Below the RLC circuit on the left side by the help of a periodic impulse generator V1 generates periodic damped oscillations with a certain frequency, and the resulting oscillating current couples to the circuit on the right side via the magnetic field. The circuit on the right side is exactly the same with the one on the left(just that it has no source). Here I also set the coupling coefficient as 1. So I(R1) and I(R2) is exactly the same as follow:

case 1 enter image description here

Now I change only the capacitance of the capacitor C2 from 10µ to 1000µ and both circuits are affected radically and they pass the same currents as follows:

case 2 enter image description here

Finally from the last configuration I reduce the coupling coefficient significantly from 1 to 0.01 and the circuit on the left is not anymore affected from the one on the right side, but the current in the right circuit still follows the same wave shape with the current in the left circuit as follows:

case3 enter image description here


1)If SPICE is correct, in case1 even though there is strong coupling, the natural frequency waveform of both circuits remain the same. In case2 however there is again strong coupling, but this time the waveform of both circuits radically changed. How can this be explained?

2)In case3 when the coupling is very loose, why does I(R2) replicate I(R1) even though the circuit on the right side has different natural frequency?

  • 1
    \$\begingroup\$ Please check the actual k used in the third case. Your schematic+ current plots hint that the actual k was not 0.001 but 0.01 \$\endgroup\$
    – user136077
    Commented Nov 26, 2017 at 23:11

1 Answer 1


The first case (k=1, C1=C2=1uF) can be considered as:

enter image description here

(See the ADDENDUM for a quick proof)

After the initial impulse there's 2 lossy capacitors in parallel with the inductor. (NOTE:the voltage source is like a short circuit after the impulse has gone) That's a resonant circuit with 2uF effective capacitance. The resonant frequency is about 1100Hz. The oscillation dies away due the losses in the resistances. C1=C2, so the current distributes equally between them.

In your middle case (k=1, C1=1uF, C2=1000uF) the equivalent circuit is the same except C2=1000uF. It practically with 10mH inductance defines the resonant frequency which is now about 35Hz. During the oscillation after the initial impulse the current mostly goes trough C2 because there's virtually the same AC voltage over the capacitors, but the reactance of C2 is much lower. During the initial impulse C2 doesn't get much voltage because the capacitive voltage division leaves most of the initial impulse voltage to C1 and the inductor does not charge much current during the impulse due the low voltage over it during the impulse. As you see the oscillation current is mostly through R2 i.e. C2 and it's much lower than the current in the first case. In the screenshot you can see about 1 pixel high amplitude.

In your third case (k=0,01 C1=1uF, C2=1000uF) we have two only loosely coupled resonant circuits with different resonant frequencies. After the initial impulse the leftmost resonator oscillates and part of the AC current is connected to the other resonator. It's far from the resonance, so no complex interactions occur between the resonators. The right side gets some AC due the coupling but the current is much lower than in the left side.

You can think that in case 3 the right side coil generates an AC voltage, which is practically shorted by C2. The current at the right side is practically k * the current in the right side, as you can easily see if you can form the mutual inductance relations with complex phasors.

ADDENDUM: The reduction of the inductors to one can be derived by using the ordinary mutual induction relations:

enter image description here

  • \$\begingroup\$ How did you calculate 5mH? Isn't the mutual inductance is M=k*sqrt(L1*L2)? \$\endgroup\$
    – user16307
    Commented Nov 27, 2017 at 19:16
  • \$\begingroup\$ @user16307 there was an error. It's fixed now. Double thanks for notice \$\endgroup\$
    – user136077
    Commented Nov 27, 2017 at 20:48

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