JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Semiconductors

For the given circuit, the power across zener diode is ______ mW.

Answer: 120

Solution

The zener holds

10

V across the load, so the

5 kΩ

load carries

I_{L} = \frac{10}{5 \times 1 0 ^{3}} = 2 mA

The

1 kΩ

series resistor drops the rest of the supply voltage, so the current drawn from the source is

I = \frac{24 - 10}{1 \times 1 0 ^{3}} = 14 mA

The zener carries the difference,

I_{Z} = 14 - 2 = 12

mA, and the power dissipated in it is

P = V_{Z} I_{Z} = 10 \times (12 \times 1 0^{- 3}) = 120 mW

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Integer answer