Animated Solution for Physics - Semiconductors: The reverse breakdown voltage of a Zener diode is $5.6\text{ V}$ in the given circuit. The current $I_z$ through the Zener is

Focus on the circuit diagram. We have a $9\text{ V}$ battery powering a network that consists of a Zener diode and two resistors. The Zener diode is connected in parallel with an $800\Omega$ load resistor, and this entire parallel combination is in series with a $200\Omega$ resistor.

Our main goal is to find the current flowing specifically through this Zener diode, denoted as $i_z$. To do this, we need to understand how the current splits in the circuit.

Kirchhoff's Current Law

To find the Zener current, we must apply Kirchhoff's Current Law (KCL) at the top junction. Imagine the total current $i_1$ coming from the battery. When it reaches the junction, it splits into two paths: one part goes through the Zener diode ($i_z$), and the rest goes through the $800\Omega$ load resistor ($i_2$).

Mathematically, this is expressed as:

Rearranging this to solve for the Zener current gives us our master equation:

First, let's find the load current. Because the Zener diode is in parallel with the $800\Omega$ resistor, the voltage across the load resistor will be exactly equal to the Zener breakdown voltage, provided the Zener is active. The problem states the breakdown voltage is $5.6\text{ V}$.

Using Ohm's law, the load current $i_2$ is this voltage divided by the resistance:

Upon calculating, we get:

Now let's determine the total current $i_1$ coming from the battery. To find this, we need the voltage drop across the $200\Omega$ series resistor. The battery provides $9\text{ V}$, and the Zener diode has locked the parallel section at $5.6\text{ V}$.

According to Kirchhoff's Voltage Law, the remaining voltage must drop across the $200\Omega$ resistor:

Applying Ohm's law again for the series resistor:

This gives us:

Finally, let's put it all together using our master equation. The Zener current will be the total current minus the load current:

So we are left with exactly $10\text{ mA}$, which is the current flowing through the Zener diode. The correct option is (a).