In the following circuit,the power dissipated | vedclass

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(C) Given: Source voltage $V_s = 12 \ V$,series resistance $R_s = 100 \ \Omega$,Zener voltage $V_z = 8 \ V$,load resistance $R_L = 800 \ \Omega$.The c

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$0.24$

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(C) Given: Source voltage $V_s = 12 \ V$,series resistance $R_s = 100 \ \Omega$,Zener voltage $V_z = 8 \ V$,load resistance $R_L = 800 \ \Omega$.The current through the series resistor $R_s$ is given by:$I_s = \frac{V_s - V_z}{R_s} = \frac{12 \ V - 8 \ V}{100 \ \Omega} = \frac{4 \ V}{100 \ \Omega} = 0.04 \ A = 40 \ mA$.The current through the load resistor $R_L$ is:$I_L = \frac{V_z}{R_L} = \frac{8 \ V}{800 \ \Omega} = 0.01 \ A = 10 \ mA$.The current through the Zener diode $I_z$ is:$I_z = I_s - I_L = 40 \ mA - 10 \ mA = 30 \ mA = 0.03 \ A$.The power dissipated in the Zener diode is:$P_z = V_z 	imes I_z = 8 \ V 	imes 0.03 \ A = 0.24 \ W$.

In the following circuit,the power dissipated in the Zener diode is (in $W$)

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