I don't understand the direction of current flow when a capacitor is wired in series with the start windings in, say, a fridge or other AC motor running at/under 120V AC.

Wiring diagrams seem to suggest that voltage energizes the hot leg of the circuit and current flows through the run windings and then returns via the neutral leg. Without a capacitor, the same thing would happen to the start windings. Problem there I guess is that both windings would be in phase and thus no spin created for the rotor.

So, a capacitor is wired in series with the start windings which I kind of understand. But I don't know how the voltage and current behave.

Does voltage energize the hot leg, flow through the start windings and then charge the capacitor?

When voltage on the hot leg oscillates down to zero, does the capacitor discharge back toward the starter windings, or does current flow through the capacitor to the neutral leg?

Basically I don't understand where the capacitor's charge goes when the hot leg is hot and when it is not.

Lastly, is the capacitor placed between (downstream from) the start windings and neutral, or could it be wired between incoming voltage and the windings?

  • 4
    \$\begingroup\$ Draw a simple circuit demonstrating your concern. This bunch of words is not helpful. \$\endgroup\$
    – Eugene Sh.
    Jan 6, 2016 at 19:07
  • \$\begingroup\$ the diagram posted \$\endgroup\$
    – Scott
    Jan 6, 2016 at 22:09
  • \$\begingroup\$ the diagram posted by the replier beneath my question should work, along with my unhelpful words \$\endgroup\$
    – Scott
    Jan 6, 2016 at 22:10
  • 2
    \$\begingroup\$ @EugeneSh. this is a question about the principle of an absolutely standard circuit, not something unique that the poster is dealing with - likely anyone capable of answering the question is already picturing it in their head. \$\endgroup\$ Aug 28, 2016 at 1:37

2 Answers 2


Trying to start a single phase motor with only one winding would be a bit like trying to start a bike with only one pedal. It's OK once you get it going but trying to get the start direction right and starting from top or bottom dead-centre is awkward.


simulate this circuit – Schematic created using CircuitLab

Induction motor with square rotor because there is no circle tool in the schematic editor.

The single phase induction motor is similar. To solve the problem an auxiliary, usually weaker, winding is added to the motor and it is offset from the main coil by, say, 30°. A capacitor is wired in series with this coil and it has the effect of causing a shift in the phase of the current in the auxiliary winding relative to that of the main winding. The result is that the magnetic field in one winding leads the other and this imparts enough rotational force into the rotor to:

  • get it to start.
  • start in the right direction.

Some motors feature a centrifugal switch which disconnects the auxiliary winding once the motor exceeds a certain speed as it is no longer required. This saves a little energy and reduces motor heating.

Understanding capacitor current

But could you just clarify this part for me? When the cap is fully charged as 120v crosses zero, what happens to the accumulated negative charge on the saturated cap plate? Does it pulse upstream against the previous voltage flow or does it just sit there? – Scott

We usually learn about capacitors in DC circuits where it is easy to visualise the capacitor charging up and then discharging and the capacitor voltage follows the RC charge / discharge curve. Usually in these scenarios the applied voltage doesn't alternate above and below zero volts. This way of thinking doesn't help us much in AC circuit analysis.

Let's consider the start winding again. For simplicity we're going to ignore the inductance of both windings and think of them as resistors. Using our simple model:

  • The current in the main winding will follow the L-N voltage and will be in-phase with it.
  • We want a phase-shift on the current in the L2-C1 branch to generate the rotation.

The capacitor current is given by the rule \$I = \frac{dQ}{dt}\$ where Q is the charge. This simply tells us that the current will be greatest when the rate of charge movement is greatest. The capacitor charge is given by \$Q = C \cdot V\$ and combining the two we get \$I = C \frac {dV}{dt}\$. All we're saying here is that the capacitor current is proportional to the rate of change of voltage.


simulate this circuit

Capacitor VI curve

Simplification: Again we're ignoring inductance and treating the windings as low value resistors (relative to the impedance of the capacitor).

At 270° the voltage (red) is at maximum negative. The capacitor is charged fully negative and since the voltage has stopped falling (going negative) the current has fallen to zero (blue curve is at zero).

From 270° to 0° the voltage will be increasing. The rate of change will get faster and faster as it approaches zero. For this reason the current will increase from zero reaching maximum current at 0°.

At the capacitor is fully discharged but the rate of change of voltage is highest (steepest on the curve). This will charge up the capacitor and, since the charge rate - the current - is proportional to the rate of change of voltage, the current reaches a maximum here.

For the next 0° to 90° the rate of change of voltage is decreasing and the current decreases to zero.

The same pattern repeats but in the opposite directions in the next 180°.


  • With this arrangement the voltage and current waveforms are always sinusoidal. There are no sudden charges / discharges or step changes in voltage or current.
  • The only 'infinitesmal pause' is when the voltage or current changes direction. It's no more of a pause than the piston of an engine reaching top of stroke. Velocity = 0 for a moment but the acceleration is highest at that point (if I'm thinking correctly).
  • What comes in on the live / hot wire on that leg must go out to the neutral on that leg.
  • C1 and the switch can go either side of L2.
  • \$\begingroup\$ I have no formal electrical training--just self-taught , so I still need further clarification, if you would: \$\endgroup\$
    – Scott
    Jan 7, 2016 at 17:40
  • \$\begingroup\$ Besides putting the start winding out of phase from the main winding (a kind of time delay I'm thinking, does the cap serve any other purpose? \$\endgroup\$
    – Scott
    Jan 7, 2016 at 17:42
  • \$\begingroup\$ What happens to its stored charge when the 120v hot leg oscillates down to zero and the cap is now fully charged? Does it discharge backwards down the hot leg (now at zero volts for an instant)? \$\endgroup\$
    – Scott
    Jan 7, 2016 at 17:45
  • \$\begingroup\$ @Scott, the current through a capacitor is a bit like a tide running in and out of an estuary. Current is highest at nominal sea level when the rate of change of height is highest. The current is lowest at high tide and low tide. Effectively the current and tide height are 90° out of phase. In the same way in the motor circuit the 'current' is sloshing back and forward through the coil at a rate determined by the rate of change of the voltage. Again it's out of phase with the voltage and hence out of phase with the primary winding. See animations.physics.unsw.edu.au/jw/AC.html. \$\endgroup\$
    – Transistor
    Jan 7, 2016 at 20:12
  • \$\begingroup\$ Thank you for this reply. But could you just clarify this part for me? When the cap is fully charged as 120v crosses zero, what happens to the accumulated negative charge on the saturated cap plate? Does it pulse upstream against the previous voltage flow or does it just sit there? \$\endgroup\$
    – Scott
    Jan 8, 2016 at 1:19

I think you are missing the point of the answers. There is no max charge of the capacitor that just "sits" there waiting for discharge. In fact, the changing voltage effectively "bleeds off" the charge that was built up as the polarity changes. This is a normal part of an AC circuit. You are thinking of a capacitor and its use in a DC circuit where it would charge up and depend either on polarity change "upstream" of the capacitor or discharge through a downstream component or series of components. The important point about the Capacitor is that it is just like a small estuary that is filling or emptying all of the time, not sitting with a full level.


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