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MHT CET 2026 (15 May, Shift 2) · Special Purpose Diodes

Special Purpose Diodes question from MHT CET 2026 (15 May, Shift 2)

Q10MHT CET 2026 · 15 MAY · SHIFT 2Medium

Special Purpose Diodes

The Zener diode has reverse breakdown voltage $V_Z$ . The current through the load resistance $R_L$ is $I_L$ . The current through Zener diode is

( $V_0$ = Input voltage, $R_S$ = Series resistance)

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Answer: (d)

Output is held at $V_Z$ , so the drop across $R_S$ is $V_0 - V_Z$ .
Current through $R_S$ : $I = \frac{V_0 - V_Z}{R_S}$
This splits into Zener and load: $I = I_Z + I_L$
$I_Z = \frac{V_0 - V_Z}{R_S} - I_L$ , so the answer is (d).

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