For the circuit shown in figure, Find
(1) the output voltage ,
(2) the voltage drop across series resistance ,
(3) the current through Zener diode.
.

Explanation

From the figure `R = 5k Omega = 5 xx 10^(3) Omega` ,
input voltage `V_("in") = 12 V` , zener voltage, `V_(z) = 50 V`
(1) Output voltage `V_(z) = 50 V`
(2) Voltage drop across series resistance `R = V_("in") - V_(z) = 120 -50 = 70 V`
(3) Load current `I_(L)=(V_(z))/(R_(L))=(50)/(10 xx 10^(3)) = 5 xx 10^(-3) A`
Current through `R = i= (V_("in") -V_(z))/( R)`
`= (70)/(5 xx 10^(3)) = 14 xx 10^(-3) A`
According to Kirchoff's first law `I = I_(L) + I_(z)`
`:.` Zener current
`I_(z)=I-I_(L) =14 xx10^(-3) -5 xx 10^(-3) = 9 xx 10^(-3) =9 mA`.