# How does $I = \mathrm{d}q/\mathrm{d}t$ work for a capacitor?

When the capacitor is charging in a circuit consisting of a resistor, a capacitor and an alternating sinusoidal generator at $t=0$, the charge across the capacitor is 0 and the current is $I =\mathrm{d}q/\mathrm{d}t$. Does this make the current zero too? While it is max across the resistor in the same circuit and they are connected in series which means that the current should be the same in all the components of the circuit.

The value of $f(t)$ at any specific value of $t$ does not automatically necessitate a value for $\frac{df(t)}{dt}$ at that value of $t$. If $q(t)\propto\sin(\omega t)$ then $\frac{dq}{dt}\propto\omega\cos(\omega t)$, which means while $q(0)=0$, you'd have $\frac{dq}{dt}|_{t=0}\propto\omega$, which would be a max for this function for $q$.

• Then if i want to calculate dq/dt i need to derive at any time not at a certain instant ??? Jan 31 '17 at 14:45

If you apply a sinusoid generator to an RC series circuit, at t=0 the sine(2pi * f * t) is indeed zero. But the voltage on the capacitor is NOT zero, rather the sum of I*R and the capacitor voltage is zero. That's because 'a sine wave' means a sine wave that started long ago, and at time t=0 it just came off a half-cycle of negative voltage excursion. The capacitor has a negative charge at that time.

For the charge to be zero at t=0, perhaps one does not supply a sine wave, but rather a MODULATED sine wave, amplitude zero at all negative times? Capacitors have stored energy, which means memory of the past excitation voltage.

The way to find the charge on the capacitor at t=0 is to do an integral over all past times using sinewave drive, or to use a different function than 'sinewave', which means your analysis will include a turnon transient.

For a capacitor $Q=CV \Rightarrow \dfrac{dQ}{dt} = I = C \dfrac{dV}{dt}$.

When the rate of change of voltage $\dfrac{dV}{dt}$ is a maximum, ie $V=0$ the current $I$ is a maximum.

When the rate of change of voltage $\dfrac{dV}{dt}$ is zero, ie $V= \pm V_{\rm max}$ the current $I$ is zero.

There is a $90^\circ$ phase difference between the current and the voltage with the voltage leading the current.