In the voltage regulator circuit shown below, the reverse breakdown voltage of zener diode is 3 V. Find the current through zener diode.

The correct option is 5.5 mA.

Step-by-Step Explanation:

Based on the provided circuit diagram, we have the following parameters:
- Input DC Voltage ($V_{\text{in}}$) = 10 V
- Series Resistor ($R_s$) = 1 kΩ
- Load Resistor ($R_L$) = 2 kΩ
- Zener Breakdown Voltage ($V_Z$) = 3 V

1. Determine the state of the Zener diode:
To check if the Zener diode is operating in the breakdown region, we calculate the open-circuit voltage ($V_{\text{open}}$) across the load resistor by temporarily removing the Zener diode:

$V_{\text{open}}=V_{\text{in}}\times \left(\frac{R_L}{R_s+R_L}\right)$

Substituting the circuit values:
$V_{\text{open}}=10\text{V}\times \left(\frac{2\text{k}\Omega }{1\text{k}\Omega +2\text{k}\Omega }\right)=10\times \frac{2}{3}\approx 6.67\text{V}$

Since $V_{\text{open}}=6.67\text{V}>V_Z=3\text{V}$, the Zener diode is indeed in the reverse breakdown state and regulates the voltage across the load resistor to exactly $V_Z=3\text{V}$.

2. Calculate the source current ($I_s$):
The voltage drop across the series resistor $R_s$ is the difference between the input voltage and the Zener breakdown voltage:

$V_{R_s}=V_{\text{in}}-V_Z=10\text{V}-3\text{V}=7\text{V}$

The current through the series resistor $R_s$ is:
$I_s=\frac{V_{R_s}}{R_s}=\frac{7\text{V}}{1\text{k}\Omega }$

3. Calculate the load current ($I_L$):
Since the voltage across the load resistor is maintained at $V_Z=3\text{V}$:

$I_L=\frac{V_Z}{R_L}=\frac{3\text{V}}{2\text{k}\Omega }$

4. Calculate the Zener current ($I_Z$):
Applying Kirchhoff's Current Law (KCL) at the junction above the Zener diode:

$I_s=I_Z+I_L$

Rearranging to solve for the Zener current:
$I_Z=I_s-I_L$

$I_Z=7\text{mA}-1.5\text{mA}=5.5\text{mA}$