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In the Fig. find out the current passing through `R_(L)` and zener diode. |
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Answer» Here, `V_(Z)=5V,` Voltage drop across R, `V_(R)= "input voltage"- V_(Z)=10-5=5V` Current though R, `I=V_(R)/R=(5V)/(80Omega)` `=6.25xx10^(-2)A.` Voltage drop across `R_(L) =V_(z)=5v`. Current through `R_(L)` is, `I_(L)=5/100=5xx10^(-2)A` Since `I=I_(L)+I_(z)`. so `I_(z)=I-I_(L).` Current through zener diode is, `I_(z)=I-I_(L)=6.25xx10^(-2)-5xx10^(-2)` `=1.25xx10^(-2)A` |
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