I am a bit confused about my derivation to get power efficiency from the coupling coefficient.

The coupling coefficient \$k=M/\sqrt{L_1L_2}\$. $$ V_2=M\frac{dI_1}{dt}=L_2\frac{dI_2}{dt}\\ V_1=M\frac{dI_2}{dt}=L_1\frac{dI_1}{dt}\\ $$ Therefore, $$ M={V_2}/{\frac{dI_1}{dt}}={V_1}/{\frac{dI_2}{dt}}\\ L_1={V_1}/{\frac{dI_1}{dt}}\\ L_2={V_2}/{\frac{dI_2}{dt}} $$ Then I substitute them into \$k\$ to get: $$ k=\sqrt{\frac{V_1dI_1}{V_2dI_2}}=\sqrt{\frac{V_2dI_2}{V_1dI_1}}\\ k^2=\frac{P_1}{P_2}=\eta_{12}=\frac{P_2}{P_1}=\eta_{21} $$ So I can say: if there are two coils that one is given power source and the other is receiving the power, then the two coils can transmit and feedback the power to each other with the efficiency of \$k^2=\eta\$. Therefore, the total power received in the receiver coil will be: $$ P_2=P_1(\eta+\eta^3+\eta^5+\dots)=P_1\sum_{n=1}^\infty\eta^{2n-1}=P_1\frac{\eta}{1-\eta^2} $$ As you can see, the efficiency of the power transfer here becomes \$\frac{\eta}{1-\eta^2}\$.

Is this correct? Or am I having any misunderstanding here?


1 Answer 1


The power into a 100% pure ideal transformer with a coupling factor of less than unity is exactly the power taken by the load. If the power out is 1 mW, then the power in will be 1 mW.

Or am I having any misunderstanding here?

Your math is wrong in that you have made assumptions about the phase relationship between V and I on the primary side and directly equated that to power in. That is not true. Imagine the situation when k is 0. All you have is an inductor with a voltage across it and some current flow determined by the inductor's reactance and operating frequency; there isn't any power because V and I are at 90° to each other.

  • \$\begingroup\$ True. I have missed the point that \$I\$ has \$90^{\circ}\$ phase lagging related to \$V\$. What I should consider is to put a load across the receiver conductor that is seen as a voltage source and a resistor series to the transmitter one so that I can evaluate the power transfer efficiency. Thank you. \$\endgroup\$
    – ONLYA
    Dec 12, 2020 at 16:13
  • 1
    \$\begingroup\$ Just use a simulator is my advice. Get a feel for things then do some math that justifies what the sim tells you. It's pretty basic bread and butter stuff for a sim. Power efficiency will always be 100% for ideal components. \$\endgroup\$
    – Andy aka
    Dec 12, 2020 at 16:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.