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Using minimum number of components, design a voltage divider which can deliver 1 W at 100 V, 2 W at − 50 V and 1.6 W at − 80 V. The voltage source has an internal resistance of 200 Ω and supplies a current of 100 mA. What is the open-circuit voltage of the voltage source ? All resistances are in ohm.

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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

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