Capacitors - 3

In response to Arth Singh . This is correct. Although I was trained an electrical engineer, it has been years that I didn't touch such problems so I decided to solve it from first principal using calculus as follows.

Let the source voltage and the instantaneous voltage across capacitor be $V_s$ and $V(t)$ respectively. From the definition of capacitance, we have charge on capacitor $q(t) = CV(t)$ , therefore current through capacitor $I(t) = C \dfrac{dV(t)}{dt}$ .

From the circuit we have:

$\begin{aligned} RI(t) + V(t) & = V_s \\ RC\dfrac{dV}{dt} + V & = V_s \\ \Rightarrow V(t) & = V_s \left(1-e^{-\frac{t}{RC}}\right) \\ \Rightarrow I(t) & = CV_s \frac{d}{dt} \left(1-e^{-\frac{t}{RC}}\right) = \frac{V_s e^{-\frac{t}{RC}}}{R} \end{aligned}$

The charge flows through the capacitor from switch on to steady state:

$\begin{aligned} Q & = \int_0^\infty I(t) dt = \int_0^\infty \frac{V_s e^{-\frac{t}{RC}}}{R} dt = \left[ -CV_se^{-\frac{t}{RC}} \right]_0^\infty = \boxed{CV_s} \end{aligned}$

Let's consider the capacitor. At the initial state, both plates are uncharged. At steady state, the left plate has $+ Q$ and the right plate has $- Q$ (where Q is positive). The total magnitude of charge that passes through the battery is $Q$ . We will now find the value of $Q$ using the following formula.

$Q = C \times V$

The capacitance of the capacitor is $5 \text{ μC} = 5 \times 10^{-6} \text{ C}$ . Since no current flows through the circuit at steady state, the potential difference across the resistor is zero and the potential difference across the capacitor is $6 \text{ V}$ . Thus, the total charge that flows through the battery is $5 \times 10^{-6} \times 6 = 30 \times 10 ^{-6} \text{ C}$ . $_\square$