Perfect Transformer, Current Discontinuity and Degeneracy

That on connecting a source in the primary circuit of a perfectly coupled transformer, the currents in both the primary and secondary coils may be discontinuous does not appear to have been widely discussed in the literature. In this discussion, we present an analysis of the general circuit and show that in general, the currents will be discontinuous, except for specific combinations of the initial currents in the two coils. Although unity coupling coefficient cannot be realized in practice, a perfectly coupled transformer is a useful concept in circuit analysis and synthesis, and the results presented here should be of interest to students as well as teachers of circuit theory.

Source: S. C. Dutta Roy, “Perfect Transformer, Current Discontinuity and Degeneracy”, IETE Journal of Education, vol. 43, pp. 135–138, July–September 2002.

Appendix

Kuo’s analysis and results for the circuit are shown in Fig. 15.1.

The loop equations for the circuit shown in Fig. 15.1 are

$$ Vu\left( t \right) = L_{1} i_{1}^{\prime } \left( t \right) + R_{1} i_{1} \left( t \right) + Mi_{2}^{\prime } \left( t \right) $$

(15.18)

$$ 0 = Mi_{1}^{\prime } \left( t \right) + R_{2} i_{2} \left( t \right) + L_{2} i_{2}^{\prime } \left( t \right) $$

(15.19)

Integrating Eqs. 15.18 and 15.19 from t = 0− to t = 0+, we get

$$ L_{1} \left[ {i_{l} \left( {0{+} } \right) - i_{1} \left( {0{-}} \right)} \right] + M\left[ {i_{2} \left( {0{+} } \right) - i_{2} \left( {0{-}} \right)} \right] = 0 $$

(15.20)

$$ M\left[ {i_{1} \left( {0{+} } \right) - i_{l} \left( {0{-}} \right)} \right] + L_{2} \left[ {i_{2} \left( {0{+} } \right) - i_{2} \left( {0{-}} \right)} \right] = 0 $$

(15.21)

Combining Eqs. 15.19 and 15.21 gives

$$ (L_{1} L_{2} -M^{2} )\left[ { {i_{1}}{( {0{+} })} - i_{1} \left( {0{-}} \right)} \right]\left[ {{i_{2}}{({0+})}-i_{2} \left( {0{-}} \right)} \right] = 0 $$

(15.22)

which, along with Eq. 15.20 or 15.21, clearly indicates that if L ₁ L ₂ > M ², i.e. k < l, then the currents are continuous at t = 0. On the other hand, if k = 1, then they need not be. In fact, in this case, Eq. 15.19 gives at t = 0+:

$$ R_{2} i_{2} \left( {0{+} } \right) = {-}(M/L_{1} )\left[ {L_{1} i_{1}^{\prime } \left( {0{+} } \right) + Mi_{2}^{\prime } \left( {0{+} } \right)} \right] $$

(15.23)

which, substituted in Eq. 15.18 with t = 0+, yields

$$ V = R_{1} i_{1} \left( {0{+} } \right) - (L_{1} /M)R_{2} i_{2} \left( {0{+} } \right) $$

(15.24)

Combining this with Eq. 15.20, one can solve for i ₁(0+) and i ₂(0+). The results are³

$$ {i_{1}}{(0{+} )} = \frac{{VL_{2} + R_{2} [L_{1} {i_{1}}{(0{-} )} + {Mi_{2} (0{-} )}]}}{{R_{1} L_{2} + R_{2} L_{1} }} $$

(15.25)

$$ {i_{2}}{(0{+} )} = \frac{{ - VM + R_{1} [Mi_{1} (0{-} ) + L_{2} {i_{2}} {(0{-} )}]}}{{R_{1} L_{2} + R_{2} L_{1} }} $$

(15.26)