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Single Phase Transformers

December 8, 2016
Published By : Pratik Kataria
Categorised in:

(1) Induced voltage:

$$ E = 4.44 f ϕ_m N $$

(2) Ratios:

Voltage ratio: $ V_1/V_2 $

Transformation ratio: $K = V_2/V_1 = E_2/E_1 = N_2 / N_1 $

Turns ratio: $N_1/N_2$

$$ I_1/I_2 = V_2 / V_1 $$

(3) Rating of Transformer:
$$ I_1 = {kVA 10^3}/{V_1} $$

(4) Copper Loss:

$$ P_{cu} = I_1_^2 R_1 + I_2_^2 R_2 $$

$$ P_{cu(HL)} = {(1/2)}^2 P_{cu(FL)} $$

(5) Iron Loss:

$P_i$ = Hysteresis Loss + eddy current loss

Hysteresis loss: $P_H = K_H B_m_^{1.67} f V$

Eddy current loss = $ I^2 $ x r

(6) Resistance transfered to primary side:

$$ R_{1T} = R_1 + R’_2 $$

where, $R’_2 = R_2/K^2$

(7) Resistance transfered to secondary side:

$$ R_{2T} = K^2 R_1 + R_2 $$

(8) Impedance:

$$ Z_1 = R_1 + j X_1 $$

where subscript 1 denotes primary, similarly 2 will denote secondary

(9) Efficiency:

$$η = {\text"Output Power"}/{\text"Input Power"} $$

$$ η= {\text"Output Power"}/{\text"Output Power + Losses"} $$

$$ η = {x kVA cosϕ_2 1000}/{(x kVA cosϕ_2 1000 ) + P_i + x^2 P_{cu}} $$

where, x = 1 for full load, 0.5 for half load, etc.


$kVA_{(load)} = kVA_{(max)} √{P_i/P_{cu}}$

(10) Voltage Regulation of Transformer:

$%Regulation = {E_2 – V_2}/{E_2} x 100 $

No load voltage = $E_2$ volts