The reciprocal relation of mutual inductance in a coupled circuit system

The integrands in Eq. (9) involve the multiplication of sine and cosine functions, and the orthogonality of trigonometric functions then requires that the time integrals of the self-inductance terms vanish: $∫−T/2T/2sin(ω t+φ1)cos(ω t+φ1)dt=0$⁠.

The product-to-sum trigonometric identity is used to reduce the mutual inductance terms in Eq. (9): $sin(ω t+φ1)cos(ω t+φ2)=12[sin(2ω t+φ1+φ2)+sin(φ1−φ2)]$⁠. Equation (11) is readily obtained by observing that $∫−T/2T/2sin(2ω t+φ1+φ2) dt=0$⁠.

In fact, the phase angles $φ1$and $φ2$ must be different from each other if Eq. (2) is to hold. To see this, simply substitute Eqs. (4) and (10) into Eq. (2) to obtain $L2ω i20cos(ωt+φ2)+M21ω i10cos(ω t+φ1)+R2i20sin(ω t+φ2)=0$⁠, where the three terms can be represented by three phasors in a phasor diagram. As can be seen from this equation, the first phasor is perpendicular to the third phasor. It follows in a straightforward manner that one must have 90° < $φ1$ − $φ2$ < 180° if the vector sum of the three phasors is to cancel out.