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(C) The Zener diode is in the breakdown region, so the voltage across it is constant at $10 \, V$.The current through the $500 \, \Omega$ load resisto

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$30$

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(C) The Zener diode is in the breakdown region, so the voltage across it is constant at $10 \, V$.The current through the $500 \, \Omega$ load resistor $(I_L)$ is:$I_L = \frac{V_Z}{R_L} = \frac{10 \, V}{500 \, \Omega} = 0.02 \, A = 20 \, mA$The current through the series resistor $(I_S)$ is:$I_S = \frac{V_{in} - V_Z}{R_S} = \frac{20 \, V - 10 \, V}{200 \, \Omega} = \frac{10 \, V}{200 \, \Omega} = 0.05 \, A = 50 \, mA$Applying Kirchhoff's Current Law at the junction, the current through the Zener diode $(I_Z)$ is:$I_Z = I_S - I_L = 50 \, mA - 20 \, mA = 30 \, mA$

$A$ Zener diode of breakdown voltage $10 \, V$ is used as a voltage regulator as shown in the figure. The current through the Zener diode is (in $ \, mA$)

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