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Question

In the circuit shown, the Zener diode is an ideal one with breakdown voltage of 5.0 volt. The values of the resistance are RS = 10 kΩ and RL = 1 kΩ. The current through the resistances, when the supply voltage is 11.0 V, is

Options

A.0.6 mA through RS and 5.0 mA through RL
B.1.0 mA through RS and 1.0 mA through RL
C.1.1 mA through RS and no current through RL
D.no current through RS and 11 mA through RL

Solution

[Key Concept: A Zener diode conducts in reverse only if the voltage across it, with the diode removed, exceeds its breakdown voltage.]

Step 1: Remove the Zener: RS and RL in series carry I = 11 V/(10 kΩ + 1 kΩ) = 1.0 mA.

Step 2: The voltage across RL (and the Zener) is then 1.0 mA × 1 kΩ = 1.0 V, which is below the 5.0 V breakdown voltage, so the Zener stays off.

Step 3: Hence both resistances carry 1.0 mA. A 5 V clamp would need 5 mA in RL but only 0.6 mA could flow in RS, which is impossible.

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