A p-n junction diode can withstand up to 20 , | vedclass

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(A) Given: Potential difference across the diode, $V_D = 0.5 \, V$.Maximum current in forward bias, $i = 20 \, mA = 20 	imes 10^{-3} \, A$.The total

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$3.0$

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(A) Given: Potential difference across the diode, $V_D = 0.5 \, V$.Maximum current in forward bias, $i = 20 \, mA = 20 	imes 10^{-3} \, A$.The total voltage $V$ of the battery is the sum of the potential drop across the resistor $R_S$ and the potential drop across the diode $V_D$.The voltage drop across the series resistor $R_S = 125 \, \Omega$ is given by $V_R = i 	imes R_S$.$V_R = (20 	imes 10^{-3} \, A) 	imes (125 \, \Omega) = 2.5 \, V$.The total voltage of the battery is $V = V_R + V_D$.$V = 2.5 \, V + 0.5 \, V = 3.0 \, V$.

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