In the circuit, Fig.he current through the zener diode is

Options

A. 10 mA
B. 6.67 mA
C. 5mA
D. 3.33 mA

Answer

D

Explanation

voltage drop across `R_(2) is
V_(2)=V_(Z)=10V`
The current through `R_(2)` is
`I_(2)=V_(2)/R_(2)=(10V)/(1500Omega)=6.67xx10^(-3)A=6.67mA`
Voltage drop across `R_(1)` is,
`V_(1)=15V-V_(2)=15V-10V=5V`
Current through `R_(1)` is,
`I_(1)=V_(1)/R_(1)=(5V)/(500Omega)=10xx10^(-3)A=10mA`
Current through the zener diode,
`I_(3)=I_(1)-I_(2)=10-6.67=3.33 mA`