Question

In the following circuit, the reading of the ammeter will be: (Take Zener breakdown voltage = 4 V)

• A. 24 mA
• B. 80 mA
• C. 10 mA
• D. 60 mA

Hint:

When dealing with Zener diodes in circuits, the voltage across the diode remains constant at the breakdown voltage. Use Ohm's law to find the current through the rest of the circuit.

Answer 1

The Correct Option is C

We are given:

• Supply voltage \( V_s = 12 \, \text{V} \)
• Series resistor \( R_s = 100 \, \Omega \)
• Zener breakdown voltage \( V_Z = 4 \, \text{V} \)
• Load resistor \( R_L = 400 \, \Omega \)

Step 1: Behavior of the Zener diode
Since the applied voltage \( V_s = 12 \, \text{V} \) is greater than the breakdown voltage of the Zener diode (\( 4 \, \text{V} \)), the Zener diode enters the breakdown region and maintains a constant voltage of \( 4 \, \text{V} \) across itself. 
Step 2: Voltage across the load resistor
The Zener diode and the load resistor are in parallel. Therefore, the voltage across the load resistor is also: \[ V_{R_L} = V_Z = 4 \, \text{V} \] 
Step 3: Current through the load resistor (measured by the ammeter)
\[ I = \frac{V}{R} = \frac{4}{400} = 0.01 \, \text{A} = 10 \, \text{mA} \] 
Step 4: Ammeter Reading
The ammeter is in series with the 400 \( \Omega \) load resistor and thus measures the current through it. 
Therefore, the reading of the ammeter is: \[ 10 \, \text{mA} \]

Answer 2

Given the circuit with a Zener diode in series with two resistors, the voltage \(V_1\) across the 100Ω resistor can be calculated as: \[ V_1 = \frac{400}{100 + 400} \times 12V = \frac{4}{5} \times 12 = \frac{48}{5} \, \text{V} \] Here, \(V_1 > V_z\), where \(V_z\) is the Zener voltage. Thus, the Zener breakdown will take place, and the voltage across the 400Ω resistor will be \(4V\). Now, the current \(I\) across the 400Ω resistor is given by Ohm’s law: \[ I = \frac{4}{400} = 10 \, \text{mA} \] \[ \boxed{I = 10 \, \text{mA}} \]