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Q. For the given circuit the power dissipated by Zener diode is
Question

NTA AbhyasNTA Abhyas 2020

Solution:

Solution
Current flowing through load resistor is given as $i=\frac{16}{1000 + 1200}=7.2mA$
Voltage drop across the load resistor, $V_{a b}=iR_{L}=7.2\times 10^{- 3}\times 1200=8.72V$ .
As $V_{a b} < 10V$
So, breakdown of zener diode does not occur and no current flows through it.
So, $I_{z}=0$ and $P_{z}=I_{Z}V_{z}=0W$