Question

In the given circuit, the breakdown voltage of the Zener diode is 3.0 V. What is the value of $I_z$?

• A. 3.3 mA
• B. 5.5 mA
• C. 10 mA
• D. 7 mA

Answer 1

Answer: (B)

5.5 mA

Answer 2

The Zener diode maintains a constant voltage of $V_z = 3.0V$ across its terminals. This sets the voltage at point A to $V_A = 3.0V$. The voltage drop across the 1k$\Omega$ resistor is $V_{R1} = 10V - 3.0V = 7.0V$. The total current through the 1k$\Omega$ resistor is $I_{total} = \frac{7.0V}{1k\Omega} = 7.0mA$. The current through the 2k$\Omega$ resistor is $I_R = \frac{3.0V}{2k\Omega} = 1.5mA$. By Kirchhoff's current law at node A, $I_z = I_{total} - I_R = 7.0mA - 1.5mA = 5.5mA$.