question_answer

The reverse breakdown voltage of a Zener diode is 5.6 V in the given circuit. The current \[{{I}_{Z}}\] through the Zener is :             [JEE Main 8-4-2019 Morning]

A) 7 mA               

B) 17 Ma

C) 10 mA

D) 15mA

Correct Answer: C

Solution :

\[9={{V}_{Z}}+{{V}_{{{R}_{1}}}}\] \[{{V}_{Z}}=5.6V\] \[{{V}_{{{R}_{1}}}}=9-5.6\] \[{{V}_{{{R}_{1}}}}=3.4\] \[{{I}_{{{R}_{1}}}}=\frac{{{V}_{{{R}_{1}}}}}{R}=\frac{3.4}{200}\] \[{{I}_{{{R}_{1}}}}=17mA\] \[{{V}_{Z}}={{V}_{{{R}_{2}}}}={{I}_{{{R}_{2}}}}({{R}_{2}})\] \[\frac{5.6}{800}={{I}_{R{{ & }_{2}}}}\] \[{{I}_{R{{ & }_{2}}}}=7mA\] \[{{I}_{Z}}=(17-7)mA\] \[=10mA\]