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(C) Given: Current $I = 10 \ mA = 10 	imes 10^{-3} \ A$,Potential drop across diode $V_d = 0.5 \ V$,Series resistance $R = 200 \ \Omega$.The total vo

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$2.5$

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(C) Given: Current $I = 10 \ mA = 10 	imes 10^{-3} \ A$,Potential drop across diode $V_d = 0.5 \ V$,Series resistance $R = 200 \ \Omega$.The total voltage $V$ supplied by the battery is the sum of the voltage drop across the resistor and the voltage drop across the diode.$V = V_R + V_d$Using Ohm's law for the resistor,$V_R = I 	imes R$.$V_R = (10 	imes 10^{-3} \ A) 	imes (200 \ \Omega) = 2 \ V$.Therefore,the total voltage $V = 2 \ V + 0.5 \ V = 2.5 \ V$.

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