Question

The breakdown voltage of a zener diode is \(10\,V\). It is used in a voltage regulator circuit shown in figure. Current through zener diode is center

center

• A. \(20\,mA\)
• B. \(30\,mA\)
• C. \(32\,mA\)
• D. \(12\,mA\)

Hint:

In zener regulator: \[ I_{\text{series}}=I_Z+I_L \] Always calculate total current first using series resistor.

Answer 1

The Correct Option is B

Step 1: Understand zener regulation.
A zener diode in breakdown holds a fixed voltage across itself. Here that is $V_Z = 10$ V, which is also the voltage across the load resistor connected in parallel with it.

Step 2: Find the voltage across the series resistor.
The supply is $18$ V and $10$ V sits across the zener, so the rest drops across the $200\,\Omega$ series resistor. \[ V = 18 - 10 = 8\,\text{V} \]
Step 3: Find the total current from the source.
This whole current flows through the series resistor. \[ I = \frac{8}{200} = 0.04\,\text{A} = 40\,\text{mA} \]
Step 4: Find the load current.
The load is $1\,k\Omega$ with $10$ V across it. \[ I_L = \frac{10}{1000} = 0.01\,\text{A} = 10\,\text{mA} \]
Step 5: Use the current split rule.
The total current splits into the load current and the zener current. \[ I_Z = I - I_L \]
Step 6: Find the zener current.
\[ I_Z = 40 - 10 = 30\,\text{mA} \] \[ \boxed{30\,\text{mA}} \]