Question:
The zener diode has a \(V_z = 30\ \mathrm{V}\). The current passing through the diode for the following circuit is ________ mA.
The zener diode has a \(V_z = 30\ \mathrm{V}\). The current passing through the diode for the following circuit is ________ mA.
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Check whether the zener breaks down (voltage across the 5 k resistor without it would exceed 30 V). If so the parallel branch is held at 30 V.
Updated On: Nov 4, 2026
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Correct Answer: 9
Solution and Explanation
Without the zener, the voltage across the \(5\ \mathrm{k\Omega}\) resistor would be \(90 \times \frac{5}{4+5} = 50\ \mathrm{V} > 30\ \mathrm{V}\), so the zener is in breakdown and holds the parallel branch at \(30\ \mathrm{V}\).
Current through the \(4\ \mathrm{k\Omega}\) resistor: \(I = \frac{90 - 30}{4\ \mathrm{k\Omega}} = 15\ \mathrm{mA}\).
Current through the \(5\ \mathrm{k\Omega}\) load: \(I_L = \frac{30}{5\ \mathrm{k\Omega}} = 6\ \mathrm{mA}\).
Current through the zener: \(I_z = 15 - 6 = 9\ \mathrm{mA}\).
Answer: 9
Current through the \(4\ \mathrm{k\Omega}\) resistor: \(I = \frac{90 - 30}{4\ \mathrm{k\Omega}} = 15\ \mathrm{mA}\).
Current through the \(5\ \mathrm{k\Omega}\) load: \(I_L = \frac{30}{5\ \mathrm{k\Omega}} = 6\ \mathrm{mA}\).
Current through the zener: \(I_z = 15 - 6 = 9\ \mathrm{mA}\).
Answer: 9
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