Transformer Voltage Regulation: Mastering the Drop

Demonstrative Video

\(\bullet\) Transformer with resistance and leakage reactance

\[\begin{aligned} Z_{1} & =\sqrt{\left(R_{1}^{2}+X_{1}^{2}\right)}\\ Z_{2} & =\sqrt{\left(R_{2}^{2}+X_{2}^{2}\right)}\\ V_{1} & =E_{1}+I_{1}\left(R_{1}+jX_{1}\right)=E_{1}+I_{1}Z_{1}\\ E_{2} & =V_{2}+I_{2}\left(R_{2}+jX_{2}\right)=V_{2}+I_{2}Z_{2} \end{aligned}\]

Leakage reactance can also be transferred from one winding to the other in the same way as resistance \[\begin{array}{ccc} X_{2}^{'}=X_{2}/K^{2} & \mbox{and} & X_{1}^{'}=K^{2}X_{1}\\ X_{01}=X_{1}+X_{2}^{'} & \mbox{and} & X_{02}=X_{2}+X_{1}^{'} \end{array}\]

\[\begin{aligned} Z_{01} & =\sqrt{\left(R_{01}^{2}+X_{01}^{2}\right)}\\ Z_{02} & =\sqrt{\left(R_{02}^{2}+X_{02}^{2}\right)} \end{aligned}\]

Voltage Drop in a Transformer

At no load: \[\begin{aligned} V_{1} & \thickapprox E_{1}\\ E_{2} & =KE_{1}=KV_{1}\\ E_{2} & =_{0}V_{2} \end{aligned}\] At load:

The approximate voltage drop is: \[I_{2}R_{02}\cos\Phi\pm I_{2}X_{02}\sin\Phi\] where \(+\) is for lagging pf and \(-\) is for leading pf

Similarly, approximate voltage drop referred to primary is \[I_{1}R_{01}\cos\Phi\pm I_{1}X_{01}\sin\Phi\] % voltage drop in secondary is \[\begin{aligned} = & \dfrac{I_{2}R_{02}cos\Phi\pm I_{2}X_{02}sin\Phi}{_{0}V_{2}}\times100\\ = & v_{r}cos\Phi\pm v_{x}sin\Phi \end{aligned}\] where \[\begin{aligned} v_{r} & =\dfrac{I_{2}R_{02}}{_{0}V_{2}}\times100=\dfrac{I_{1}R_{01}}{V_{1}}\times100=\mbox{percentage resistive drop}\\ v_{x} & =\dfrac{I_{2}X_{02}}{_{0}V_{2}}\times100=\dfrac{I_{1}X_{01}}{V_{1}}\times100=\mbox{percentage reactive drop} \end{aligned}\]

Equivalent Circuit of a Transformer

Equivalent circuit is basically a diagram in which the resistance and leakage reactance of the transformer are imagined to be external to the winding

The equivalent circuit diagram of transformer is given below:-

The secondary circuit and its equivalent primary value

The total equivalent circuit is obtained by adding in the primary impedance

It can be simplified

At last, the circuit is simplified by omitting \(I_0\) altogether

The total impedance between the input terminal: \[\begin{aligned} Z & =Z_{1}+Z_{m}||\left(Z_{2}^{'}+Z_{L}^{'}\right)\\ & =Z_{1}+\dfrac{Z_{m}\left(Z_{2}^{'}+Z_{L}^{'}\right)}{Z_{m}+\left(Z_{2}^{'}+Z_{L}^{'}\right)} \end{aligned}\] Therefore the input voltage is given by \[V_{1}=I_{1}\left[Z_{1}+\dfrac{Z_{m}\left(Z_{2}^{'}+Z_{L}^{'}\right)}{Z_{m}+\left(Z_{2}^{'}+Z_{L}^{'}\right)}\right]\]