Experimental Setup:

Part – A:

Part – B:

Calculations:

R1 = 3.9 ohm R2 = 1.2 ohm

Cos θ1 = W1

V11 I1

= 0.115

Θ1 = 83.370

V1’ = √ [(V11 – I1R1Cos θ1)2 + (I1R1Sin θ1)2]

= 199. 94 V

α1 = Tan-1 I1R1 Sin θ1

V1 – I1R1 Cos θ1

= 0.1430

Im = I1Sin (θ1 +α1)

= 0.129 A

L1 =V1’

2πf Im = 4.94 H

M12 = V2

2πf Im

= 2.52 H

Cos θ2 = W2

V22 I2

= 0.076

Θ2 = 85.58 0

V2’ = √[ ( V22 – I2R2Cos θ2 )2 + (I2R2Sin θ2 )2 ]

= 101.9 V

α2 = Tan-1 I2R2Sin θ2

V2 – i2R2Cos θ2

= 0.169 0

Im = I2Sin (θ2 + α2)

= 0.254 A

L2 = 𝐕2’

2πf Im

= 1.278 H

M21 = 𝑉1

2πf Im

= 2.51 H

S No. (R in ohm)

Cos θ Theta V’ (In V) α Im (Amp) L (In H) M(In H)

1 (R1=3.9)

CosΘ1=0.115 Θ1 = 83.370

V1’=199.94 α1=0.1430 Im= 0.129

L1= 4.94 M12= 2.52

2 (R2=1.2)

CosΘ2=0.076 Θ2 = 85.580

V2’=101.90 α2=0.1690 Im= 0.254

L2= 1.28 M21= 2.51

M avg = (M12 +M21) /2

= 2.51 H

Coupling Factor (K) = [M avg / √(L1L2)] = 0.99

Conclusions:

The mutual inductance of coil 1 due to coil 2 is approximately same as that of

coil2 due to coil 1.

Coupling coefficient is coming out to be approximately 1.

Result:

1. Graph for V/I characteristic was linear.

2. L1 = 4.94 H L2 = 1.28 H

M21 = M12 = 2.51 H

Coupling Factor (K) = 0.99

3. A and C have the same polarity in the given circuit. In the 2nd case the bulb

glow more brightly as compared to the 1st case. As per the Dot convention

when the current enter the dotted terminals, flux from the two coils get

added and when the current enters from dot terminal in one coil and in

other coil not from dot terminal, the flux gets subtracted. Hence we see

change in brightness of the bulb during the experiment.

4. Equation: W1’/V1 = W2/V2

Precautions:

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) It should be known that circuit is operating in linear region of the

magnetization curve that is not magnetically saturated.

6) The test should be conducted at proper level of current and frequency for

two connected windings.

7) The effect of the power loss due to hysteresis and eddy current should be

suitable accounted for.

Questions And Answers

1. We choose the coil with higher voltage rating as coil 1 to ensure that B-H

curve does not lie in saturation region as flux is a function of operating

voltage (for determining voltage current characteristics ).

2. We plot the V/I characteristics for increasing value of applied voltage to

find out the voltage from where saturation region starts. So, below this

voltage we can perform our experiment to ensure that B-H curve is linear.

3. Leff = L1+L2+2M or Leff = L1+L2-2M

Impedance = jw(L1+L2(+-)2M)

For series addition, impedance = jw(L1+L2+2M). So, impedance is higher,

hence less current flows, so bulb glows dim. But for opposite connections

impedance = L1+L2-2M , So current increases and bulb glows brightly.

4. I1 is maintained same so that applied voltage in coil 2 remains equal to V2 so

that flux level remains same in both the cases and also to ensure that setup

have not moved in saturation region of B-H curve.

5. When magnetic core is present, because of its permeability, it offers some

reluctance and So, flux would be surely limited due to flux leakage.

In absence of magnetic core, there will be no flux leakage and thus W12 and

W21 will be zero, otherwise we have finite value of current and finite

wattmeter readings.

ESO210 Lab Report by:

AKSHAY AJMERA

Y8055

LAB PARTNERS : KAMAL SAHNI (Y8232) , SANDEEP SINGH ADHIKARI (Y8443)