A p-n junction diode is connected to a batter | vedclass

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(B) Given:Battery emf,$V = 5.7 \ V$Barrier potential of the diode,$V_B = 0.7 \ V$Series resistance,$R_S = 5 \ k\Omega = 5 	imes 10^3 \ \Omega$In a fo

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$1 \ mA$

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(B) Given:Battery emf,$V = 5.7 \ V$Barrier potential of the diode,$V_B = 0.7 \ V$Series resistance,$R_S = 5 \ k\Omega = 5 	imes 10^3 \ \Omega$In a forward-biased $p-n$ junction diode,the effective voltage across the resistor is the difference between the battery emf and the barrier potential.Effective voltage,$V_{eff} = V - V_B = 5.7 \ V - 0.7 \ V = 5.0 \ V$Using Ohm's law,the current $I$ in the circuit is:$I = \frac{V_{eff}}{R_S} = \frac{5.0 \ V}{5 	imes 10^3 \ \Omega} = 1 	imes 10^{-3} \ A$$I = 1 \ mA$

$A$ $p-n$ junction diode is connected to a battery of emf $5.7 \ V$ in series with a resistor of $5 \ k\Omega$ such that it is forward biased. If the barrier potential of the diode is $0.7 \ V$,neglecting the diode resistance,the current in the circuit is:

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