In the following circuit, the current through the Zener diode is (a) 5.00 mA (b) 6.25 mA (c) 11.25 mA (d) 16.25 mA

Answer: (b). Zener holds the node at 9 V, so R₁ has 18 - 9 = 9 V across it. I = 9/800 = 11.25 mA Current in R₂: I₂ = 9/1800 = 5 mA IZ = I - I₂ = 11.25 - 5 = 6.25 mA So the answer is (b) 6.25 mA.

In the following circuit, the current through the Zener diode is | MHT CET 2026 (19 May, Shift 1)

Q9MHT CET 2026 · 19 MAY · SHIFT 1Medium

Special Purpose Diodes

In the following circuit, the current through the Zener diode is

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Answer: (b)

Zener holds the node at $9$ V, so $R_1$ has $18 - 9 = 9$ V across it.
$I = \frac{9}{800} = 11.25$ mA
Current in $R_2$ : $I_2 = \frac{9}{1800} = 5$ mA
$I_Z = I - I_2 = 11.25 - 5 = 6.25$ mA
So the answer is (b) $6.25$ mA.

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