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 \[{{\operatorname{I}}_{zener}}=I-{{I}_{l}}\] \[{{\operatorname{I}}_{zener}}=34-2.5=31.5\operatorname{mA}\]You need to login to perform this action.
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