From the zener diode circuit shown in the figure, find (1) the output voltage (2) the voltage drop across series resistance and (3) the current through zener diode.

Explanation

From the figure
`R=5kOmega=5xx10^(3)Omega`
Input voltage =220V
Load resistance `R_(L)=20kOmega=20xx10^(3)Omega`
Zener voltage, `V_(z)=50V`
(1) Output voltage, `V_(z)=50V`
(2) Voltage drop across, series resistance,
R=input voltage `-V_(z)=220-50=170V`
(3) Load current,
`I_(L)=(V_(z))/(R_(L))=(50)/(20xx10^(3))2.5xx10^(-3)A`
Current through R,
`I=("Voltage drop acrossR")/(R)=(170)/(5xx10^(3))`
`=34xx10^(-3)A`
According to Kirchoff.s first `I=I_(L)+I_(Z)`
`:.` Zener current, `I_(Z)=I-I_(L)`
`=34xx10^(-3)-2.5xx10^(-3)` `=31.5xx10^(-3)A=31.5mA`