A dc source of emf , E_1 = 100 , V and intern | vedclass

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(C) For no current to pass through the battery $E_2$,the potential difference across the external resistor $R$ must be equal to the $emf$ of the batte

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(C) For no current to pass through the battery $E_2$,the potential difference across the external resistor $R$ must be equal to the $emf$ of the battery $E_2$.Let $I$ be the current flowing through the circuit containing $E_1$,$r$,and $R$.According to Ohm's law,the current in the circuit is $I = \frac{E_1}{R + r}$.The potential difference across the resistor $R$ is $V = I 	imes R = \frac{E_1 	imes R}{R + r}$.For no current to flow through the branch containing $E_2$,we must have $V = E_2$.Substituting the given values: $\frac{100 	imes R}{R + 0.5} = 90$.Dividing both sides by $10$: $\frac{10 	imes R}{R + 0.5} = 9$.$10R = 9(R + 0.5)$.$10R = 9R + 4.5$.$R = 4.5 \, \Omega$.

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