In a circuit with just a resistor and a capacitor I'm trying to figure out which voltage is being referred to that is lagging the current(which is the same current throughout the entire SERIES AC circuit). The voltage which leads or lags is the same voltage referred to by $$I(t)=C\frac{dV(t)}{dt}$$ which comes from the derivative of $$V(t) = \frac{Q(t)}{C} = \frac{1}{C}\int_{t_0}^t I(\tau) \mathrm{d}\tau + V(t_0)$$ Therefore the voltage which lags is the voltage drop across the capacitor because the charges are added above to the plates of the capacitor so the voltage refers to it. In an AC circuit, the voltage source is forced to alternate with a cosine wave and the phase difference between the source current driven by $$V_0\cos(\omega t)\tag{1}$$ and the voltage which I'm asking about comes from: $$I = C \frac{dV}{dt} = -\omega {C}{V_\text{0}}\sin(\omega t)\tag{2}$$ which is the same as $$I = {I_\text{0}}{\cos({\omega t} + {90^\circ})}$$

The voltage used in this formula was the source voltage not the voltage drop across the capacitor which defines Capacitance. The voltage across the capacitor is not instantaneous and in fact exponentially decays up to the applied voltage as shown by a constant DC voltage circuit where : $$V_0 = v_\text{resistor}(t) + v_\text{capacitor}(t) = i(t)R + \frac{1}{C}\int_{t_0}^t i(\tau) \mathrm{d}\tau$$ Taking the derivative: $$RC\frac{\mathrm{d}i(t)}{\mathrm{d}t} + i(t) = 0$$ Solving the first order: $$I(t) = \frac{V_0}{R} \cdot e^{\frac{-t}{\tau_0}}$$ Assuming initially the resitor is $V_0$ the voltage of capacitor: $$V(t) = V_0 \left( 1 - e^{\frac{-t}{\tau_0}}\right)$$

Thus I'm confused about where the $90^\circ$ voltage lag comes from. If it's because of the derivative of the source voltage why is formula 2 even applicable to the source voltage. Second question: What is the formula for the voltage reached by the capacitor in an ac circuit. It appears as if it is the source max voltage but I don't believe/understand that. Here is an identically solved using a sin source Voltage:

enter image description here $$I_C+I_{max}\sin(\omega t +90^\circ)$$ In the above derivation, the source voltage is again mixed with the formula for the voltage stored across a capacitor or I'm to believe the maximum source voltage is somehow reached on the exponential decay to the voltage on a capacitor during a cycle.

Could someone either explain why the source voltage is used as if it was the capacitor voltage or the identically reversed inductor or refer me to a source that explains it?




  • $\begingroup$ Note that, for AC analysis, sinusoidal steady state is assumed, i.e, all transients have decayed away. You seem to be mixing AC analysis and transient analysis and this is most likely causing some confusion. $\endgroup$ Commented Jul 6, 2017 at 10:50

2 Answers 2


The defining equation for a capacitor is $Q=CV_{\rm C}$ and when that equation is differentiated with respect to time one gets $\dfrac{dQ}{dt} = I = C\dfrac{dV_{\rm C}}{dt}$
So the current is proportional to the rate of change of voltage across the capacitor Applying a sinusoidal voltage to a capacitor results it the following current and voltage graphs.

enter image description here

Notice that the current is determined by the gradient of the voltage against time graph. being a maximum at time $a$ and zero at times $b$and $d$.
Whatever the current is doing the voltage does a quarter of a period (equivalent to $90^\circ$) later.
So the current is a maximum at time $a$ and the voltage is a maximum at a later time $b$.
We say that current leads the voltage across a capacitor by $90^\circ$.
In the graph $V_{\rm C}(t)=V_{\rm max} \sin \omega t$ and so the current is $I(t) =\omega CV_{\rm max} \cos\omega t$ with a peak current $I_{\rm max}=\omega CV_{\rm max}$.

When you add a series resistor to the circuit the current is the same in all parts of the circuit.
The voltage across the capacitor still lags the current by $90^\circ$ and the voltage across the resistor will be in phase with the current.
The (applied) voltage across both components will lag the current through the circuit at some value between $0^\circ$ and $90^\circ$ depending on the values of the capacitance of the capacitor, the resistance of the resistor and the frequency of the applied voltage.

Here I have considered what are called steady state conditions and so there are no transients which would be characterised by an exponential function and a time constant.

The difference for an inductor is that the defining equation is $V_{\rm L} = L \dfrac{dI}{dt}$ and the voltage across an inductor leads the current by $90^\circ$.

Update as a result of a comment

I think that what you are asking about is the transient behaviour which occurs when you first connect a capacitor to the voltage source. If by chance you make this connection to an uncharged capacitor when the voltage of the supply is zero then there is no transient and the circuit currents and voltages are as per the graph shown above. If on the other that is not so you will have a combination of the transient (the exponential function you have described) and the steady state. After about 10 time constants (10CR) the transients would have decayed away and all that is left is steady state

Now with an "ideal" circuit with no resistance the time constant is zero and the circuit settles down to steady state behaviour "instantly". However with a finite resistance in the circuit then there will be a transient behaviour which you tend to to see because it decays away.

I can show you this idea of a transient in action by using the "Circuit Sandbox" which is available in the edX Circuits and Electronics course, a course I thoroughly recommend even if it just to be able to used the circuit simulator.

Here is the result of a simulation where there is dc voltage of 1 V across a capacitor and after one second a sinusoidal voltage of peak value 1 V and frequency 1 Hz is applied across a resistor and a capacitor connected in series.
The graph is voltage across the capacitor in volts against time in seconds.

You can see very clearly the transient behaviour (the exponential decay) and then the steady state behaviour.

enter image description here

The 10RC s just a rule of thumb where $e^{-10} \approx 4.5 \times 10^{-5}$ and the decay has effective finished.
Others use 5RC which corresponds to a decrease of $e^{-5} \approx 6.7 \times 10^{-3}$.

Update 2
Here is the supply voltage shown in red and the voltage across the capacitor shown in cyan.
The supply voltage and voltage across the capacitor start at $+1 \, \rm V$ and then a $\pm 1 \, \rm V$ sinusoidal voltage is added after 1 second.
It clearly shows the $90^\circ$ phase shift.

enter image description here

  • $\begingroup$ This is very good but it explains everything that I already learned and identically skips the critical information that all the text books I consulted also skip. You say "So the current is proportional to the rate of change of voltage across the capacitor". The voltage of the capacitor is from the charge on the plates which has thus far accumulated over time which is different from the sinusoidal source voltage which is driven by some form of EM generator. When the applied voltage is say 1, the voltage caused by accumulated electrons on the capacitor could start out at 0. $\endgroup$ Commented Jul 6, 2017 at 9:38
  • $\begingroup$ By the time the applied voltage equals zero(if frequency=4), enough electrons are on the capacitor to still be less than (1-e^{-1}) in a 1 ohm 1F circuit had the voltage been constant. The question is why is the sinusoidal voltage applied, plugged into the formula which represents the voltage caused by electrons on the plates of the capacitor $I=CV'$ They are two different values from two unrelated processes. $\endgroup$ Commented Jul 6, 2017 at 9:38
  • $\begingroup$ I see, so the capacitor will have enough electrons on it that if I put a voltmeter across it (eventually) during one time cycle the voltage of the electrons on the plates WILL equal the max voltage applied by the source at exactly $1/4$ and $3/4$ of the cycle through? That means the electrons on the capacitor get on one plate and then on the other plate at the same rate as the voltage cycles, regardless of how great the voltage is that is applied. You said "After about 10 time constants (10CR)" do you know of a book that does these obscure calculations of that nature that I can read? $\endgroup$ Commented Jul 6, 2017 at 9:56
  • 1
    $\begingroup$ Because the ac has a +1 V dc offset. $\endgroup$
    – Farcher
    Commented Jul 6, 2017 at 10:44
  • 1
    $\begingroup$ $\pm 1 \,\rm V$ and remember that it is the voltage across the capacitor which has been graphed. I can add the supply voltage graph if you wish? $\endgroup$
    – Farcher
    Commented Jul 6, 2017 at 10:59

When we have a series RC circuit we know that:


We now can use:

  • $$\text{V}_\text{R}\left(t\right)=\text{I}_\text{R}\left(t\right)\cdot\text{R}\tag2$$
  • $$\text{I}_\text{C}\left(t\right)=\text{V}_\text{C}'\left(t\right)\cdot\text{C}\tag3$$

So, we get:


And for the current we can write:


So, we end up with:


Now, when we have an input voltage that looks like:

$$\text{V}_\text{in}\left(t\right)=\hat{\text{u}}\cdot\cos\left(\omega t+\varphi\right)\tag7$$

We find:

$$\text{V}_\text{in}'\left(t\right)=\frac{\partial}{\partial t}\left(\hat{\text{u}}\cdot\cos\left(\omega t+\varphi\right)\right)=\text{I}_\text{in}'\left(t\right)\cdot\text{R}+\text{I}_\text{in}\left(t\right)\cdot\frac{1}{\text{C}}\tag8$$

You can use, for example, Laplace transform to solve $\left(8\right)$ in the general case.

Assuming that $\varphi=0$, $\text{I}_\text{in}\left(0\right)=0$ and $\hat{\text{u}}=1$:

$$\text{I}_\text{in}\left(t\right)=\frac{\frac{\text{W}}{\text{R}}\cdot\left(\text{W}\cdot\left(\cos\left(\omega t\right)-\exp\left(-t\cdot\frac{\omega}{\text{W}}\right)\right)-\sin\left(\omega t\right)\right)}{1+\text{W}^2}\tag9$$

Where $\text{W}=\text{RC}\omega$

  • $\begingroup$ This is very interesting. I wonder why the current is different from $\frac{û\cdot \cos(\omega t+\phi)}{\sqrt{R^2+(\omega L-\frac{1}{\omega C})^2}}$ $\endgroup$ Commented Jul 6, 2017 at 20:39

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