In the given circuit the current through Zener Diode is close to :
A
4.0mA
B
6.7mA
C
0.0mA
D
6.0mA
Text Solution
Verified by Experts
The correct Answer is:
C
Voltage across `R_2`=10 V So, `i(R_2)=10/1500=2/3xx10^(-2)`A So, current through `500 Omega =(4/3xx10^(-3)+ i_z)` So , voltage across `500 Omega =(20/3+500 i_z)` So, `12-10=20/3 + 50 i_z` This gives a negative value of `i_z`. So, `i_z=0`
