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14.Semiconductor Electronics
medium
In the circuit shown below, maximum zener diode current will be $.....mA$
A
$9$
B
$90$
C
$95$
D
$45$
(JEE MAIN-2022)
Solution
$I =\frac{(120-60)\,V }{4000\,\Omega}=0.015\,A$
Thus $\quad I _{2}= I – I _{ L }$
$=0.015-0.006=0.009 A =9\,mA$
Standard 12
Physics
