Physics - Semiconductors Question with Solution | TestHub

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Current through load resistor is $I_L=\frac{60 \mathrm{V}}{10000 \mathrm{\Omega }}=0.006 \mathrm{A}$ and current through $4000 \mathrm{\Omega }$ resistor is $I=\frac{\left(120-60\right) \mathrm{V}}{4000 \mathrm{\Omega }}=0.015 \mathrm{A}$

Thus, maximum current through Zener diode is $I_Z=I-I_L$

$=0.015-0.006=0.009 \mathrm{A}=9 \mathrm{mA}$