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For the circuit shown in Fig. find 1) the output voltage, 2) the voltage drop across series resistance, 3) the current through Zener diode |
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Answer» Solution :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` 
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