Diagrams used:

Formulas used:

Observations:

Measurement of Active Power:

Serial No. V (Volts)

I (Amps)

W1 (Watts)

W2 (Watts)

Total Power (W1+W2) Watts

Part-1 231 0.9 180 160 360 Part-2 224 2.15 420 420 640

Part-3 226 0.9 -60 120 60

Measurement of Reactive Power:

V (Volts)

I (Amps)

Reactive Power (VAr)

231 0.9 0

224 2.15 0

226 0.92 311.76

Calculations:

Part 1: Star connected load

W1 = 180 Watt W2 = 180 Watt

Experimental Total Power = W1+ W2

= 360 Watt

Ө = Tan-1[ {(W1- W2)√3} / (W1 +W2)]

= 00

Power Factor (Cos Ө) = 1

Phase Current (I) = 0.9 Amp

Line Voltage (VL) = 231 V

Theoretically, Active Power = √3 * VR * I *Cos Ө

= √3*231*0.9*1

= 360.1 W

Active Power Theoretical = Active Power Experimental

Also total reactive power = √3 * (W2-W1) =0

=

Part 2: -Delta Connected Loads

W1 = 420 W W2 = 420 W

Experimental Total Active Power = W1+W2

= 840 W

Experimental Total Reactive Power = √3 * (W2-W1)

= 0 VAr

Now Ө = Tan-1* {√3*(W2-W1)} / (W1+W2) ]

= 00

Power Factor (Cos Ө) = 1

Line Current (IL) = 2.15 Amp Line Voltage (VL) = 224 V

Theoretically, Total Active Power = √3 * VL * IL

= 834 W

Power Dissipated Theoretically = Active Power Experimental

Part 3:- Delta connected load of Lamp & Capacitor combination

a) W1 = -60W W2 = 120 W

Experimental, Total Active Power = W1+W2

= 60 W

Experimental, Total Reactive Power = √3 * (W2-W1)

= 311.76 VAr

Ө = Tan-1 * {√3(W2-W1)} / (W1+W2) ]

= 79.110

Power Factor (Cos Ө) = 0.189

Total Active Power theoretical = √3 * VL * IL * Cos Ө

Here, Line Voltage (VL) = 226 V Line Current (IL) = 0.92 Amp

Active Power theoretical = √3 * 226 * 0.92 * 0.189

= 68.06 W

Active Power Theoretical = Active Power Experimental

b) Total Reactive Power = √3 * Wattmeter Reading

= 311.76 VAr

Reactive Power (In part (a)) = Reactive Power (In part (b))

Results:

1) In case of Star connected load with no Inductor/Capacitor

a) Power Factor = 1

b) Total Active Power = 360 W

c) Total Reactive Power = o VAr

2) In case of a Delta connected load with no Capacitor/ Inductor

a) Power Factor = 1

b) Total Active Power = 840 W

c) Total Reactive Power = 0 VAr

3) In case of a Delta connected load with Capacitor

a) Power Factor = 0.189

b) Total Active Power = 60 W

c) Total Reactive Power = 311.76 VAr

Conclusions:

1) In case of a Star connected load without any capacitor or inductor,

Power Factor (P.F.) =

2) In case of balanced load, Total Active and Reactive Power can be given by

only two wattmeter, where

Total Active Power = √3 * Difference of two Wattmeter reading

3) In case of balance load, Total Reactive Power can be given by only one

Wattmeter,

Total Reactive Power = √3 * Wattmeter Reading

Precaution:

1) We must wear shoes while doing experiment.

2) No loose connections should be there in the circuit.

3) Circuit should be checked before taking the reading.

4) Voltmeter reading should be taken carefully.

5) Multiplying factor should be taken into account, whenever necessary.

6) Wattmeter reading should be added or subtracted with their sign taken

into account.

ESO210 Lab Report by:

AKSHAY AJMERA

Y8055

LAB PARTNERS : KAMAL SAHNI, SANDEEP SINGH ADHIKARI