To find out turn ratio and flux density of a single-phase transformer.

Experiment No.:01

NAME OF THE EXPERIMENT: –

                                To find out the turn ratio and flux density of a single-phase transformer.

Objective of the Experiment:

To be written by student.

APPARATUS REQUIRED: –

Sl. No.NameSpecificationQuantity
1Single Phase Auto transformer5 KVA, (0-270) V, 10A1 Nos.
2Three winding transformers1 KVA, iron core 1 Nos.
3Voltmeter(0-250) V, MI1 Nos.
4Voltmeter(0-150) V, MI1 Nos.
5Voltmeter(0-15) V, MI1 Nos.
6Connecting WiresPVC Insulated copperAs per requirement.
Table No. 1.1

CIRCUIT DIAGRAM: –

fig.1.1

1: Single phase Auto-transformer.

2: Three winding transformer.

3: Single-phase AC supply.

THEORY: –

                 The turns ratio of a single-phase transformer can be identified as the ratio of number of turns of the secondary coil (NS) to the number turns of the primary coil (NP). 

\[The \;turns\; ratio\; is\; defined\; as\; k= \dfrac{N_S}{N_P}\]

As the frequency of the alternating current remains the same while flowing through the transformer, then the voltage per unit turn remains the same for both primary and secondary coils, 

so that

\[\dfrac{V_P}{N_P}=\dfrac{V_S}{N_S}\] \[=\dfrac{N_S}{N_P} =\dfrac{ V_S}{ V_P}\]
\[Thus, \;the \;turns\; ratio\] \[ k = \dfrac{N_S}{N_P} = \dfrac{V_S}{ V_P}\] \[where,\; V_S \;and\; and\; V_P \;are\; the\] \[corresponding\; voltage \;across\; the\] \[ secondary\; and\; primary\; coils\; respectively.\]

Further the power on both sides of the transformer remains the same, then the power across the primary (PP) side will be equal to the power across the secondary (PS) side.

\[i.e.,\; V_PI_P = V_SI_S\] \[=\dfrac{I_p}{I_S} = \dfrac{V_S}{V_P}\]
\[So, \;the \;turns\; ratio\; can \;be\; related\; as\] \[k = \dfrac{V_S}{V_P} = \dfrac{I_P}{I_S} = \dfrac{N_S}{N_P}\]

Where, Ip and IS are the respective currents in the primary and secondary coils.

From the expression of the induced voltage in the transformer E = 4.44* f* N*ɸm, the flux density (Bm) can be determined while the number of turns of primary coil (NP) or secondary coil (NS) and ɸm and cross-sectional area of the core are known.

To determine the number of turns of primary and secondary coils an additional 10 turns are wound on the core, which is termed as tertiary coil. With the help of tertiary coil the primary and secondary turns can be specified as stated below.

Let, 

VT is voltage across the tertiary coil,

VS is voltage across the secondary coil,

NT is number of turns of the tertiary coil,

NS is number of turns of the secondary coil,

Where NT, VT, and VS are known and the unknown quantity Ns can be calculated by the ratios

\[\dfrac{N_S}{N_T} = \dfrac{V_S}{ V_T}\] \[N_S = \dfrac{V_S}{ V_T}* N_T\] \[Similarly,\; by\; the\; expression\; given\; N_P \] \[\;can\; be\; calculated\;\] \[\dfrac{N_S }{N_P} = \dfrac{V_S}{ V_P}\] \[N_P = \dfrac{V_P}{V_S}*N_S\] \[After\; determining\; N_S \;and \;N_P \;the \;turns\] \[ ratio\; can \;be \;determined \;as\] \[k = \dfrac{N_S}{N_P} = \dfrac{V_S}{ V_P}\] \[Thus,\; to\; determine \;ɸ_m,\; we\; may\; use\; the\] \[expression \;as\] \[ɸ_m= \dfrac{(E_S \;or\; V_S)}{ (4.44*f*N_S)}\; and\] \[ the\; flux\; density \;B_m = \dfrac{ɸ_m}{(a*s)}\]

where a is the cross-sectional area of the core in m2 and s is the stacking factor = 0.9

Stacking Factor: It is the ratio of effective area to the overall cross-sectional area  of the core. Typically, its value lies between 09 to 0.95. 

PROCEDURES: –

  1. Connect all the instruments as per circuit diagram.
  2. Now slowly increase the input voltage from the auto-transformer to the circuit and take the readings of voltmeters.
  3. Calculate the unknown quantity as per given formulae.

PRECAUTIONS: –

  1. Before connecting to the supply check the auto transformer and voltmeter reading/dial is at zero position.
  2. Care should be taken for the connections to be tightened properly.

OBSERVATION TABLE: –

Sl. No.VP in volts.Vs in volts.VT in volts.NT in turnsNP in turnsNS in turnsTurns ratio
VS/VP		Turns ratio
NS/NP		ɸm in milli WbBmax in Wb/m2
1
2
3
4
5
6
7
Table No. 1.2

CONCLUSION:

To be written by student.

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