Question

In the circuit shown in figure, Zener diode is properly biased. Power dissipated in diode (in $mW$ ) is:

Answer 1

Answer: 40

Power can be calculated as
$P=V{I}_{zener}P=4\times I_{zener}$
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 $P=4\times \frac{1}{100}=\frac{1}{25}=40\times 10^{-3}watt=40mW$