Short circuit and a capacitor

[Electronman]

Hi,

Today a friend asked me to explain the relationship between the voltage and the current of a capacitor when connected to an AC (sinusoidal) source.
After doing so, and showing him the 90 degrees phase difference between the voltage and the current, I came across a problem in my mind.
At the peak point of the current (which means the current is feed backed to the source) the voltage is zero. This means the current will pass through the power source when the voltage of the source is at 0V!

The question:
Why the current flows at 0V in this condition?

I was thinking of a short circuit where the voltage tends to be zero and in this situation we replace the voltage source with a wire. But I am not sure if this the key to solve my problem? I have heard the short and open circuits for DC circuits, but I think they can be used here for an AC source!

Now that I am thinking of a short circuit, I am wondering why a voltage source is assumed to be a short circuit when the voltage is zero?

Thanks in advance

 

[dknguyen]

I was thinking of a short circuit where the voltage tends to be zero and in this situation we replace the voltage source with a wire. But I am not sure if this the key to solve my problem? I have heard the short and open circuits for DC circuits, but I think they can be used here for an AC source!

Now that I am thinking of a short circuit, I am wondering why a voltage source is assumed to be a short circuit when the voltage is zero?

Thanks in advance

A voltage source with zero voltage is a component which forces the voltage across it's terminals to zero. A short circuit (a resistor with zero ohms) does the same thing. So a 0V voltages source behaves like a short circuit. This goes for any voltage source...AC or DC.

Remember that there is a difference between steady state and transient analysis.

 

[Diver300]

The question:
Why the current flows at 0V in this condition?

The current is the capacitance multiplied by the rate of change of voltage. With as sinewave, the rate of change of voltage is the largest when the voltage is zero. The capacitor is charging (or discharging) the fastest at that point so the current is largest.

Now that I am thinking of a short circuit, I am wondering why a voltage source is assumed to be a short circuit when the voltage is zero?

A perfect voltage source has zero impedance. That means that the voltage does not depend on the current taken by the load.

(Real voltage sources have impedance, but in the case of main supplies they are very low and can usually be ignored)

When the voltage is zero, it still has zero impedance, so it can be considered a short.

 

[Electronman]

Thanks for replies,

My problem is just at 0V. Why the current is following at that point? Is the voltage source shorted at 0V? Is there any thing related to the short circuits here?

When the voltage is zero, it still has zero impedance, so it can be considered a short""
Ok an Ideal voltage source has a zero impedance., but can you talk me why the source becomes a short circuit at 0V???

Thanks.

 

[Willbe]

The AC equivalent circuit of an ideal voltage source is a short. It's a little tricky.

 

[Hayato]

Thanks for replies,

My problem is just at 0V. Why the current is following at that point? Is the voltage source shorted at 0V? Is there any thing related to the short circuits here?

When the voltage is zero, it still has zero impedance, so it can be considered a short""
Ok an Ideal voltage source has a zero impedance., but can you talk me why the source becomes a short circuit at 0V???

Thanks.

Well. Simple ohm's law.

U = Ri.
If you have a short circuit, then R will be near 0 ohms, and "i", will be maximum:
U = (near zero)*(very very big) = 0 V.

You'll have the potentials there, but not the difference of potentials. Imagine a simple wire, if you connect a multimeter between the extremes of a wire feeding a shower, you'll read someting near 0V, but even though you have current flowing in the wire (feeding the shower, generating heat).

 

[Electronman]

Hayato,

I know those things, but maybe you are misunderstood of my explanation?

I know when the power supply is shorted the voltage is 0. but Why we have a such situation at AC power supply connected to an cap(please look at my first post aging)?
Here is my questions:
why at 0V AC the power supply is shorted?
According to the story of the cap connected to the sine wave, If you are agree that when the voltage is zero then the current will came back to the cap, please let me know why the power supply is shorted at 0V to allows the current to backs? If I am misunderstood, and the supplier would not be shorted at 0V so please let me know how the current is followed at 0V?

 

[Roff]

Hayato,

I know those things, but maybe you are misunderstood of my explanation?

I know when the power supply is shorted the voltage is 0. but Why we have a such situation at AC power supply connected to an cap(please look at my first post aging)?
Here is my questions:
why at 0V AC the power supply is shorted?
According to the story of the cap connected to the sine wave, If you are agree that when the voltage is zero then the current will came back to the cap, please let me know why the power supply is shorted at 0V to allows the current to backs? If I am misunderstood, and the supplier would not be shorted at 0V so please let me know how the current is followed at 0V?

In order for the capacitor's voltage to change, current must flow through it. The current is proportional to the rate of change of the voltage:

I=C*dv/dt

Are you thinking that a voltage source can't supply any current when its output is zero? This is not true.

 

[Electronman]

Roff, if we short a power supplier with a wire then the WHOLE current will pass through the wire not through the circuit (like the cap in our case)
I think you have found what my problem is though.

Can you tell me if the current of the caps comes back to the power supply at 0V or not?

 

[Hayato]

I really can't understand your doubt.

Capacitor (and inductors) are at imaginary dominium (ideally they do not have real behavior) components.

Why is the current 90° out-of-phase? Because they store energy. Those components have to charge, store energy.

In case of a capacitor, yes, it starts as a short, after it is fully charged, it is a open circuit. But this is in DC/Transient.

As Roff said the current in a capacitor is C*dv/dt.
But in permanent sinoidal state, (the circuit is stabilized, afer turning it on in a sine wave AC), a capacitor is an impedance, that is, a "resistance" an real part and a imaginary part.

Zc = 1/(jwC) ohms and phase = arctan(Xc/R) rads. Xc = 1/(wC) ohms

As you can see, if you have a resistance in parallel/series with the capacitor, the phase will not be 90° anymore.

In AC, there is a bit useless to think in charge/discharge of capacitors/inductors, you consider it as impedances.
But YES, when the AC = Peak, the supply charges the capacito, when AC = 0V, capacitor discharges back to the supply.

 

[duffy]

Electronman - think of the capacitor as a spring. The voltage is how far it is stretched or compressed. The current is how fast it is moving. As you stretch it and compress it, it passes through an instant right in the middle where it is completely relaxed, but it is also moving at its fastest.