In the diagram shown,the Zener diode has a re | vedclass

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(D) From the circuit diagram,the total current $I$ flowing through the series resistor $R_S$ is given by the voltage drop across it divided by its res

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$\left(\frac{V_0-V_Z}{R_S}\right)-I_L$

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(D) From the circuit diagram,the total current $I$ flowing through the series resistor $R_S$ is given by the voltage drop across it divided by its resistance:$I = \frac{V_0 - V_Z}{R_S}$According to Kirchhoff's current law at the junction point,the total current $I$ splits into the Zener diode current $I_Z$ and the load current $I_L$:$I = I_Z + I_L$Rearranging this equation to solve for the Zener diode current $I_Z$:$I_Z = I - I_L$Substituting the expression for $I$:$I_Z = \left(\frac{V_0 - V_Z}{R_S}\right) - I_L$

In the diagram shown,the Zener diode has a reverse breakdown voltage of $V_Z$. The current through the load resistance $R_L$ is $I_L$. The current through the Zener diode is

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