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

• A

  $\frac{V_0-V_Z}{R_S}$
• B

  $\frac{V_0-V_Z}{R_L}$
• C

  $\frac{V_Z}{R_L}$
• Correct answer
  D

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

Solution

From the diagram,
$I=I_Z+I_L$ $\Rightarrow \quad I_Z=I-I_L=\frac{V_O-V_Z}{R_S}-I_L$

Hence, the correct answer is option: 'D'.