A Zener diode of breakdown voltage 20 V is c | vedclass

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(C) The circuit consists of a $40 \ V$ $DC$ source,a series resistor $R_s = 2 \ k\Omega$,a Zener diode with breakdown voltage $V_z = 20 \ V$,and a loa

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$6$

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(C) The circuit consists of a $40 \ V$ $DC$ source,a series resistor $R_s = 2 \ k\Omega$,a Zener diode with breakdown voltage $V_z = 20 \ V$,and a load resistor $R_L = 5 \ k\Omega$.First,calculate the total current $I$ flowing through the series resistor $R_s$:$I = \frac{V_{source} - V_z}{R_s} = \frac{40 \ V - 20 \ V}{2 \ k\Omega} = \frac{20 \ V}{2 	imes 10^3 \ \Omega} = 10 	imes 10^{-3} \ A = 10 \ mA$.Next,calculate the load current $I_L$ flowing through the load resistor $R_L$:$I_L = \frac{V_z}{R_L} = \frac{20 \ V}{5 \ k\Omega} = \frac{20 \ V}{5 	imes 10^3 \ \Omega} = 4 	imes 10^{-3} \ A = 4 \ mA$.The current through the Zener diode $I_z$ is given by the difference between the total current and the load current:$I_z = I - I_L = 10 \ mA - 4 \ mA = 6 \ mA$.Therefore,the correct option is $C$.

$A$ Zener diode of breakdown voltage $20 \ V$ is connected as shown in the given circuit. The current through the Zener diode is (in $mA$)

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