In the given circuit, the value of current $$I_L$$ will be ______ mA. (When $$R_L = 1$$ k$$\Omega$$)

Based on the provided circuit diagram, the value of the load current $$I_L$$ (represented as $$I_2$$ in the image) is calculated as follows:

1. Circuit Analysis

The circuit shows a Zener diode connected in parallel with a load resistor $$R_L$$.

• The Zener voltage ($$V_Z$$) is given as 5 V.
• The supply voltage is 10 V.
• The load resistance ($$R_L$$) is 1 k$$\Omega$$.

2. Determining Voltage across $$R_L$$

Since the Zener diode is in its breakdown region (as the supply voltage is higher than the Zener voltage), it maintains a constant voltage across itself and any component connected in parallel to it.

Therefore, the voltage across the load resistor $$V_L$$ is:

$$V_L = V_Z = 5\text{ V}$$

3. Calculation of Load Current ($$I_L$$)

Using Ohm's Law ($$I = \frac{V}{R}$$):

$$I_L = \frac{V_L}{R_L}$$

Substituting the values ($$R_L = 1\text{ k}\Omega = 1000\ \Omega$$):

$$I_L = \frac{5\text{ V}}{1000\ \Omega}$$

$$I_L = 0.005\text{ A}$$

To convert to milliamperes (mA):

$$I_L = 0.005 \times 1000\text{ mA}$$

Final Result:

$$\boxed{I_L = 5\text{ mA}}$$