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Q.
In the circuit shown in figure, Zener diode is properly biased. Power dissipated in diode (in $mW$ ) is:
NTA AbhyasNTA Abhyas 2022
Solution:
Power can be calculated as
$P=VI_{z e n e r}P=4\times I_{z e n e r}$
Now lets find the current through Zener
Potential difference across $400ohm$ Connected in parallel with zener will be $4volt$
Hence, potential difference across another $400ohm$ Will be $12-4=8V$
So current in this resistor is $I_{1}=\frac{V}{R}=\frac{8}{400}=\frac{1}{50}$
And current in Resistance which in parallel with zener will be $I_{1}-I_{2}=\frac{V}{R}=\frac{4}{400}=\frac{1}{100}I_{2}=\frac{1}{50}-\frac{1}{100}=\frac{1}{100}$
Hence power will be $P=4\times \frac{1}{100}=\frac{1}{25}=40\times 10^{- 3}watt=40mW$
