \, \Omega$Induced emf, $e = 120 \, V$Current drawn, $I = 50 \, A$For a generator, the relationship between the induced emf $(e)$, terminal voltage $(V)$, and armature resistance $(R)$ is given by the equation:$e = V + I R$Rearranging the formula to solve for the terminal voltage $(V)$:$V = e - I R$Substituting the given values:$V = 120 - (50 	imes 0.1)$$V = 120 - 5$$V = 115 \, V$Thus, the terminal voltage is $115 \, V$.

$A$ generator has an armature resistance of $0.1 \, \Omega$ and develops an induced emf of $120 \, V$ when driven at its rated speed. Its terminal voltage when a current of $50 \, A$ is being drawn is ................. $V$.

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