$I_L=10$ mA; $R_L = 500 \Omega$
Zener Diode Regulator Analysis
This problem involves calculating the load current ($I_L$) and load resistance ($R_L$) for a Zener diode voltage regulator circuit.
Circuit Parameters
- Input Voltage ($V_{in}$): 25 V
- Zener Voltage ($V_Z$): 5 V
- Series Resistor ($R_S$): 400 $\Omega$
- Zener Current ($I_Z$): 4 times Load Current ($I_L$), i.e., $I_Z = 4 I_L$
Step-by-Step Solution
-
Calculate Voltage Across Series Resistor ($R_S$)
The voltage across the series resistor $R_S$ is the difference between the input voltage $V_{in}$ and the Zener voltage $V_Z$.
$V_{RS} = V_{in} - V_Z$
$V_{RS} = 25 \text{ V} - 5 \text{ V} = 20 \text{ V}$ -
Calculate Total Current Through $R_S$
Using Ohm's Law, the total current ($I_{total}$) flowing through $R_S$ is:
$I_{total} = \frac{V_{RS}}{R_S}$
$I_{total} = \frac{20 \text{ V}}{400 \Omega} = 0.05 \text{ A} = 50 \text{ mA}$ -
Determine Load Current ($I_L$)
The total current is the sum of the Zener current ($I_Z$) and the load current ($I_L$):
$I_{total} = I_Z + I_L$Given the relationship $I_Z = 4 I_L$, substitute this into the equation:
$I_{total} = (4 I_L) + I_L = 5 I_L$Now, equate this with the calculated total current:
$5 I_L = 50 \text{ mA}$
$I_L = \frac{50 \text{ mA}}{5} = 10 \text{ mA}$ -
Calculate Load Resistance ($R_L$)
The voltage across the load resistance $R_L$ is equal to the Zener voltage $V_Z$. Using Ohm's Law:
$R_L = \frac{V_Z}{I_L}$
$R_L = \frac{5 \text{ V}}{10 \text{ mA}} = \frac{5 \text{ V}}{10 \times 10^{-3} \text{ A}}$
$R_L = \frac{5}{0.01} \Omega = 500 \Omega$
Therefore, the load current ($I_L$) is 10 mA and the load resistance ($R_L$) is 500 $\Omega$. This corresponds to Option 3.
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