1.

In the Fig. find out the current passing through `R_(L)` and zener diode.

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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