A generator has an e.m.f. of 440,V and an int | vedclass

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Question

(D) The total resistance of the circuit is the sum of the internal resistance $(r)$ and the load resistance $(R)$.$R_{total} = r + R = 400\,\Omega + 4

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$400$

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(D) The total resistance of the circuit is the sum of the internal resistance $(r)$ and the load resistance $(R)$.$R_{total} = r + R = 400\,\Omega + 4000\,\Omega = 4400\,\Omega$.Using Ohm's law,the current $(I)$ flowing through the circuit is given by $I = \frac{E}{R_{total}}$,where $E$ is the e.m.f.$I = \frac{440\,V}{4400\,\Omega} = 0.1\,A$.The voltage across the load $(V_{load})$ is given by $V_{load} = I 	imes R$.$V_{load} = 0.1\,A 	imes 4000\,\Omega = 400\,V$.

$A$ generator has an e.m.f. of $440\,V$ and an internal resistance of $400\,\Omega$. Its terminals are connected to a load of $4000\,\Omega$. The voltage across the load is $...........\,V$.

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