Question

In the diagram shown, the Zener diode has a reverse breakdown voltage of VZ . The current through the load resistance RL is IL The current through the Zener diode is

• A. $\frac{v_0-v_z}{R_s}$
• B. $\frac{v_0-v_z}{R_L}$
• C. $\frac{v_z}{R_L}$
• D. $(\frac{v_0-v_z}{R_S})-I_l$

Answer 1

Answer: (D)

$(\frac{v_0-v_z}{R_S})-I_l$

Answer 2

The correct answer is Option (D) : $(\frac{v_0-v_z}{R_S})-I_l$