TS EAMCET 10-Sep-2020 Shift 2 Solved Paper

The circuit given is as shownPotential drop across RL=VL=VZ=10V
Potential drop across 4kΩ resistance is
VR=60−10=50V
Current through 4kΩ resistance is I=‌

VR

R

=‌

50

4×103

=12.5×10−3A=12.5mA
Current through load resistance =IL
=‌

VL

RL

=‌

10

2×103

=5mA
Also, IZ=I−IL=12.5−5=7.5mA