Will the capacitors in the circuits shown in Fig. Q26.18 charge at the same rate when the switch S is closed? If not, in which circuit will the capacitors charge more rapidly? Explain.

In each of the two circuits below, two capacitors and a resistor are used, in the first two capacitors are linked in series, so the equivalent capacitance is:

$\frac{1}{C_{series}}=\frac{1}{C_1}+\frac{1}{C_2}$

Here both capacitors have the same capacitance C then:

width="130">1Cseries=1C+1C=2CCseries=C2

And in the second circuit, the capacitors are connected in parallel combination, the equivalent capacitance is:

$C_{parallel}=C_1+C_2$

Here both capacitors have the same capacitance C then:

$$\begin{gathered} C_{parallel}=C_1+C_2 \\ =C+C \\ =2C \end{gathered}$$

The charge over the capacitor in the RC circuit is;

$q=C\epsilon \left(1-e^{-\frac{t}{RC}}\right)$

The time derivative of the charge is the rate at which the capacitor charges, so the charge is:

$$\begin{gathered} i=\frac{dq}{dt} \\ =\frac{d}{dt}\left(C\epsilon \left(1-e^{-\frac{t}{RC}}\right)\right) \\ =C\epsilon \left(1-e^{-\frac{t}{RC}}-\frac{1}{RC}\right) \\ i=\frac{\epsilon }{R}{e}^{-\frac{t}{RC}} \end{gathered}$$

For the series combination:

$i_{series}={\frac{\epsilon }{Re}}^{\frac{-2t}{RC}}$

For the parallel combination:

$i_{parallel}=\frac{\epsilon }{R}{e}^{-\frac{2t}{RC}}$

The rate of charge flow declines slower in the parallel combination than in the series combination, implying that the parallel capacitor will be charged slower than the series capacitor. The higher capacitance will lead to a slower rate of charging.

Capacitors in series combinations are like resistors in parallel combinations. With the same current, they charge faster.

The current is inversely proportional to the charging process, peaking at t=0 and vanishing at $t\to \infty$. Thus, the equation of charge can be used to calculate the amount of charge in the capacitor in both arrangements after time t.