JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Semiconductors

The reverse breakdown voltage of a Zener diode is 5.6 V in the given circuit .

The current

I_{z}

through the Zener is:

(A)10 mA✓ Correct
(B)15 mA
(C)7 mA
(D)17 mA

Answer: (A)

Solution

Read the two resistor currents separately and then apply the junction rule at the Zener. Across the series resistor the drop is the supply minus the Zener voltage,

9 - 5.6 = 3.4

V, so the current there is

\frac{3.4}{200} = 17

mA. The parallel resistor sits across the Zener and carries

\frac{5.6}{800} = 7

mA. By Kirchhoff's current law the Zener takes the difference,

I_{z} = 17 - 7 = 10

mA.

JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Single correct option

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