Kerala Engineering Entrance (KEAM) 2011 Question (MCQ) on Zener Diode

“Everything should be made as simple as possible, but not simpler”

– Albert Einstein

The following multiple choice question on zener diode was included in Kerala Engineering Entrance (KEAM) 2011 question paper. If you identify that R₁ is the current limiting resistor and R₂ is the load resistor in the simple shunt voltage regulator making use of a zener diode, you won’t have any confusion. Here is the question:

In the circuit given the current through the zener diode is

(a) 10 mA

(b) 6.67 mA

(c) 5 mA

(d) 3.33 mA

(e) 0 mA

Since the zener breakdown voltage (across the zener diode) is 10 V, the voltage across the resistor R₂ is 10 V. The current through R₂ is 10 V/1500 Ω = 0.00667 A = 6.67 mA.

The voltage drop across R₁ is 15 V – 10 V = 5 V.

Therefore, the current through R₁ is 5 V/500 Ω = 0.01 A = 10 mA

The above current gets divided between the resistor R₂ and the zener diode. Therefore, the current through the zener diode is 10 mA – 6.67 mA = 3.33 mA.

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