Capacitors
12.0 $R$-$C$ Circuits
12.0 $R$-$C$ Circuits
Charging of a capacitor in $R$-$C$ Circuit
To understand the charging of a capacitor in $R$-$C $ circuit, let us first consider the charging of a capacitor without resistance.
In such a circuit when we close the switch the capacitor gets charged immediately. Charging takes no time. A charge $${q_ \circ } = CV$$ comes on the capacitor as soon as switch is closed and the $q-t$ graph in this case is a straight line parallel to $t$-axis as shown in figure.
But if there is some resistance in the circuit charging takes some time. Because resistance oppose the charging (or current flow in the circuit). Final charge ( called steady state charge) is still ${q_\circ}$ but it is acquired after a long period of time.
This charge can be calculated by the following method.
Let consider the circuit shown in the figure. Suppose the switch is closed at time $t=0$. At some instant of time, let charge
on the capacitor is $q(<{q_\circ})$ and it is still increasing and hence current is flowing in the circuit. Applying loop law in circuit shown in figure, we get
on the capacitor is $q(<{q_\circ})$ and it is still increasing and hence current is flowing in the circuit. Applying loop law in circuit shown in figure, we get$$\begin{equation} \begin{aligned} - \frac{q}{C} - iR + V = 0 \\ i = \frac{{dq}}{{dt}} \\ - \frac{q}{C} - \left( {\frac{{dq}}{{dt}}} \right)R + V = 0 \\ \frac{{dq}}{{V - \frac{q}{C}}} = \frac{{dt}}{R} \\ \int_0^q {\frac{{Cdq}}{{CV - q}} = \int_0^t {\frac{{dt}}{R}} } \\ - RC{[\ln (CV - q)]_0}^q = t \\ \ln \left( {CV - q} \right) - \ln (q) = - \frac{t}{{RC}} \\ \ln \left( {1 - \frac{q}{{CV}}} \right) = - \frac{t}{{RC}} \\ 1 - \frac{q}{{CV}} = {e^{ - \frac{t}{{RC}}}} \\ q = CV(1 - {e^{ - \frac{t}{{RC}}}}) \\ \\\end{aligned} \end{equation} $$
Substituting $CV = {q_o}$ and $RC = \tau $, we have $$q = {q_o}(1 - {e^{ - \frac{t}{\tau }}})$$ $q-t$ graph is an exponentially increasing curve.the charge $q$ increases exponentially from 0 to ${q_o}$. From the graph and the equation, we see that at $$t = 0,q = 0$$ $$t = \infty ,q = {q_o}$$
Definition of ${\tau}$ At $t = {\tau}$,$$q = {q_{{o_{}}}}(1 - {e^{ - 1}}) = 0.632{q_o}$$Hence, ${\tau}$ can be defined as the time in which 63.2% charging is completed in a $C - R$ circuit. Note that ${\tau}$ is the time. Hence, the dimension of ${\tau}$is$$[{\tau _o}] = [CR] = [{M^0}{L^0}T]$$
Current flows in a $C$ - $R$ circuit during the charging of capacitor. Once charging of the capacitor is over or the steady state is achieved, the current becomes zero. The current at any time $t$. Hence,
$$i = \frac{{dq}}{{dt}} = \frac{d}{{dt}}\{ {q_o}(1 - {e^{ - \frac{t}{{{\tau}}}}})\} $$$$i = \frac{{{q_o}}}{{{\tau }}}{e^{ - \frac{t}{{{\tau}}}}}$$
Substituting ${q_o} = CV$ and ${\tau } = CR$, we have
$$i = \frac{V}{R}{e^{ - \frac{t}{{{\tau}}}}}$$ $$\frac{V}{R} = {i_o}$$ $$i = {i_o}{e^{ - \frac{t}{{{\tau }}}}}$$ i.e., current decreases exponentially with time.
Discharging of a capacitor in the $C$-$R$ circuit
For understanding of discharging of the capacitor in a $C$ -$ R$ circuit again we first consider the discharging without resistance. Suppose a capacitor has a charge ${q_o}$. The positive plate has a charge +${q_o}$ and negative -${q_o}$. It implies that the positive plate has deficiency of electrons. When the switch is closed, the extra electrons on negative plate
immediately comes to the positively plate and net charge on both plates becomes zero. So we can say that discharging of the capacitor takes place immediately.
immediately comes to the positively plate and net charge on both plates becomes zero. So we can say that discharging of the capacitor takes place immediately.In case of a $C$ -$ R$ circuit discharging of capacitor also takes time. As we know that final charge will be zero but after long period of time. The $q - t$ equation in this case is,$$q = {q_o}{e^{ - \frac{t}{{{\tau}}}}}$$
Thus, $q$ decreases exponentially from ${q_o}$ zero, as shown in the figure. From the curve and the equation, we see that
- at $t = 0,q = {q_o}$
- at $t = \infty ,q = 0$
Definition of time constant$({\tau})$
In case of discharging of capacitor definition of ${\tau }$ is charged.
At time $t = {\tau }$,$$q = {q_o}{e^{ - 1}} = 0.368{q_o}$$
Hence, in this case ${\tau }$ can be defined as it is the time when charge reduces to $36.8\%$ of its maximum value ${q_o}$.
During the discharging of the capacitor, current flows in the circuit till $q$ becomes zero. This current can be found by differentiating $q$ with respect to $t$ but along with negative sign because the charge is decreasing with the time. So,$$i = \left( { - \frac{{dq}}{{dt}}} \right) = - \frac{d}{{dt}}({q_o}{e^{ - \frac{t}{{{\tau }}}}})$$$$ = \frac{{{q_o}}}{{{\tau }}}{e^{ - \frac{t}{{{\tau }}}}}$$$$\frac{{{q_o}}}{{{\tau}}} = {i_o}$$$$i = {i_o}{e^{ - \frac{t}{{{\tau }}}}}$$
The above equation of current is mathematically exponentially decreasing equation. Thus, $i - t$ graph decreases exponentially with time from ${i_o}$ to zero. The $i - t$ curve is shown in the figure.
