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(D) Given: Series resistance $R = 4 \, k\Omega = 4 	imes 10^3 \, \Omega$,Input voltage $V_i = 60 \, V$,Zener voltage $V_Z = 10 \, V$,Load resistance

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$12.5 \, mA, 7.5 \, mA, 5 \, mA$

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(D) Given: Series resistance $R = 4 \, k\Omega = 4 	imes 10^3 \, \Omega$,Input voltage $V_i = 60 \, V$,Zener voltage $V_Z = 10 \, V$,Load resistance $R_L = 2 \, k\Omega = 2 	imes 10^3 \, \Omega$.$1$. The load current $I_L$ is determined by the Zener voltage across the load resistance:$I_L = \frac{V_Z}{R_L} = \frac{10 \, V}{2 	imes 10^3 \, \Omega} = 5 	imes 10^{-3} \, A = 5 \, mA$.$2$. The total current $I$ flowing through the series resistor $R$ is determined by the voltage drop across it:$I = \frac{V_i - V_Z}{R} = \frac{60 \, V - 10 \, V}{4 	imes 10^3 \, \Omega} = \frac{50 \, V}{4000 \, \Omega} = 12.5 	imes 10^{-3} \, A = 12.5 \, mA$.$3$. Applying Kirchhoff's Current Law at node $A$:$I = I_Z + I_L$$I_Z = I - I_L = 12.5 \, mA - 5 \, mA = 7.5 \, mA$.Thus,the currents are $I = 12.5 \, mA$,$I_Z = 7.5 \, mA$,and $I_L = 5 \, mA$.

$A$ Zener diode is connected to a battery and a load as shown below. The currents $I$,$I_Z$,and $I_L$ are respectively:

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