1.

In the circuit given, the current through the Zener diode is

Answer»

10 mA
6.67 mA
5 mA
3.33 mA

Solution :The VOLTAGE drop across `R_(2)` is
`V_(R_(2))=V_(Z )=10V`
The current through `R_(2)` is
`I_(R_(2))=(V_(R_(2)))/(R_(2))=(10V)/(1500Omega)`
`=0.667xx10^(-2)A`
`=6.67xx10^(-3)A=6.67mA`
The voltage drop across `R_(1)` is `V_(R_(1))=15V-V_(R_(2))=15V-10V=5V`

The current through `R_(1)` is
`I_(R_(1))=(V_(R_(1)))/(R_(1))=(5)/(500OMEGA)=10^(-2)A=10xx10^(-3)A=10mA`
The currentthrough the zener diode is
`I_(Z)=I_(R_(1))-I_(R_(2))=(10.6.67)mA=3.33 mA`


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