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(C) The circuit consists of a generator $G$ with terminal voltage $V_G = 120 \ V$,a resistor $R$,and a battery being charged with $emf$ $E = 100 \ V$

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(C) The circuit consists of a generator $G$ with terminal voltage $V_G = 120 \ V$,a resistor $R$,and a battery being charged with $emf$ $E = 100 \ V$ and internal resistance $r = 1 \ \Omega$.When charging a battery,the current $I$ flows into the positive terminal of the battery. The potential difference across the battery is given by $V_{battery} = E + Ir$.Substituting the given values:$V_{battery} = 100 + (10 	imes 1) = 110 \ V$.The total voltage provided by the generator is $120 \ V$. This voltage is dropped across the resistor $R$ and the battery being charged.Applying Kirchhoff's Voltage Law to the loop:$V_G = I 	imes R + V_{battery}$$120 = 10 	imes R + 110$Rearranging to solve for $R$:$10 	imes R = 120 - 110$$10 	imes R = 10$$R = 1 \ \Omega$.

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