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We were taught about single phase series circuits today and I had a bit of a problem understand this little thing. Let's look at this case

1) purely capacitive circuit

It's said that current leads the voltage by an angle of 90°. So at t= 0 voltage is zero, but current is non zero. I cannot imagine this happening - it's wierd.

2) purely inductive circuit

Voltage leads the current by an angle of 90°, so there's a small period of time where there's no current inspite of a potential difference. And the current comes up only when the voltage is at wt = 90°. Again, I wonder how this is possible.

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    \$\begingroup\$ Did you learn that Capacitors and Inductor store energy? \$\endgroup\$ – Tyler Oct 21 '17 at 17:42
  • \$\begingroup\$ But what does energy have to do with this? \$\endgroup\$ – user406653 Oct 21 '17 at 19:02
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Isn't it clear that in AC circuits, the voltage swings both positive and negative continually, and so does the current?

So, there always has to be an instant between positive voltage and negative voltage when the voltage is zero, no matter how brief. Same thing for current.

With resistors, voltage and current are always proportional to each other, so they have to be zero at the same time. Inductors and capacitors store energy, so the voltage zero and the current zero do NOT have to occur at the same time — and in most cases, they cannot occur at the same time.

The governing equations are

$$V(t) = L\frac{dI(t)}{dt}$$

and

$$I(t) = C\frac{dV(t)}{dt}$$

Loosely translated, the voltage on a coil is proportional to the rate of change of the current through it, and the current through a capacitor is proportional to the rate of change of voltage across it. Continuous sine waves (and cosine waves) constitute just one of many possible solutions to this set of differential equations, and their zeros are 90° apart.

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the I-V relation for inductors and capacitors is: V= L (dI/dt) for inductor I= C (dV/dt) for capacitor

Note that in those expressions you have a derivative, not as Resistors where V=R*I. In resistors you can infer about the voltage value if you know the instantaneous value of I.

In the case of inductors you will get voltage if the current changes over time (i.e. dI/dt is not zero). If your current is I=cos(wt) then your voltage will be V=L(-sin(wt))=L(cos(w*t+90°)) Then we say that Voltage leads the current by an angle of 90°.

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This a non-mathematical approach.

1) purely capacitive circuit

It's said that current leads the voltage by an angle of 90°. So at t= 0 voltage is zero, but current is non zero. I cannot imagine this happening - it's wierd.

You are correct. The statement omits to mention that t = 0 is not the instant the circuit is switched on. It is a reference point sometime after the circuit has become stable.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. A simple capacitor circuit fed at 1 kHz.

enter image description here

Figure 2. The simulation of Figure 1 shows that at t = 0 both V and I are zero.

  • At t = 0 we see that V = 0 and I = 0.
  • For this setup the circuit seems to have settled down to normal by the second cycle.
  • Note that at \$ I_{MAX} \$ the rate of change of voltage, \$ \frac {dV}{dt} \$ is also greatest. This is what we would expect. If we were filling a tank of water the rate of change of height would be when the inward flow is maximum.
  • Note that at \$ V_{MAX} \$ the current has fallen to zero.

2) purely inductive circuit

Voltage leads the current by an angle of 90°, so there's a small period of time where there's no current in spite of a potential difference. And the current comes up only when the voltage is at wt = 90°. Again, I wonder how this is possible.

Again in this situation the

schematic

simulate this circuit

Figure 3. A simple inductor circuit fed at 1 kHz.

enter image description here

Figure 4. In the inductor V leads I.

Unfortunately the simulator doesn't start properly with an inductor but the principle is similar but complimentary to that of the capacitor.

  • As you stated, at maximum voltage the current is zero (first arrow in Figure 4). This is correct but notice that the rate of change of current is maximum.
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  • \$\begingroup\$ "The statement omits to mention that t = 0 is not the instant the circuit is switched on. It is a reference point sometime after the circuit has become stable" - yes this is what I wanted to hear. Thanks! For the graphs too! :) \$\endgroup\$ – user406653 Oct 21 '17 at 18:59
  • \$\begingroup\$ So in the inductance case does the inductance jump from 0 to max in a really small amount of time? \$\endgroup\$ – user406653 Oct 21 '17 at 19:01
  • \$\begingroup\$ No, the inductance is constant. Only V and I change. I think the simulator is doing the \$ I \$ calculation first, seeing the maximum rate of change as it crosses zero and since \$ V = L \frac {dI}{dt} \$ the voltage must jump. In a real-life situation the I and V values depend on what part of the mains cycle you switch on at. I'll try to think of an analogy ... \$\endgroup\$ – Transistor Oct 21 '17 at 19:23

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