I am looking at a circuit diagram like this:


simulate this circuit – Schematic created using CircuitLab

Switch 1 and switch 2 never are on at the same time (they alternate) I turn on S2 and shut off S1, then after a while, I turn on S1 and shut off S2. At that point I think I can redraw the circuit like this:


simulate this circuit

But if I do that, I think the current through the switch would then be zero? And then the switch would be considered "off?" Or maybe I could arbitrarily define the current through the switch as being from either the left or right hand side circuits? But that seems to be not possible because originally the switch was shared between both of the circuits.

Can someone figure out what is wrong with my reasoning? What should the current through the circuit be in the "on" position?

Note: ignore the values of the circuit elements, they are not important.

  • \$\begingroup\$ @IgnacioVazquez-Abrams Edited to hopefully fix that. \$\endgroup\$ – AAC Oct 5 '14 at 4:22
  • \$\begingroup\$ The on position will give you the difference of the currents flowing through the two LC tanks. Or sum, depending on how you specify your signs. \$\endgroup\$ – Ignacio Vazquez-Abrams Oct 5 '14 at 4:22
  • \$\begingroup\$ I don't get how that could be nonzero, because if for example L1 and C1 share the same current, then the current flowing out of node1 is the same as the current flowing into it. \$\endgroup\$ – AAC Oct 5 '14 at 4:33
  • \$\begingroup\$ What I mean is, I do think that the currents in each LC circuit are different. However, I don't get how the current across the switch is then defined as the difference between the two. \$\endgroup\$ – AAC Oct 5 '14 at 4:49
  • \$\begingroup\$ Model the "closed" switch with a small-value resistance, say 0.1ohm, to model the current through the switch. Ideal switch Ron=0, practical switch Ron is non-zero. \$\endgroup\$ – MarkU Oct 5 '14 at 4:52

Do KCL at the node between C1 and L2:


simulate this circuit – Schematic created using CircuitLab

Then the current through the switch is just the difference (or sum, depending on sign definition) of the current in each of the two LC circuits drawn in the second diagram. It is clear in diagram 1 what is going on but possibly less clear in diagram 2.


If you turn S1 on and wait long enough, the currents will eventually decay to zero (assuming non-ideal components). In the meantime, the S1 current is equal to the sum of the two inductor currents. Redrawing the circuit like you did removes S1 altogether, which is okay unless you want to know the switch current. :-)

This circuit looks like a fancy switching regulator, so probably you wouldn't wait a long time. You would switch frequently, and vary the duty cycle to control the output voltage. It's easiest to understand if you look at it over two cycles:

1a. S2 on, V1 boosts the current in L1.

1b. S1 on, L1 pulls current through S1 and C1, charging up C2 (positive end away from V1).

2a. S2 on, C1 discharges through L2, boosting its current and feeding the load.

2b. S1 on, L2 pulls current through S1, feeding the load.

C2 acts as a filtering capacitor to reduce voltage ripple at the load. You can see that when S1 is on (steps 1b and 2b), both inductors are pulling current from ground through S1.


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