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I'm really confused about the capacitive current leads the voltage, inductive current lags the voltage. The reason for this confusion is the graphs I've seen in different sources trying to explain this, the inconsistancy has really made me question everything I know about voltage and current.

One source used $$ V_R ; V_L ; V_C $$ to show that Vc leads i, VL lags i and VR will be in phase. Another source used V_S and showed the currents respectively leading/lagging the input voltage.

This probably sounds stupid, but it has confused me a lot now, is the voltage across the capacitor in phase with the voltage source, but the current through it is out of phase? Or is it something else?

In my scope readin, I have an RL circuit with the output voltage leading the input voltage, but in an RLC series circuit (below resonance when -X_C > X_L right?) I have a similar output with the output voltage leading the input.

When people say I_c leads V, is the V the voltage across the capacitor or the voltage source?

Edit

Firstly I really apprieciate the answers, all have been a lot of help and its helped me to understand the concept a lot more. I have still some doubts, particularly when it is applied to a real example. I made up an LCR circuit in series. I checked out the maths for frequencies at 4100Hz and 5500Hz (resonance was about 4.81kHz. At 4100Hz, the angle of impedance is \$-82^{\circ}\$, this is what I expected since the \$-X_C > X_L\$ giving \$\theta = \arctan(\frac{X_L - X_C}{R})\$ so \$\theta\$ is negative.I got the info from this site, the current/voltage plots are towards the bottom of the page

Similarly for 5500Hz, \$-X_C < X_L\$ and results in a postive angle of impedance ( \$75^{\circ}\$) as expected.

What I do not understand is how I have a negative angle of impedance, but the phase shift is positive and above resonance, a positive angle but negative impedance. Am I misunderstanding or should I indeed have a lag where it is leading and a lead where I have a lag in the readings below?

Scope readings of bandpass filter at various frequencies

Or should I be taking this a step further and calculating the voltages across the components, as shown in the first example here? using this method, I get a positive angle of \$86^{\circ}\$, however I the capacitor voltage is greater than V across the inductor, which would surely result in a negative phase angle? Likewise, at 5530Hz, I calculate (using the same method) a phase angle of \$76^{\circ}\$, which \$V_L > V_C\$, which surely means a positive phase shift?

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    \$\begingroup\$ If the impedance of the complete circuit is inductive the angle of \$ Z \$ is positive and the current lags the voltage at the supply terminals. \$\endgroup\$ – skvery Feb 28 '17 at 13:14
  • \$\begingroup\$ Ok, this is the case when \$f=5530Hz\$, therefore, when I compare the voltage across V_R and V_s (the resistor is placed after the L and C), I should expect to see an output voltage lagging the input voltage (as in the 3rd scope reading)? Since current and voltage are in phase for R, this has to be the case under these conditions. Have I got that right? \$\endgroup\$ – Maitiu Feb 28 '17 at 13:27
  • \$\begingroup\$ You are correct if the source trace is blue and the resistive trace yellow. If you are saying that the voltage and current in the resistor must be in phase true. (Watch out for high frequencies though.) \$\endgroup\$ – skvery Feb 28 '17 at 13:32
  • \$\begingroup\$ Yes yellow is the Vr and blue Vs. I think I finally understand this concept now. Each post here was a massive help! \$\endgroup\$ – Maitiu Feb 28 '17 at 13:41
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Imagine a capacitor with an existing, stable, unchanging voltage across it. It might be a DC power supply placed across the capacitor, for example, where it's been a long time and the capacitor has "charged up." In this case, there is no current because... well... there's no need for any. The circuit has reached equilibrium. It just sits there.

Now, you turn a knob and the DC power supply changes its voltage. The capacitor must also change, too. (You can't have a power supply with one voltage and a capacitor with a different voltage when they are tied together like this.) But it can't change instantly because the capacitor is a large reservoir of charge, in effect, and to change its voltage you must change that reservoir's "level of charge." To change that, you have to supply (or remove) some charge. But moving charge requires time and together, charge motion and time, you must have current to get there.

So, if you change the voltage then that must stimulate some charge to flow onto, or off of, the capacitor. If you change the voltage slowly, then the rate of change of charge in the capacitor's reservoir over time is less. If you change the voltage rapidly, then the rate of change of charge in the capacitor's reservoir must be more. To achieve a faster rate of change of voltage across the capacitor, you must supply a higher current in order to fill (or drain) the capacitor's storage of charge.

The voltage across a capacitor is: \$V=\frac{Q}{C}\$. So with the capacitance held fixed, to get a higher voltage \$V\$ you need more charge \$Q\$.

Now, when you look at the equation:

$$I = C\cdot \frac{\textrm{d}\:V}{\textrm{d}\:t}$$

You can see all that wonderful hand-waving tied up in a package with nice bow. The current has to be larger if the capacitor has a larger capacitance. Why? Because it is a bigger reservoir and it needs more charge to achieve the same voltage. The current has to be larger also if the rate of change of voltage is more. For reasons just discussed above. That equation puts it all in one place.

Now, what does this mean regarding lagging or leading currents and voltages? Well, take a look at a sine wave centered on \$y=0\$ with voltage on the \$y\$ axis. Then tell me at what value of \$y\$ is the rate of change of the sine wave at its most rapid. It will be when the voltage is itself at zero. In other words, the current into or out of the capacitor will have to be at its maximum value when the voltage across the capacitor is itself at zero (for the sine wave case, anyway.)

The only thing left to worry about is lagging vs leading, and which is which in the case of the capacitor. This is just a matter of sign. In the case of a capacitor, when the voltage is rapidly rising away from zero in the positive going direction, the conventional current into one side of the capacitor is also conventional current away from the other side. You want to look at this as a current "through" the capacitor, even though physical charges don't actually leap through the insulator of the capacitor. So the sign is taken as positive as the above equation suggests.

Now go back and look at those curves you mentioned. You will see that the shape of the current (which obeys all of the above discussion) through the capacitor will "look like" it is \$90^\circ\$ earlier than the voltage curve. You could also claim that it is \$270^\circ\$ later. But to keep things simple everything should be seen as \$-180^\circ \lt \theta \lt 180^\circ\$. (The special case of \$\theta = 180^\circ=-180^\circ\$ is reserved a special term: antiphase.)

So the current is said to "lead" voltage in a capacitor. Or else voltage is said to "lag" current in a capacitor. Either way means the same thing. That it happens to do so by exactly \$90^\circ\$ is only true when you aren't taking into account other "parasitics" such as Ohmic resistance in the wires. Resistors develop a voltage drop across them in strict accordance with the current through them. So when the voltage change attempts to cause a current to flow, the resistor immediately opposes this by developing a voltage drop across it hindering the voltage that is actually then applied to the capacitor. And this fact shifts the lead/lag calculation so that it is no longer \$90^\circ\$, anymore.

There's a lot of algebraic tools of the trade you learn to simplify the work you have to do, just like learning to perform long-hand multiplication is a trick that helps you multiply big numbers without having to perform lots and lots of additions, over and over. These tricks are based upon good theoretical ideas. But if you just learn them and apply them without really understanding where they come from, they will still work for you. Just like you don't need to understand why long-hand multiplication works in order to use it.


The capacitor is easier to describe because it works with attributes that are easier to imagine. We can count them. They are units of charge. An inductor works with equivalent units of charge, but as a matter of magnetism. These units are in Webers, instead of charge. And it's hard to "imagine" a Weber and count them (these are "volt-seconds," or \$\int V_t\:\textrm{d}t\$), for us normal folks. Some people have no problem. Others do. But these are a symmetrical unit for charge.

I'll do a short derivation to explain why, taken from an energy perspective (if there is a founding bedrock principle in physics it is the conservation of energy.) Let's follow here:

$$\begin{split} W &= \frac{1}{2}\:C\: V^2\\\\ \textrm{d} W &= C\: V\:\textrm{d}V + \frac{1}{2}\: V^2\:\textrm{d}C\\\\ \frac{\textrm{d} W}{V} &= C\: \textrm{d}V + \frac{1}{2}\: V\:\textrm{d}C \end{split} \quad\leftrightarrow\quad \begin{split} W &= \frac{1}{2}\:L\: I^2\\\\ \textrm{d} W &= L\: I\:\textrm{d}I + \frac{1}{2}\: I^2\:\textrm{d}L\\\\ \frac{\textrm{d} W}{I} &= L\: \textrm{d}I + \frac{1}{2}\: I\:\textrm{d}L \end{split}$$

noting,

$$\textrm{where } I=\frac{\textrm{d}Q}{\textrm{d}t}\textrm{ and } V=\frac{\textrm{d}W}{\textrm{d}Q}\textrm{ and d}L=0 \textrm{ and d}C=0$$

resulting in,

$$\begin{split} \frac{\textrm{d} W}{\frac{\textrm{d}W}{\textrm{d}Q}} &= C\: \textrm{d}V \\\\ \frac{\textrm{d} W}{\textrm{d}W}\textrm{d}Q &= C\: \textrm{d}V \\\\ \textrm{d} Q &= C\: \textrm{d}V\\\\ \int\textrm{d} Q &= \int C\: \textrm{d}V\\\\ Q &= C\: V \end{split} \quad\leftrightarrow\quad \begin{split} \frac{\textrm{d} W}{\frac{\textrm{d}Q}{\textrm{d}t}}&= L\: \textrm{d}I\\\\ \frac{\textrm{d} W}{\textrm{d}Q}\textrm{d}t &= L\: \textrm{d}I\\\\ V\textrm{d}t &= L\: \textrm{d}I\\\\ \int V\textrm{d}t &= \int L\: \textrm{d}I\\\\ V\:t &= L\: I \end{split}$$

And there you are. Countable things on the left. But weird units on the right. Volt-seconds (Webers) are to inductors as Coulombs of charge are to capacitors.

Another bit of slight of hand from the above can be had, as well:

$$\begin{split} \textrm{d} Q &= C\: \textrm{d}V\\\\ \frac{\textrm{d} Q}{\textrm{d} t} &= C\: \frac{\textrm{d}V}{\textrm{d} t}\\\\ I &= C\: \frac{\textrm{d}V}{\textrm{d} t} \end{split} \quad\leftrightarrow\quad \begin{split} V\textrm{d}t &= L\: \textrm{d}I\\\\ \frac{V\textrm{d}t}{\textrm{d} t} &= L\: \frac{\textrm{d}I}{\textrm{d} t}\\\\ V &= L\: \frac{\textrm{d}I}{\textrm{d} t} \end{split}$$

Sorry about the diversion. But I thought it may help some (and you?)

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When one says "the current in a resistor is in phase with the voltage" he thinks only a resistor that is connected to an AC voltage supply (ideal or not, no matter). He does not think a filter, amp or other circuit that has an input voltage and the resulted output voltage. The only thing that he tells is "the changes in the current follow the changes in the voltage without any timing difference, the minimum or the maximum are in both at the same time"

When one says that inductive current lags the voltage, he thinks an AC voltage source and an inductor, no inputs and no outputs. He tells that the current is somehow late when compared to the voltage. The voltage can already have passed its peak and is falling, but the current is still growing. If the voltage is sinusoidal, then the current is excactly 1/4 of the cycle duration behind the voltage or as we say "lagging 90 degrees "

In capacitor the scenario is opposite. The current seems somehow to rush before the voltage. If the voltage is sinusoidal, then the current is sinusoidal, too and its timing is exactly 1/4 cycle in front of the voltage.

These can be expressed exactly in math. We have differential equations for arbitary changing voltages and currents. The complex phasor and impedance notations are derived from those differential equations for sinusoidal voltages and currents.

In filtering circuits that have input and output voltages, one can compare the timing difference between the voltages. If the input is sinusoidal and the circuit makes no distortion, the output is also sinusoidal, but often its timing is in front (=leading) or behind (=lagging) the timing of the input voltage and the difference does not change as long as the frequency doesn't change. The lead or lag is usually expressed as degrees and called "phase shift".

How much a filtering circuit causes phase shift at a given frequency? That can be solved by the means of the "circuit analysis".

Your example RL filter where R is between the input and output, the phase shift is positive. It means thet the output voltage is in front of the input voltage. The filter cause leading. How much - that depends of the frequency and the component values.

ADDENDUM the phase shift calculation between resistor's and inductor's voltages in the RL series circuit

Let's assume voltage V1 to be sinusoidal => the current I is sinusoidal, the complex phasors can be used.

enter image description here

This calculation shows a basic fact that nearly all experienced electronics enthusiasts know without any calculations.

A little more useful would be to continue and find the phase shift and amplitude ratio between V1 and VL because VL can be considered as the output voltage of a RL filter.

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  • \$\begingroup\$ This makes sense, but for a couple of things. So to speak, is it as if the current is 'universal' throughout the circuit, i.e. the current through a resistor will be in the same phase as that through the inductor, but it is the voltages across them that change? In other words there is never a phase shift in current? In my RL circuit, I understand from the maths the phase shift should be positive (as it is an inductor) and from this, the output of an RL circuit should always lead the input voltage? \$\endgroup\$ – Maitiu Feb 28 '17 at 0:32
  • \$\begingroup\$ @Maitiu If the same current (=no branching between!) goes through a resistor and an inductor, the voltage over the inductor is 90 degrees to forward phase shifted when compared to the voltage over the resistor. PROOF: no phase shift between the current and voltage in the resistor, 90 degrees phase shift between the voltage and current in the inductor, the current is the same. I'll add the math to my answer. \$\endgroup\$ – user287001 Feb 28 '17 at 0:48
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The "lag" and "lead" are relative to the currents and voltages at the terminals of the component. Let us start with the plain basic differential equations for each:

$$i(t) = C{d v(t)\over dt}$$ and $$v(t) = L{d i(t)\over dt}.$$

The first equation says that if the voltage is proportional to \$\sin(\omega t)\$ the current is proportional to \${{d \sin(\omega t)}\over d t} \propto \cos(\omega t)\$ and the current leads the voltage.

The second equation says that if the current is proportional to \$\sin(\omega t)\$ the voltage is proportional to \${{d \sin(\omega t)}\over d t} \propto \cos(\omega t)\$ and the voltage leads the current, or the current lags the voltage.

These differential equations are universal for any voltage and current at the component, also for distorted currents and voltages.

What happens in a circuit depends on the connections and any circuit can be analysed by following these two simple rules with normal circuit analysis.

For completeness for sinusoidal currents and voltages (and edited as per comment by @user287001) :

$$i(t) = I\cos(\omega t) ={C{{d V\sin(\omega t)}\over d t}} = \omega C\cdot V\cos(\omega t) $$ and $$v(t) = V \cos(\omega t) = {L{{d I \sin(\omega t)}\over d t}} = \omega L \cdot I\cos(\omega t)$$ proving that \$ X_L = \omega L\$ and \$X_C = {1\over{\omega C}}\$.

(Oops? Where does the \$\sqrt{-1}\$ fit in? :-)

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    \$\begingroup\$ Sorry, but your "for completeness" part is not a proof. It's a symbol manipulation sequence . The step j=cos(wt)/1 is something out of the math. In math j is the square root of minus 1. \$\endgroup\$ – user287001 Feb 27 '17 at 23:56
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Start by imagining a capacitor connected to a voltage source. The voltage of the voltage source doesn't change no matter what load you present to it, including that load being a capacitor. The only question is what the current will be.

The current thru a capacitor is always the derivative of the voltage across it. That's what capacitors do. There is one degree of freedom, which is how much current you get for any particular voltage slope. That scale factor to go from Volts/second to Amps is the capacitance. In fact, one Farad is defined:

  Farad = Amp / (Volt/second) = Amp second / Volt

Conversely, the current thru a capacitor (as we are trying to find above) is:

  Amps = Farad Volt / second

Therefore, if you have a 1 F capacitor in the setup described above, and the voltage is ramped up by 1 V/s, then the capacitor will draw 1 A.

So what does this have to do with leading and lagging voltage? Using the terms "leading" and "lagging" implies a periodic waveform. From Fourier analysis we know that those can be decomposed into a series of sines. It is therefore useful to consider what happens when the voltage source in our example produces a sine voltage.

The derivative of a sine is a cosine. The capacitor current will be a cosine while the voltage across it is a sine. Since a cosine can also be thought of as a sine with -90° phase shift, we can say that in the case of a sine, the current thru a capacitor leads the voltage across it by 90°.

Unfortunately, especially for beginners, it is usually not pointed out that the signals are assumed to be sinusoids. The current thru a cap doesn't "lead" the voltage for arbitrary signals. This is a special case that applies only to sinusoids.

Think of the voltage being a triangle wave instead of a sine. The current thru the cap will be two separate fixed values, one for each of the two fixed slopes of the triangle. In other words, the current will be a square wave.

Now try putting a square wave across the cap. The equations blow up since the derivative at the steps are infinite. In theory, with a ideal voltage source and ideal capacitor, the current really would be infinite. However, in real life the non-idealness will get in the way, and the voltage steps will be somewhat smushed out, and the current will be high blips at the edges.

Inductors work similarly, except that the voltage is the derivative of current, instead of the other way around. If you do the math with a fixed sine voltage, you will see that the current is -cosine, or has a +90° phase shift from the voltage, or in other words lags the voltage by 90°. Again, this only applies to sinusoids.

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