Semiconductor Electronics Part 1

Given, Zener diode voltage, Vz=30V
Supply voltage, V=90V
According to figure, Let I be supply current, then
I2 be Zener current (=I−I1)
and I1 be current through 5kΩ.
∴‌‌V1=V2=30V
By Ohm's law,
‌‌V1=V2=I1R
⇒‌‌I1=V2∕R=‌

30

5000

=6mA
If V′ be the voltage across 4kΩ resistor.
∴‌‌V′=90−30=60V
and I=‌

60

4000

=15mA
Therefore, I2=I−I1
=15−6=9mA