question_answer

From the zener diode circuit shown in figure, the current through the zener is:

A) 34mA                           

B) 31.5mA

C) 36.5mA            

D) 2.5mA

Correct Answer: B

Solution :

Voltage drop across \[{{\operatorname{R}}_{L}}\]is 50V.  Voltage across \[5k\Omega \,\,is\,\,(220-50)V\]apply ohm?s law. V=IR \[170=1\times 5000\] \[1=34mAacross\,\,5k\Omega \] \[{{\operatorname{I}}_{1}} across {{R}_{L}}\] \[50={{I}_{l}}\times 20,000\] \[{{\operatorname{I}}_{l}}=2.5mA\] apply junction law at point P wet get \[{{\operatorname{I}}_{zener}}=I-{{I}_{l}}\] \[{{\operatorname{I}}_{zener}}=34-2.5=31.5\operatorname{mA}\]