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(C) The circuit consists of a source voltage $V = 24 	ext{ V}$, a series resistor $R_s = 1 	ext{ k}\Omega$, a Zener diode with breakdown voltage $V_

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$120$

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(C) The circuit consists of a source voltage $V = 24 	ext{ V}$, a series resistor $R_s = 1 	ext{ k}\Omega$, a Zener diode with breakdown voltage $V_z = 10 	ext{ V}$, and a load resistor $R_L = 5 	ext{ k}\Omega$.$1$. The current through the series resistor $R_s$ is given by:$I_s = \frac{V - V_z}{R_s} = \frac{24 	ext{ V} - 10 	ext{ V}}{1 	imes 10^3 	ext{ }\Omega} = \frac{14 	ext{ V}}{1000 	ext{ }\Omega} = 14 	ext{ mA}$.$2$. The current through the load resistor $R_L$ is:$I_L = \frac{V_z}{R_L} = \frac{10 	ext{ V}}{5 	imes 10^3 	ext{ }\Omega} = 2 	ext{ mA}$.$3$. The current through the Zener diode $I_z$ is:$I_z = I_s - I_L = 14 	ext{ mA} - 2 	ext{ mA} = 12 	ext{ mA}$.$4$. The power dissipated across the Zener diode $P_z$ is:$P_z = V_z 	imes I_z = 10 	ext{ V} 	imes 12 	ext{ mA} = 120 	ext{ mW}$.

For the given circuit, the power across the Zener diode is $....... \text{mW}$.

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