R-C Circuits

An RC circuit is shown in fig.7.1. Since, in practical circuits, power is always switched on at certain time, a switch is provided here. This switch closes at time $t=0$.

Figure 7.1: An RC Circuit

We are interested in finding how voltage across capacitor $v_c (t)$ changes with time? We can also assume that voltage across the capacitor is zero $t < 0$. Using Kirchoff's voltage law across the only loop in circuit we can find the equation relating $v_c$, $v_r$ and $V$.

Using the characterstic equations of capacitors, resistors i.e.,

$$C\frac{dv}{dt}=i$$

$$iR=v_r$$

and using KVL

$$v_r + v_c=V$$

$$\frac{dV}{dt}=\frac{dv_r}{dt}+\frac{dv_c}{dt}=R\frac{di}{dt}+\frac{dv_c}{dt}$$

$$\frac{dV}{dt}=R\frac{di}{dt}+\frac{i}{C} t>0$$

$$t<0 i(t)=0$$

$$RC\frac{di}{dt}+i=c\frac{dv}{dt}$$

$$RC e^{\frac{t}{RC}}\frac{di}{dt}+e^{\frac{t}{RC}}i=C e^{\frac{t}{RC}}\frac{dV}{dt}$$

$$t\geq 0 V=$$

$$\frac{dV}{dt}=0$$

$$\frac{d(i e^{\frac{t}{RC}})}{dt}=0\Rightarrow ie^{\frac{t}{RC}}=K$$

$$i(t)=Ke^{- \frac{t}{RC}} K$$

At $t=0$, capacitor voltage will be 0. Hence $i(t=0)=\frac{V}{R}e^{-\frac{t}{RC}}$

Alternatively,

$$V_c=V-iR$$

$$V_c=V-RK e^{- \frac{t}{RC}}$$

$$t=0, v_c=0$$

$$K=\frac{V}{R}$$

Thus,

$$V_c=V(1- e^{-\frac{t}{RC}})$$ (7.1)

The curves showing $v_c (t)$ and $i(t)$ are shown in the figures 7.2 and 7.3.

Figure 7.2: i vs t

Figure 7.3: $V_c$ vs t

These show the exponetially decaying (growth) nature of current (voltage across capacitor).

Consider the figure shown in 7.1. The switch is closed at $t=0$.

Now,

$$V=v_c+iR$$

$$V=\frac{1}{C}\int_{-\infty}^{t}i\;dt+iR$$

$$\frac{dV}{dt}=\frac{1}{C}i+R\frac{di}{dt}$$

$$e^{\frac{t}{RC}}\frac{dv}{dt}=e^{\frac{t}{RC}}\frac{i}{C}+Re^{\frac{t}{RC}}\frac{di}{dt}$$

$$0=e^{\frac{t}{RC}}\frac{dv}{dt}=\frac{d(Rie^{\frac{t}{RC}})}{dt}$$

$$Integrating\;with\;proper\;limits,\;we\;get\;i(t)=i_0e^{-\frac{t}{RC}}$$

$$and\;\;v_C(t)=V-Ri(t)=V-Ri_0e^{-\frac{t}{RC}}$$

$$v_C(t)=V-Ri_0$$

$$Ri_0=V-V_c(0)$$

$$\Rightarrow v_C(t)=V_F-(V_F-v_c(0))e^{-\frac{t}{RC}}$$

For RC circuit with source voltage zero, and an initial capacitor voltage of $v(0)$, this expression reduces to $v(t)=v(0)e^{-\frac{t}{RC}}$.

For constant current charging of a capacitor, as shown in 7.4, the analysis:

Figure 7.4: Constant current excitation of a capacitor

$$C\frac{dv}{dt}=i=I$$ (7.2)

$$\Rightarrow V=\frac{I}{C}t$$ (7.3)

That is, voltage varies linearly with time on constant current charging.

Figure 7.5:

$$R_1i_1 + \frac{1}{C_1}\int_0^t (i_1-i_2)dt\; = \;0$$

$$\frac{1}{C_1}\int_0^t(i_2-i_1)dt\;+\;i_2R_2\;+\;\frac{1}{C_2}\; = \;0$$

$$R_1\frac{di_1}{dt}\;+\frac{1}{C_2}(i_i)-\frac{1}{C_1}i_2\; = \;0$$

$$\frac{1}{C_1}(i_1-i_2)\;+\;\frac{1}{C_2}i_2\;+\;\frac{di_2}{dt}\; = \;0$$

$$\mathrm{From the above, we have}$$

$$R_1C_1\frac{di_1}{dt} + i_1\; = i_2$$

$$\frac{1}{C_1}( R_1C_1\frac{di_1}{dt} ) + \frac{1}{C_2}(R_1C_1\frac{di_1}{dt} + i_1) + R_2R_1C_1\frac{d^ti_1}{dT^2} + R_2\frac{di_1}{dt}\; =$$ 0

$$R_1R_2C_1\frac{d^2i_1}{dt^2} + (\frac{R_1C_1}{C_1} + \frac{R_1C_1}{C_2} + R_2)\frac{di_1}{dt} + \frac{1}{C_2}i_1\; = \;0$$

Now consider the circuit shown in figure 7.6

Figure 7.6:

The switch is turned off at $t = 5$ sec. There is no charge on the capacitor initially. Therefore, after $t = 0$ and before $t = 5$, the circuit is equivalent to figure 7.7

Figure 7.7:

Taking thevenin equivalent in the direction of the arrow leads to figure 7.8

Figure 7.8:

Therefore ,

$$V_F =5\mathrm{ V}\quad,\quad RC\; = \;2.5\times10$$

$$= 25$$

For $v_c(r\mathrm{sec})$, we have the following equation

$$\Rightarrow\qquad v_c(5\mathrm{sec})\; = 5 + (0 - 5)e^{-\frac{t}{25}}$$

$$= \;5(1-e^{-\frac{5}{25}})$$

$$= \;0.906 \mathrm{ volts}$$

After $t = 5 \mathrm{sec}$, the switch is once again thrown open and the equivalent circuit is shown in figure 7.9

Figure 7.9:

Now,

$$v_F\;\;=\;\;0\qquad,\qquad v_I\;\;=\;\;0.906$$

$$v_c(t)\; = \;0 + (0.906-0) e^{-\frac{t-5}{30}}$$

$$= \;0.906 e^{-\frac{t-5}{30}}$$

Therefore,

$$v_c(t)\; = \;\;5(1-e^{-\frac{t}{25}})\qquad0<t\leq 5$$

$$\;\;0.906e^{-\frac{t-5}{30}}$$

The graph of $v_c (t)$ with time is shown in figure 7.10

Figure 7.10: