The reverse break down voltage of a Zener diode is 5.6 V in the given circuit.

The current I$$_z$$ through the Zener is:

Problem Analysis

• Source Voltage ($$V$$): $$9\text{ V}$$
• Series Resistance ($$R_s$$): $$200\ \Omega$$
• Zener Breakdown Voltage ($$V_z$$): $$5.6\text{ V}$$
• Load Resistance ($$R_L$$): $$800\ \Omega$$
• Correct Option: C ($$10\text{ mA}$$)

Since the Zener diode operates in its reverse breakdown region, it maintains a constant potential drop of $$5.6\text{ V}$$ across itself and the parallel load resistor ($$R_L$$).

Step-by-Step Solution

Step 1: Calculate Total Current ($$I_1$$)

The voltage drop across the series resistor ($$200\ \Omega$$) is the difference between the source voltage and the Zener voltage:

$$V_{200\Omega} = 9\text{ V} - 5.6\text{ V} = 3.4\text{ V}$$

Using Ohm's Law, the total source current ($$I_1$$) is:

$$I_1 = \frac{3.4\text{ V}}{200\ \Omega} = 0.017\text{ A} = 17\text{ mA}$$

Step 2: Calculate Load Current ($$I_2$$)

Since the load resistor ($$800\ \Omega$$) is in parallel with the Zener diode, the voltage across it is exactly $$5.6\text{ V}$$:

$$I_2 = \frac{5.6\text{ V}}{800\ \Omega} = 0.007\text{ A} = 7\text{ mA}$$

Step 3: Calculate Zener Current ($$I_z$$)

Applying Kirchhoff's Current Law (KCL) at the junction:

$$I_1 = I_z + I_2$$

$$I_z = I_1 - I_2$$

$$I_z = 17\text{ mA} - 7\text{ mA} = 10\text{ mA}$$