JEE Main · 2024 · 29 Jan 2024 · Shift 1 · Current Electricity

In the given circuit, the breakdown voltage of the Zener diode is 3.0 V. What is the value of

I_{z}

Answer: (B)

Solution

The reverse-biased Zener conducts, clamping the voltage at its cathode node to

V_{Z} = 3 V

.

The branch supplying current

I

runs from the

10 V

node to this clamped node through a

1 k Ω

resistor:

I = \frac{10 - 3}{1000} = 7 mA

A second,

2 k Ω

resistor also connects to the clamped node, carrying

I_{1} = \frac{3}{2000} = 1.5 mA

By Kirchhoff's current law at that node, the Zener current is

I_{Z} = I - I_{1} = 7 - 1.5 = 5.5 mA

Key confirmed: B.

JEE Main · 2024 · 29 Jan 2024 · Shift 1 · Single correct option