\[\begin{aligned}
E_1 & = 3000~\mathrm{V} \qquad E_2 = 200~\mathrm{V} \qquad f =
50~\mathrm{Hz}\\
A & = 150~\mathrm{cm}^2=150 \times
10^{-4}~\mathrm{m}^2\qquad N_2 = 80
\end{aligned}\]
Given data:
Core Losses & No-load operation
Problem-4
A \(50 ~\mathrm{kVA}, 2300 / 230~
\mathrm{V}, 50 \mathrm{~Hz}\) transformer takes \(200 ~\mathrm{W}\) and \(0.3~ \mathrm{A}\) at no load, when \(2300 ~\mathrm{V}\) are applied to the
high-voltage side. The primary resistance is \(3.5~ \Omega\). Determine: core loss, and
no-load pf.
\[
\begin{aligned}
W_i & =200 \mathrm{~W} \\
I_0 & =0.3 \mathrm{~A} \\
V_1 & =2300 \mathrm{~V} \\
R_1 & =3.5 \Omega
\end{aligned}
\]
(i) Core loss \[\begin{aligned}
W_i & =V_1 I_0 \cos \phi_0 \\
\Rightarrow~200 & =2300 \times 0.3 \times \cos \phi_0 \\
\Rightarrow~\cos \phi_0 & =0.29(\text { lagging) }
\end{aligned}\]
\[\begin{aligned}
\text{Copper loss in primary}~ &=I_0^2 R_1=(0.3)^2
\times 3.5=0.315 \mathrm{~W}\\
\text{Core loss}~&=\text { Input power }- \text { Copper
loss }\\
&=200-0.315=199.685 \mathrm{~W}
\end{aligned}\]
Problem-5
A 230/110 V, single-phase transformer takes an input of 350 VA at
no load and at rated voltage. The core loss is \(110 \mathrm{~W}\). Find (i) no-load power
factor; (ii) the iron loss component of no-load current, and (iii)
magnetizing component of no-load current.
\[\begin{aligned}
S & =350 \mathrm{VA} \qquad
W_i =110 \mathrm{~W} \qquad
E_1 =230 \mathrm{~V}
\end{aligned}\]
\[\begin{aligned}
S & =V_1 I_0=E_1 I_0 \\
\Rightarrow~350 & =230 \times I_0 \\
\Rightarrow~I_0 & =1.52 \mathrm{~A} \\
W_i & =V_1 I_0 \cos \phi_0 \\
\Rightarrow~110 & =350 \cos \phi_0 \\
\Rightarrow~\cos \phi_0 & =0.314
\end{aligned}\]
Given data :
\[I_\omega=I_0 \cos \phi_0=1.52 \times 0.314=0.48
\mathrm{~A}\]
\[I_\mu=I_0 \sin \phi_0=1.52 \times \sin
\left(\cos ^{-1} 0.314\right)=1.44 \mathrm{~A}\]
(ii) Iron loss component of no-load current
Transformer on Load, Copper Loss & Equivalent Parameters
Problem-6
\[
\begin{aligned}
E_1 & =6600 \mathrm{~V} \\
E_2 & =400 \mathrm{~V} \\
R_1 & =2.5 \Omega \\
X_1 & =3.9 \Omega \\
R_2 & =0.01 \Omega \\
X_2 & =0.025 \Omega
\end{aligned}
\]
\[K =\dfrac{E_2}{E_1}=\frac{400}{6600}=0.06\]
Given data:
\[R_{01}=R_1+\frac{R_2}{K^2}=2.5+\frac{0.01}{(0.06)^2}=5.28
\Omega\]
(i) Equivalent resistance referred to primary
\[X_{01}=X_1+\frac{X_2}{K^2}=3.9+\frac{0.025}{(0.06)^2}=10.84
\Omega\]
\[X_{02}=K^2 X_{01}=(0.06)^2 \times 10.84=0.04
\Omega\]
\[R_{02}=K^2 R_{01}=(0.06)^2 \times 5.28=0.02
\Omega\]
(ii) Equivalent reactance referred to primary
Problem-7
A \(30~ \mathrm{kVA}, 2400 / 120
\mathrm{~V}, 50 \mathrm{~Hz}\) transformer has high-voltage
winding resistance of \(0.1 \Omega\)
and leakage reactance of \(0.22
\Omega\). The low voltage winding resistance is \(0.035 \Omega\) and leakage reactance is
\(0.012 \Omega\). Calculate equivalent
resistance as referred to primary and secondary, equivalent reactance as
referred to primary and secondary, equivalent impedance as referred to
primary and secondary, copper loss at full load and at \(75 \%\) of full load.
\[
\begin{aligned}
\mathrm{kVA} \text { rating } &=30 \mathrm{kVA} \\
E_1 &=2400 \mathrm{~V} \\
E_2 &=120 \mathrm{~V} \\
R_1 &=0.1 \Omega \\
X_1 &=0.22 \Omega \\
R_2 &=0.035 \Omega\\
X_1 &= 0.012 \Omega
\end{aligned}
\]
\[K =\dfrac{E_2}{E_1}=\frac{120}{2400}=0.05\]
Given data:
\[\begin{aligned}
& R_{01}=R_1+\frac{R_2}{K^2}=0.1+\frac{0.035}{(0.05)^2}=14.1
\Omega \\
& R_{02}=K^2 R_{01}=(0.05)^2 \times 14.1=0.035 \Omega
\end{aligned}\]
\[\begin{aligned}
& Z_{01}=\sqrt{R_{01}^2+X_{01}^2}=\sqrt{(14.1)^2+(5.02)^2}=14.97
\Omega \\
& Z_{02}=K^2 Z_{01}=(0.05)^2 \times 14.97=0.037 \Omega
\end{aligned}\]
\[\begin{aligned}
& X_{01}=X_1+\frac{X_2}{K^2}=0.22+\frac{0.012}{(0.05)^2}=5.02
\Omega \\
& X_{02}=K^2 X_{01}=(0.05)^2 \times 5.02=0.013 \Omega
\end{aligned}\]
(i) Equivalent resistance as referred to primary and secondary
\[E_2 \simeq V_2=120
\mathrm{~V}\]
of full load (iv) Copper loss at full load and at
\[\begin{aligned}
I_2 & =\dfrac{\mathrm{kVA} \text { rating } \times 1000}{V_2}\\
&=\dfrac{30 \times 1000}{120}=250 \mathrm{~A}
\end{aligned}\]
Copper loss
\[\begin{aligned}
W_{\mathrm{Cu}} & =I_2^2 R_{02}=(250)^2 \times 0.035=2.18
\mathrm{~kW} \\
\text{At 75\% of F.L.} &
=x^2 W_{\mathrm{Cu}}=(0.75)^2 \times 2.18=1.23 \mathrm{~kW}
\end{aligned}\]
Problem-8
A single-phase, 440/220 V, \(10
~\mathrm{kVA}, 50 \mathrm{~Hz}\) transformer has a resistance of
\(0.2 \Omega\) and reactance of \(0.6 \Omega\) on H.V. side. The
corresponding values of L.V. side are \(0.04
~\Omega\) and \(0.14~ \Omega\).
Calculate the percentage regulation on full load for (i) 0.8 lagging pf
(ii) 0.8 leading pf, (iii) unity pf.
\[\begin{aligned}
\mathrm{kVA} \text { rating } & =10 \mathrm{~kVA} \quad
E_2 =220 \mathrm{~V} \quad
E_1=440 \mathrm{~V} \\
R_2 & =0.04 \Omega \quad
R_1=0.2 \Omega \quad
X_2=0.14 \Omega \quad
X_1=0.6 \Omega \\
\end{aligned}\]
Given data:
\[\begin{aligned}
V_2 & \approx E_2=220 \mathrm{~V}\\
I_2 & =\frac{\mathrm{kVA} \text { rating } \times
1000}{V_2}=\frac{10 \times 1000}{220}=45.45 \mathrm{~A} \\
K & =\frac{E_2}{E_1}=\frac{220}{440}=0.5
\end{aligned}\]
\[\begin{aligned}
& R_{02}=R_2+K^2 R_1=0.04+(0.05)^2 \times 0.2=0.09 \Omega \\
& X_{02}=X_2+K^2 X_1=0.14+(0.5)^2 \times 0.6=0.29 \Omega
\end{aligned}\]
(i) Percentage regulation on full load for 0.8
lagging
\[\begin{aligned}
\% \text { regulation } & =\frac{I_2\left(R_{02} \cos
\phi-X_{02} \sin \phi\right)}{E_2} \times 100 \\
& =\frac{45.45(0.09 \times 0.8-0.29 \times 0.6)}{220} \times 100
\\
& =-2.11 \%
\end{aligned}\]
(ii) Percentage regulation on full load for 0.8 leading pf
\[\begin{aligned}
\cos \phi & =1 \qquad
\sin \phi =0 \\
\% \text { regulation } & =\frac{I_2\left(R_{02} \cos \phi \pm
X_{02} \sin \phi\right)}{E_2} \times 100 \\
& =\frac{45.45(0.09 \times 1-0.29 \times 0)}{220} \times 100 \\
& =1.86 \%
\end{aligned}\]
(iii) Percentage regulation on full load for unity pf
Problem-9
Iron loss of \(80 \mathrm{kVA}, 1000 /
250 \mathrm{~V}\), single-phase, \(50
\mathrm{~Hz}\) transformer is \(500
\mathrm{~W}\). The full-load copper loss is \(400 \mathrm{~W}\). Find (i) area of cross
section of limb if working flux density is \(1
\mathrm{~T}\) and there are 1000 turns on the primary, (ii)
efficiency at full load and pf 0.8 lagging, and (iii) efficiency at
\(75 \%\) of full load and unity
pf.
\[\begin{aligned}
\text { Full load kVA } & =80 \mathrm{~kVA} \quad
E_1 =1000 \mathrm{~V} \quad
E_2 =250 \mathrm{~V} \quad
f =50 \mathrm{~Hz} \\
W_i & =500 \mathrm{~W}=0.5 \mathrm{~kW} \quad
W_{C u} =400 \mathrm{~W}=0.4 \mathrm{~kW} \\
B_m & =1 \mathrm{~T} \quad
N_1 =1000
\end{aligned}\]
\[\begin{aligned}
E_1 & =4.44 f \phi_m N_1=4.44 f B_m A N_1 \\
\Rightarrow~1000 & =4.44 \times 50 \times 1 \times
\mathrm{A} \times 1000 \\
\Rightarrow~A & =4.5 \times 10^{-3} \mathrm{~m}^2
\end{aligned}\]
Given data:
\[\begin{aligned}
x & =1 \quad
\mathrm{pf} =0.8 \\
\% \eta & =\frac{x \times \text { full-load } \mathrm{kVA}
\times \mathrm{pf}}{x \times \text { full-load } \mathrm{kVA} \times
\mathrm{pf}+W_i+x^2 W_{C u}} \times 100 \\
& =\frac{1 \times 80 \times 0.8}{1 \times 80 \times
0.8+0.5+(1)^2 \times 0.4} \times 100 \\
& =98.61 \%
\end{aligned}\]
\[\begin{aligned}
x & =0.75 \quad
\mathrm{pf} =1 \\
\% \eta & =\frac{x \times \text { full-load } \mathrm{kVA}
\times \mathrm{pf}}{x \times \text { full-load } \mathrm{kVA} \times
\mathrm{pf}+W_i+x^2 W_{C u}} \times 100 \\
& =\frac{0.75 \times 80 \times 1}{0.75 \times 80 \times
1+0.5+(0.75)^2 \times 0.4} \times 100 \\
& =98.81 \%
\end{aligned}\]
\(75 \%\)(ii) Efficiency at full load and 0.8 lagging pf
Problem-10
\[\begin{aligned}
\text{Full load kVA} & =150 \mathrm{~kVA}\quad
W_i =1.4 \mathrm{~kW} \quad
W_{cu} =1.6 \mathrm{~kW}
\end{aligned}\]
Given Data:
\[\begin{aligned}
\text { Load kVA } & =\text { Full load kVA } \times
\sqrt{\frac{W_i}{W_{C u}}}\\
&=150 \times \sqrt{\frac{1.4}{1.6}}=140.31~ \mathrm{kVA}
\end{aligned}\]
(i) Load kVA for maximum efficiency and the maximum efficiency
\[\begin{aligned}
W_i & =W_{C u}=1.4 \mathrm{~kW} \\
\mathrm{pf} & =0.8 \\
\% \eta_{\max } & =\frac{\text { load } \mathrm{kVA} \times
\mathrm{pf}}{\text { load } \mathrm{kVA} \times \mathrm{pf}+W_i+W_i}
\times 100 \\
& =\frac{140.31 \times 0.8}{140.31 \times 0.8+1.4+1.4}
\times 100 \\
& =97.57 \%
\end{aligned}\]
\[\begin{aligned}
x & =0.5 \qquad
\mathrm{pf} =0.8 \\
\% \eta & =\frac{x \times \text { full-load kVA } \times
\mathrm{pf}}{x \times \text { full-load kVA } \times \mathrm{pf}+W_i+x^2
W_{C u}} \times 100 \\
& =\frac{0.5 \times 150 \times 0.8}{0.5 \times 150 \times
0.8+1.4+(0.5)^2 \times 1.6} \times 100 \\
& =97.08 \% .
\end{aligned}\]
\(0.8 \mathrm{pf}\)For maximum efficiency,
Problem-11
A \(5 ~\mathrm{kVA}\), \(200 / 400 \mathrm{~V}\), \(50 \mathrm{~Hz}\), single-phase transformer
gives the following test results:
\[\begin{aligned}
\text{OC test (LV side)} & ~~200 \mathrm{~V} ~~ 0.7 \mathrm{~A} ~~ 60 \mathrm{~W} \\
\text{SC test (HV side)} & ~~22 \mathrm{~V} ~~ 16 \mathrm{~A} ~~ 120 \mathrm{~W}
\end{aligned}
\]
Draw the equivalent circuit of the transformer and insert all
parameter values.
Find efficiency and regulation at \(0.9
\mathrm{pf}\) (lead) if operating at rated load.
Find current at which efficiency is maximum.
\[\begin{aligned}
& W_i=60 \mathrm{~W} \quad V_1=200 \mathrm{~V} \quad I_0=0.7
\mathrm{~A} \\
& \cos \phi_0=\frac{W_i}{V_1 I_0}=\frac{60}{200 \times 0.7}=0.43
\\
& \sin \phi_0=0.9 \\
& I_w=I_0 \cos \phi_0=0.7 \times 0.43=0.3 \mathrm{~A} \\
& R_0=\frac{V_1}{I_w}=\frac{200}{0.3}=666.67 \Omega \\
& I_\mu=I_0 \sin \phi_0=0.7 \times 0.9=0.63 \mathrm{~A} \\
& X_0=\frac{V_1}{I_\mu}=\frac{200}{0.63}=317.46 \Omega
\end{aligned}\]
From OC test (meters are connected on LV side, i.e., primary), (i) Equivalent circuit of the transformer
\[\begin{aligned}
& W_{\mathrm{sc}}=120 \mathrm{~W} \quad V_{\mathrm{sc}}=22
\mathrm{~V} \quad I_{\mathrm{sc}}=16 \mathrm{~A} \\
&
Z_{02}=\frac{V_{\mathrm{sc}}}{I_{\mathrm{sc}}}=\frac{22}{16}=1.375
\Omega \\
&
R_{02}=\frac{W_{\mathrm{sc}}}{I_{\mathrm{sc}}^2}=\frac{120}{(16)^2}=0.47
\Omega \\
&
X_{02}=\sqrt{\left(Z_{02}\right)^2-\left(R_{02}\right)^2}=\sqrt{(1.375)^2-(0.47)^2}=1.29
\Omega
\end{aligned}\]
test (meters
are connected on HV side, i.e., secondary), From
\[\begin{aligned}
K & = \dfrac{400}{200}=2\\
R_{01} & = \dfrac{R_{02}}{K^2} = \dfrac{0.47}{2^2} =
0.12~\Omega \\
X_{01} & = \dfrac{X_{02}}{K^2} = \dfrac{1.29}{2^2} =
0.32~\Omega \\
\end{aligned}\]
\[W_i=60
\mathrm{~W}=0.06 \mathrm{~kW}\]
\[\begin{aligned}
I_2 & =\frac{5 \times 1000}{400}=12.5 \mathrm{~A} \\
W_{C u} & =I_2^2 R_{02}\\
&=(12.5)^2 \times 0.47=73.43 \mathrm{~W}=0.073 \mathrm{~kW} \\
x & =1 \qquad
\mathrm{pf} =0.9 \\
\% \eta & =\frac{x \times \text { full load kVA } \times
\mathrm{pf}}{x \times \text { full-load kVA } \times \mathrm{pf}+W_i+x^2
W_{C u}} \times 100 \\
& =\frac{1 \times 5 \times 0.9}{1 \times 5 \times 0.9+0.06+(1)^2
\times 0.073} \times 100 \\
& =97.13 \%
\end{aligned}\]
leading (ii) Efficiency at rated load and
\[\begin{aligned}
\cos \phi & =0.9 \qquad
\sin \phi =0.44 \\
\% \text { regulation } & =\frac{I_2\left(R_{02} \cos
\phi-X_{02} \sin \phi\right)}{E_2} \times 100 \\
& =\frac{12.5(0.47 \times 0.9-1.29 \times 0.44)}{400} \times 100
\\
& =-0.45 \%
\end{aligned}\]
\[\begin{aligned}
W_i & =I_2^2 R_{02} \\
I_2 & =\sqrt{\frac{W_i}{R_{02}}}=\sqrt{\frac{60}{0.47}}=11.3
\mathrm{~A}
\end{aligned}\]
lead Regulation at rated load and