MCQ
In the circuit shown below, maximum zener diode current will be $.....mA$
- ✓$9$
- B$90$
- C$95$
- D$45$
✓
Answer
Correct option: A.
$9$
a
$I =\frac{(120-60)\,V }{4000\,\Omega}=0.015\,A$
$I =\frac{(120-60)\,V }{4000\,\Omega}=0.015\,A$
Thus $\quad I _{2}= I - I _{ L }$
$=0.015-0.006=0.009 A =9\,mA$
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