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Solution via Differential Equations

Let's write v=v_C for short, and

$$i =\frac{v_g-v}{R}$$

KCL (node at top of capacitor) gives

$$\frac{v_g-v}{R} = C \frac{dv}{dt}$$

since the current through the resistor (left) equals the current through the capacitor (right).

Rewriting, we get the differential equation  (for t>0):

$$\frac{dv}{dt} + \frac{v}{RC} = \frac{v_g}{RC}$$ (24)

with initial condition v(0-)=2V.

Let's solve this differential equation. There are many ways of doing this, and you may use your favourite method; here we use a method which applies also to higher-order equations, and is quite general. The solution is the sum

v(t) = v_h(t) + v_p(t)

where

$$v_h(t)=Ke^{-\frac{t}{RC}} \quad\quad {\rm homogeneous \ solution}$$

  and

$$v_p(t) = A \quad {\rm (a\ constant)} \quad\quad {\rm particular \ solution}$$

 

The homogeneous solution follows because the characteristic equation is

$$0 = s + \frac{1}{RC}$$ (25)

 

The constant A defining the particular solution is found by plugging it into the differential equation:

$$0 + \frac{A}{RC} = \frac{5}{RC}$$

since dA/dt = 0 (derivative of a constant is zero) and v_g=5. Solving gives A=5, and so the particular solution is v_p(t)=5 V for $t \geq 0$, and hence the solution at this stage of the computation is

$$v(t)= 5 + Ke^{-\frac{t}{RC}}, t \geq 0 .$$

We can now find the constant K using the initial condition. Set t=0 and note that by the continuity principle for capacitors,

v(0+) = v(0-) = 2V

and so

2 = 5 + K

giving K = -3. Therefore the complete capacitor voltage waveform is

$$v(t) = 5 - 3e^{-\frac{t}{RC}} \ {\rm V}, \ t\geq 0 .$$ (26)

This is illustrated in Figure 26. Looking at the graph we see that the voltage rises from its initial value of 2V to its final steady state value of 5V. The response is transient in nature.

  

Figure 26: First-order RC circuit.

The current can be obtained in either of two ways. The first method uses Ohm's law for capacitors:

$$i(t)=C \frac{dv(t)}{dt} = 3 e^{-\frac{t}{RC}} \quad {\rm mA}, \quad t \geq 0 \eqno{(13)}$$

Alternatively, we can apply KVL:

$$i(t) = \frac{v_g(t)-v(t)}{R} = 3 e^{-\frac{t}{RC}} \quad {\rm mA}, \quad t \ge 0 .$$

The current, whose value at time t=0^- is 0 mA, rises suddenly to 3 mA at time t=0⁺, and then decays to zero as the capacitor is charged.

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