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Q.
In the figure, the reverse breakdown voltage of a Zener diode is $5.6V$ , then the current $I_{z}$ through the diode is
NTA AbhyasNTA Abhyas 2022
Solution:
For zener break down potential difference across $800\,\Omega$ resistor will be $5.6\,V$
$V_{z}=5.6\,V$
$i_{2}=\frac{V_{2}}{800}=\frac{5 . 6}{800}=7\,mA$
$ΔV$ across $200\,\Omega=9-5.6=3.4\,V$
$i_{1}=\frac{3 . 4}{200}=17\,mA$
$i_{1}=i_{2}+i_{z}$
$i_{z}=17\,mA-7\,mA=10\,mA$
