Question
In the circuit shown below, maximum zener diode current will be $.....mA$

Answer

$I =\frac{(120-60)\,V }{4000\,\Omega}=0.015\,A$

Thus $\quad I _{2}= I - I _{ L }$

$=0.015-0.006=0.009 A =9\,mA$

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