Transformer Equivalent Circuits: Solved Problems for Mastery
Demonstrative Video
VIDEO
Problem-1
\[\begin{aligned}
R_1 & = 0.286~\Omega ~~ R_2^{\prime} = 0.319~\Omega ~~ R_0 =
250~\Omega \\
X_1 & = 0.73~\Omega ~~ X_2^{\prime} = 0.73~\Omega ~~ X_0 =
1250~\Omega
\end{aligned}\]
\(z_{L}=0.387+j 0.29\) transformer are given below: The parameters of a
Using the exact equivalent circuit with normal voltage across the
primary, determine
Solution-1
\[\begin{aligned}
K &=230 / 2300=1 / 10 \\
Z_{L}&=0.387+j 0.29 \\
\mathrm{Z}_{L}^{\prime} &=\mathrm{Z}_{L} / K^{2}\\
&=100(0.387+j 0.29)=38.7+j 29=48.4 \angle 36.8^{\circ} \\
\end{aligned}\]
\[\begin{aligned}
\mathrm{Z}_{2}^{\prime}+\mathrm{Z}_{L}^{\prime}
&=(38.7+0.319)+j(29+0.73)\\
&=39.02+\mathrm{j} 29.73=49.0 \angle 37.3^{\circ} \\
Y_{m} &=(0.004-j 0.0008) \\
Z_{m}&=1 / \mathrm{Y}_{m}=240+j 48=245 \angle 11.3^{\circ} \\
\mathrm{Z}_{m}+\left(Z_{2}^{\prime}+Z_{L}{ }^{\prime}\right)
&=(240+j 48)+(39+j 29.7)=290 \angle 15.6^{\circ}
\end{aligned}\]
\[\begin{aligned}
I_{1}
&=\frac{V_{1}}{Z_{1}+\frac{Z_{m}\left(Z_{2}^{\prime}+Z_{L}\right)}{Z_{m}\left(Z_{2}^{\prime}+Z_{L}^{\prime}\right)}}=\left[\frac{2300
\angle 0^{\circ}}{0.286+j 0.73+41.4 \angle 33^{\circ}}\right] \\
&=\frac{2300 \angle 0^{\circ}}{42 \angle 33.7^{\circ}}=54.8
\angle-33.7^{\circ} \\
I_{2}^{\prime} &=I_{1} \times
\frac{Z_{m}}{\left(Z_{2}^{\prime}+Z_{L}^{\prime}\right)+Z_{m}}=54.8
\angle-33.7^{\circ} \times \frac{245 \angle 11.3^{\circ}}{290 \angle
15.6^{\circ}} \\
&=54.8 \angle-33.7^{\circ} \times 0.845 \angle-4.3^{\circ}=46.2
\angle-38^{\circ} \\
I_{0} &=I_{1} \times
\frac{\left(Z_{2}^{\prime}+Z_{L}^{\prime}\right)}{Z_{m}+\left(Z_{2}^{\circ}+Z_{L}\right)}=54.8
\angle-33.7^{\circ} \times \frac{49 \angle 37.3^{\circ}}{290 \angle
15.6^{\circ}} \\
&=54.8 \angle-33.7^{\circ} \times 0.169 \angle 21.7^{\circ}=9.26
\angle-12^{\circ}
\end{aligned}\]
\[\begin{aligned}
\text { Input power factor } &=\cos 33.7^{\circ}=0.832 \text {
lagging }\\
\text { Power input } & =V_{1} I_{1} \cos \phi_{1}=2300 \times 54.8
\times 0.832=105 \mathrm{~kW} \\
\text { Power output } & =46.2^{2} \times 38.7=82.7 \mathrm{~kW} \\
\text { Primary Cu loss } & =54.8^{2} \times 0.286=860 \mathrm{~W}
\\
\text { Secondary Cu loss } & =46.2^{2} \times 0.319=680 \mathrm{~W}
\\
\text { Core loss } & =9.26^{2} \times 240=20.6 \mathrm{~kW} \\
\eta & =(82.7 / 105) \times 100=78.8 \% \\
V_{2}^{\prime}&=I_{2}^{\prime} Z_{L}^{\prime}=46.2 \times 48.4=2,240
\mathrm{~V}\\
\text { Regulation }&=\frac{2300-2240}{2240} \times 100=2.7 \%
\end{aligned}\]
Problem-2
In no-load test of single-phase transformer, the following test data
were obtained:
\[\begin{aligned}
\text{Primary voltage}: & 220 \mathrm{~V} \quad \text{Secondary
voltage}: 110 \mathrm{~V}\\
\text{Primary curren}t: & 0.5 \mathrm{~A} \quad \text{Power
input}: 30 \mathrm{~W}
\end{aligned}\]
Find the following
The turns ratio
the magnetizing component of no-load current
working (or loss) component
the iron loss
Given: Resistance of the primary winding \(=0.6\) ohm
Solution-2
\[\begin{aligned}
\text{Turn ratio},~ N_{1} / N_{2}&=220 / 110=2 \\
W&=V_{1} I_{0} \cos \phi_{0}\\
\Rightarrow \cos \phi_{0} &=30 / 220 \times 0.5=0.273 \\
\sin \phi_{0}&=0.962 \\
I_{\mu} &=I_{0} \sin \phi_{0}=0.5 \times 0.962=0.48
\mathrm{~A} \\
I_{w}&=I_{0} \cos \phi_{0}=0.5 \times 0.273=0.1365 \mathrm{~A} \\
\text{Primary Cu loss} &=I_{0}^{2} R_{1}=0.5^{2} \times 0.6=0.15
\mathrm{~W} \\
\text{Iron loss} &=30-0.15=29.85 \mathrm{~W}
\end{aligned}\]
Problem-3
Obtain the equivalent circuit parameters of a \(200 / 400 \mathrm{~V}, 50 \mathrm{~Hz}, 1\)
-phase transformer from the following test data:
\[\begin{aligned}
\text{OC Test}:~~& 200 \mathrm{~V}~~ 0.7 \mathrm{~A}~~ 70
\mathrm{~W} --\mathrm{LV side} \\
\text{SC Test}: ~~& 15 \mathrm{~V} ~~ 10 \mathrm{~A} ~~ 85
\mathrm{~W}--\mathrm{HV side}
\end{aligned}\]
Calculate the secondary voltage when delivering \(5 \mathrm{KW}\) at \(0.8 \mathrm{pf}\) lagging, the primary
voltage being \(200 \mathrm{~V}\)
Solution-3
\[\begin{aligned}
V_{1} I_{0} \cos \phi_{0} &=W_{0} \\
200 \times 0.7 \times \cos \phi_{0} &=70 \\
\cos \phi_{0} &=0.5 \\
\sin \phi_{0}&=0.866 \\
I_{w} &=I_{0} \cos \phi_{0}\\
&=0.7 \times 0.5=0.35 \mathrm{~A} \\
I_{\mu} &=I_{0} \sin \phi_{0}\\
&=0.7 \times 0.866\\
&=0.606 \mathrm{~A} \\
R_{0} &=V_{1} / I_{\mathrm{w}}=200 / 0.35=571.4 \Omega \\
X_{0} &=V_{1} / I_{\mathrm{u}}=200 / 0.606=330 \Omega
\end{aligned}\]
From O.C.Test
\[\begin{aligned}
Z_{02} &=V_{sc} / I_{2}=15 / 10=1.5 \Omega \\
K&=400 / 200=2 \\
Z_{01} &=Z_{02} / K^{2}=1.5 / 4=0.375 \Omega \\
I_{2}^{2} R_{02} &=W \\
\Rightarrow R_{02}&= W/I_{2}^{2} = 85 / 100=0.85 \Omega \\
R_{01} &=R_{02} / K^{2}=0.85 / 4=0.21 \Omega \\
X_{01} &=\sqrt{Z_{01}^{2}-R_{01}^{2}}=\sqrt{0.375^{2}-0.21^{2}}=0.31
\Omega
\end{aligned}\]
\[\begin{aligned}
\text { Output } \mathrm{kVA}&=5 / 0.8 \\
\text { Output current } I_{2}&=5000 / 0.8 \times 400=15.6
\mathrm{~A}
\end{aligned}\]
From S.C. Test : instruments have been placed in the secondary i.e.
high-voltage winding whereas the low-voltage winding i.e, primary has
been short-circuited.
\[\begin{aligned}
&=I_{2}\left(R_{02} \cos \phi_{2}+X_{02} \sin \phi_{2}\right)\\
&=15.6(0.85 \times 0.8+1.24 \times 0.6)=22.2 \mathrm{~V} \\
V_{2}&=400-22.2=377.8 \mathrm{~V}
\end{aligned}\]
Total transformer drop as referred to secondary
