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

Explanation

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