Question:

The reverse breakdown voltage of a Zener diode is 5.6 V in the given circuit. The current IZ through the Zener is?

7 mA

17 mA

15 mA

10 mA

Correct option is C.
10 mA
9=VZ+VR1
VZ=5.6 V
VR1=9−5.6
VR1=3.4
IR1=VR1/R=3.4/200
IR1=17 mA
VZ=VR2=IR2(R2)
5.6/800=IR2
IR2=7 mA
IZ=(17−7)mA=10 mA.