A zener diode, having breakdown voltage equal to `15 V` is used in a voltage regulator circuit shown in the figure. The current through the diode is

Options

A. 5 mA
B. 10 mA
C. 15 mA
D. 20 mA

Answer

A

Explanation

The voltage drop across `1kOmega=V_(Z)=15V`
The current through `1kOmega` is
`I.=(15V)/(1xx10^(3)Omega)=15xx10^(-3)=A=15mA`
`I.=(15V)/(1xx10^(3)Omega)=15xx10^(-3)A=15mA`
The voltage drop across `250Omega=20V-15V=5V`
The currect through `250Omega` is
`I=(5V)/(250Omega)=0.02A=20mA`
The current through the zener diode is
`I_(Z)=I-I.=(20-15)mA=5mA`