Objectives:

Components & Specification:

1. 3-phase, AC supply at f = 50 Hz.2. Potentiometer with max rating: 1000 ohm, 500 W.3. A resister with RL = 10 ohm, P = 10W.4. Capacitor with C = 15 µF.5. Inductor with L = 0.3 1h and Ir = 3.5 ohm.6. Ammeter range: (a) 0-0.5A

(b) 0-1 A7. Voltmeter range: (a) 0-150 V

(b) 0-300 V

(c) 0-600 V

Circuit diagram

Calculation:

R = 150 ohm

L = 0.3 H

C = 15 µF

ZR = 150 ohm

ZL = (3.5+j94.2) ohm

ZC = (-j212.3) ohm

Calculation of current using Thevenin Method :

Calculation of Z Thevenin (ZT):

1/ZT = 1/ZR + 1/ZC + 1/ZL + 1/ZR

So, ZT = (61.52 + j26.4) ohm

Calculation of V Thevenin (VT):

VT = VAB.(ZC || Zr || Zl)/(Zc+Zr+Zl) + VA’B’/(1+( ZC + (Zr || Zr || Zr) ).

VAB = 69 V & VA’B’ = (-35 - j60.62) V.

So, VT = (28.29 – j12.14) + (21.92 – j2.63) V

VT = 50.21 – j14.77 V

Calculating I through RL:

ZL = 10 ohm

So, IL = VT / (ZT + ZL)

= 0.55 - j0.4 A

= 0.68 Ampere

By KVL Method :

Vab – Zri1 – (i1 –i2)Zl = 0

(i2- i1)Zl + (i2 –i3)Zl’ = 0

(I3 –i2)Zl’ +(i3 –i4)Zr = 0

(i4-i3)Zr +i4*Zc + Va’b’ = 0

I = i2-i3

Solving the above Equations we get

I = I2 –i3 = 0.696 – 0.153j

|I| = |i2 –i3| = 0.71 Ampere

RESULTS:

|VT| (Experimentally) = 51 V

|IL| (Experimentally) = 0.76 A

|ZT| = |VI/I| = 67.1 ohm

|VT| (Estimated) = 52.33 V

|ZT| (Estimated) = 63.33 ohm

|IL| (Estimated by Thevenin) = 0.68 A

|IL| (Estimated by KVL) = 0.71 A

Conclusion:

As you can see results theoretically obtained matches with the experimentalones. The variation may be because of the following reasons:

(a) Change in resistance due to heating.(b) Non-ideal behavior of components.

Since, both these factors increase resistance and so, decreased current. Hence our expresults has increased from the estimated one.

Precaution:

(a) Don’t switch on the circuit until it’s complete and verified by instructor.(b) While using multimeter to measure resistance, we should switch off the

circuit.(c) We should wear shoes while doing experiment.