Question:
The reverse breakdown voltage of a Zener diode is 5.6 V in the given circuit.
The current \(I_Z\) through the Zener is :
The reverse breakdown voltage of a Zener diode is 5.6 V in the given circuit.
The current \(I_Z\) through the Zener is :
The current \(I_Z\) through the Zener is :
Show Hint
The Zener clamps the voltage across the 800 ohm load to 5.6 V; find the total current through the 200 ohm series resistor and subtract the load current.
Updated On: Nov 5, 2026
- 7 mA
- 17 mA
- 15 mA
- 10 mA
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The Correct Option is D
Solution and Explanation
The Zener holds the voltage across the parallel combination at \(V_Z = 5.6\ \mathrm{V}\).
Voltage across the \(200\ \Omega\) series resistor \(= 9 - 5.6 = 3.4\ \mathrm{V}\), so total current \(I = \frac{3.4}{200} = 17\ \mathrm{mA}\).
Current through the \(800\ \Omega\) load \(= \frac{5.6}{800} = 7\ \mathrm{mA}\).
\(I_Z = 17 - 7 = 10\ \mathrm{mA}\).
Answer: (4) 10 mA
Voltage across the \(200\ \Omega\) series resistor \(= 9 - 5.6 = 3.4\ \mathrm{V}\), so total current \(I = \frac{3.4}{200} = 17\ \mathrm{mA}\).
Current through the \(800\ \Omega\) load \(= \frac{5.6}{800} = 7\ \mathrm{mA}\).
\(I_Z = 17 - 7 = 10\ \mathrm{mA}\).
Answer: (4) 10 mA
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