Question:
In the following circuit, the current through the Zener diode is
In the following circuit, the current through the Zener diode is
Show Hint
Check whether the Zener is in breakdown, find the total current through the series resistor, subtract the current in the parallel resistor.
Updated On: Nov 12, 2026
- \(5.00\text{ mA}\)
- \(6.25\text{ mA}\)
- \(11.25\text{ mA}\)
- \(16.25\text{ mA}\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
The Zener diode in breakdown holds the voltage across itself and the parallel resistor at \(V_z\).
Step 2: Check Breakdown
Without the Zener, the voltage across \(R_2\) would be \(18 \times \dfrac{1800}{800 + 1800} \approx 12.5\text{ V}\), which is more than 9 V. So the Zener conducts and holds 9 V.
Step 3: Currents
Total current \(I = \dfrac{18 - 9}{800} = 11.25\text{ mA}\)
Current in \(R_2\): \(\dfrac{9}{1800} = 5\text{ mA}\)
Step 4: Zener Current
\(I_z = 11.25 - 5 = 6.25\text{ mA}\)
Final Answer:
The current through the Zener diode is 6.25 mA. \[ \boxed{\text{(B) }6.25\text{ mA}} \]
The Zener diode in breakdown holds the voltage across itself and the parallel resistor at \(V_z\).
Step 2: Check Breakdown
Without the Zener, the voltage across \(R_2\) would be \(18 \times \dfrac{1800}{800 + 1800} \approx 12.5\text{ V}\), which is more than 9 V. So the Zener conducts and holds 9 V.
Step 3: Currents
Total current \(I = \dfrac{18 - 9}{800} = 11.25\text{ mA}\)
Current in \(R_2\): \(\dfrac{9}{1800} = 5\text{ mA}\)
Step 4: Zener Current
\(I_z = 11.25 - 5 = 6.25\text{ mA}\)
Final Answer:
The current through the Zener diode is 6.25 mA. \[ \boxed{\text{(B) }6.25\text{ mA}} \]
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