Question

From the zener diode circuit shown in figure, the current through the zener diode is:

• A. 34 mA
• B. 31.5 mA
• C. 36.5 mA
• D. 2.5 mA

Answer 1

The correct option is B 31.5 mA


Voltage drop across $R_L$ is 50V. Voltage across $5k\Omega$ is (220-50)V. Apply Ohm's law.
$V=IR170=1\times 5000$
$I=34mA$ across $5k\Omega$
$I_1$ across $R_L$
$50=I_1\times 20000I_1=2.5mA$
Apply junction law at point P wet get
$I_{zener}=I-I_1{I}_{zener}=34-2.5=31.5mA$