Rate of charging the capacitor

Question

Why does the rate at which a capacitor in an RC circuit charges, only depend on $R$ and $C$? We know that for power $$P = \frac{V_{\text{emf}}^2}{R} e^{-t/RC}=\frac{QI}{C} $$ so why are we ignoring the effect of changing $V_\text{emf}$ on the current and power?

Answer 1

Of course, in an ideal RC circuit the capacitor never fully charges. This is because the factor of $1-e^{-t/RC}$ can never be exactly equal to $1$ in the expression for the charge on the capacitor $$q(t)=CV_\text{emf}\left(1-e^{-t/RC}\right)$$

Therefore, we need to define some fraction $\rho$ of $CV_\text{emf}$ as the threshold of "the capacitor is charged". Then we can determine the time it takes for us to reach this threshold starting with a completely uncharged capacitor: $$t_\text{charged}=RC\ln\left(\frac{1}{1-\rho}\right)$$

So we do see that this charging time only depends on $R$ and $C$ for a given charging threshold. Notice how the dependence of $V_\text{emf}$ dropped out because the threshold for charging also depends on $V_\text{emf}$ in the same exact way. In other words, $V_\text{emf}$ determines the maximum amount of charge that will end up on the capacitor. So while the power and current will indeed be lessened for a smaller $V_\text{emf}$, the amount of charge to reach the threshold is also lessened.

Of course, the current in the circuit does depend on $V_\text{emf}$, since $$i(t)=\dot q=\frac{V_\text{emf}}{R}e^{-t/RC}$$ So, if you are taking "rate of charging" to just mean the current, then the rate does depend on $V_\text{emf}$. But I would argue that the rate of charging is considering the rate at which you get closer to being fully charged. And, as we have shown above, this does not change.

Answer 2

Why does the rate at which capacitor in an RC circuit charges only depend on R and C

But the rate at which a capacitor charges does depend on $V_\mathbf{emf}$ if, by this, you mean the rate of change of $Q$. For a series RC circuit with zero initial voltage, we have

$$\frac{dQ}{dt} = i_C(t) = \frac{V_\mathbf{emf}}{R}e^{-t/RC}$$

Clearly, the rate at which the capacitor charges depends on $V_\mathbf{emf}$, $R$, and $C$.

Now, the time constant depends only on the product $RC$. One can see this on dimensional grounds - the time constant must have dimension of time and, sure enough, the product of resistance and capacitance has dimension of time. If $V_\mathbf{emf}$ were to be a factor in the time constant, we would need some physical parameter to cancel the units of volts.

The voltage across the capacitor is also a function of the applied voltage:

$$v_C(t) = V_\mathbf{emf}\left(1 - e^{-t/RC}\right)$$

But, if you're interested in is how long will it take for the capacitor voltage to reach some fraction of the final value, then you would of course divide this equation by the final value of $v_C(t)$:

$$\frac{v_C(t)}{v_C(\infty)} = \frac{V_\mathbf{emf}}{V_\mathbf{emf}}\left(1 - e^{-t/RC}\right)$$

and then solve. Clearly, this does not depend on the applied voltage since we're interested in a dimensionless number - the fraction of the final voltage rather than the absolute voltage.

For a simple exercise, how long does it take for the capacitor voltage to reach $(1 - 1/e) \approx 63.2\%$ of the final value? Note that we're not asking for the time it takes to get to some voltage - that answer would depend on the applied voltage.