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(C) The input voltage is $V_{in} = 20 \ V$ and the series resistance is $R_s = 2 \ k\Omega$. The Zener diode maintains a constant voltage of $V_z = 5

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$2.5$

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(C) The input voltage is $V_{in} = 20 \ V$ and the series resistance is $R_s = 2 \ k\Omega$. The Zener diode maintains a constant voltage of $V_z = 5 \ V$ across the load resistor $R_L = 1 \ k\Omega$.The total current flowing from the source is:$I_{in} = \frac{V_{in} - V_z}{R_s} = \frac{20 \ V - 5 \ V}{2 	imes 10^3 \ \Omega} = \frac{15 \ V}{2000 \ \Omega} = 7.5 \ mA$.The current flowing through the load resistor $R_L$ is:$I_L = \frac{V_z}{R_L} = \frac{5 \ V}{1 	imes 10^3 \ \Omega} = 5 \ mA$.Applying Kirchhoff's Current Law at the node, the current through the Zener diode $I_Z$ is:$I_Z = I_{in} - I_L = 7.5 \ mA - 5 \ mA = 2.5 \ mA$.

Find the current through the Zener diode in the given circuit. (in $mA$)

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