From the given load conditions, the load currents are as follows :
IL1 = 1/100 = 10 mA,
IL2 = 2/50 = 40 mA,
IL3 = 1.6/80 = 20 mA
For economising the number of components, the internal resistance of 200 Ω can be used as the series dropping resistance. The suitable circuit and the ground connection are shown in Fig.
Applying Kirchhoff’s laws to the closed circuit ABCDA, we have
V − 200 × 100 × 10−3 − 100 − 80 = 0 or V = 200 V
I1 = 100 − 10 = 90 mA
∴ R1 = 100 V/90mA = 1.11 kΩ
I3 = 100 − 20 = 80 mA;
voltage drop across R3 = − 50 − (− 80) = 30 V
∴ R3 = 30 V/80 mA = 375 Ω
I2 + 40 = 80
∴ I2 = 40 mA; R2 = 50 V/40 mA = 1.25kΩ