Capacitor
Other Topics
RELATION BETWEEN VOLTAGE AND CURRENT IN A CAPACITOR
Consider a capacitor connected to a battery as shown below,
Once battery is connected, capacitor begins to charge.
- the positive terminal of the battery begins to attract electrons from metal plate $1$, which makes the plate as positively charged.
- the negative terminal of the battery begins to supply electrons to metal plate$2$, which makes the plate as negatively charged.
Now we are going analyze this charging process time-to-time, from $t=0^-$ to $t=\infty$.
- $t=0$ – time at which battery is connected to the capacitor.
- $t=0^-$ – instant before the battery connection is made.
- $t=0^+$ – instant after the battery connection is made.
- $t=\infty$ – time at which the capacitor will be fully charged.
At time $t=0^-$
$t=0^-$ is nothing but instant before the battery connection is made (i.e.,) no source is connected.
Before any source is connected, the capacitor may or may not have any residual charge. If there is some residual charge $(Q_0)$ present in the capacitor, there will be an initial voltage in the capacitor $(V_0)$.$$V_c(0^-)=V_0$$
If there is no residual charge at $t=0^-$, then the initial voltage of the capacitor will be zero. $$V_c(0^-)=0$$
When battery is connected (at $t=0$),
At time $t=0$, Battery is connected to the capacitor. Because of huge potential difference between the battery (say $V_b=5V$) and the capacitor (say $V_c=0V$), $$V_b-V_c=5V$$it can attract maximum number of electrons from metal plate $1$ and supplies same number of electrons to metal plate $2$.
This flow of maximum number of electrons depends on the battery and wire we are using.
If we use $1\mu\;Amps$ rating battery and wire, then the current flow at $t=0$ will be $1\mu\;Amps$. If we use $100\mu\;Amps$ rating battery and wire, then the current flow will be $100\mu\;Amps$.
This looks like the capacitor does not resist any current flow at that instant.
No resistance = short circuit
Hence we can say,
Capacitor behaves as a short circuit at $t=0^+$
Practice Question
When a series RC circuit is connected to a constant voltage at $t=0$, the current passing through the circuit at $t=0^+$ is
a) $\infty$
b) $0$
c) $\frac{V}{R}$
d) $\frac{V}{\omega C}$
Correct Answer:Option C
At $t=0^+$, Capacitor acts as an short circuit. Hence the current in the circuit depends on the external circuit. Hence $$I_c(0^+)=V/R$$
At time interval $\Delta t_1$ (from $t=0$ to $t=t_1$),
At time $t=0$, Battery is connected to the capacitor. Because of huge potential difference between the battery (say $V_b=5V$) and the capacitor (say $V_c=0V$), $$V_b-V_c=5V$$it can attract large number of electrons from metal plate$1$ and supplies same number of electrons to metal plate $2$, hence those metal plates get positively charged and negatively charged respectively.
Charge induced at this time interval:$$Q_{c1}=100Q$$
Current flow at this time interval:$$I_{c1}=\frac{Q_{c1}}{\Delta t_1}$$$$I_{c1}=\frac{100Q}{\Delta t_1}$$
Total charge accumulated on the plates:$$\sum Q_c=Q_{c1}+Q_0$$$$\sum Q_c=100Q+0Q$$$$\sum Q_c=100Q$$
Voltage of the capacitor at this time interval:$$V_c=0V$$
At time interval $\Delta t_2$ (from $t=t_1$ to $t=t_2$),
As time goes, due to the accumulation of charge on capacitor plates, the voltage of the capacitor increases.
At time $t=t_1$, the potential difference between battery (say $V_b=5V$) and capacitor (say $V_c=1V$) is somewhat less than the above case $(V_b-V_c=4V)$.
Hence it will attract only less number of electrons than the previous case (say $70Q$ electrons).
Charge induced at this time interval:$$Q_{c2}=70Q$$
Current flow at this time interval:$$I_{c2}=\frac{Q_{c2}}{\Delta t_2}$$$$I_{c2}=\frac{70Q}{\Delta t_2}$$
Total charge accumulated on the plates:$$\sum Q_c=Q_{c2}+Q_{c1}+Q_0$$$$\sum Q_c=70Q+100Q+0Q$$$$\sum Q_c=170Q$$
Voltage of the capacitor at this time interval:$$V_c=1V$$
$$Q_c\downarrow\;\;\; \sum Q_c \uparrow \;\;\; V_c \uparrow \;\;\; I_c \downarrow$$
At time interval $\Delta t_3$ (from $t=t_2$ to $t=t_3$),
At time $t=t_2$, the potential difference between battery (say $V_b=5V$) and capacitor (say $V_c=3V$) is somewhat less than the above case $(V_b-V_c=2V)$.
Hence it will attract only less number of electrons than the previous case (say $30Q$ electrons).
Charge induced at this time interval:$$Q_{c3}=30Q$$
Current flow at this time interval:$$I_{c3}=\frac{Q_{c3}}{\Delta t_3}$$$$I_{c3}=\frac{30Q}{\Delta t_3}$$
Total charge accumulated on the plates:$$\sum Q_c=Q_{c3}+Q_{c2}+Q_{c1}+Q_0$$$$\sum Q_c=30Q+70Q+100Q+0Q$$$$\sum Q_c=200Q$$
Voltage of the capacitor at this time interval:$$V_c=3V$$
$$Q_c\downarrow\;\;\; \sum Q_c \uparrow \;\;\; V_c \uparrow \;\;\; I_c \downarrow$$
At time $t=\infty$
$t=\infty$ represents the time at which capacitor is fully charged.
As capacitor is fully charged, then the voltage of the capacitor will be equal to the voltage of the battery. $$V_{c}(\infty)=V_{b}$$ Hence, there will be no current flow. $$I_c(\infty)=0$$
As current flow cease to zero at $t=\infty$, we can say, Capacitor behaves as open circuit at $t=\infty$
Summarizing,
Behaviour of Capacitor from $t=0$ to $t=\infty$, when DC source is connected,
- At $t=0$, Capacitor behaves as a short circuit.
From $t=0^+$ to $t=\infty$, Capacitor behaves as an opposing battery.
At $t=\infty$, Capacitor behaves as an open circuit.
Relation between capacitor voltage and current:
From the above discussion, we can say,
As charge gets accumulated on the capacitor plates, the voltage of the capacitor increases.
The capacitor voltage at time $t$ will be proportional to how much charge gets accumulated on the capacitor plates, $$V_c(t) \propto \sum_{0}^{t} Q_c(t)$$ As we know $C=Q/V$, $$V_c(t) =\frac{1}{C} \sum_{0}^{t} Q_c(t)$$As our period of observation is discrete, we are using summation here.
Dividing by $\Delta t$ on both sides, $$\frac {V_c(t)}{\Delta t} =\frac{1}{C} \sum_0^t \frac{Q_c(t)}{\Delta t}$$$${V_c(t)}=\frac{1}{C} \sum_0^t \frac{Q_c(t)}{\Delta t} \Delta t$$$${V_c(t)} =\frac{1}{C} \sum_0^t I_c(t) \; \Delta t$$ As $\Delta t \to 0$ (reducing the period of observation as small as possible), summation can be replaced by integration, $${V_c(t)} =\frac{1}{C} \int_0^t I_c(t) \;dt$$ Rearranging the above equation, $${I_c(t)}=C \frac{dV_c(t)}{dt}$$
In a capacitor,
the relation between voltage and current is$${V_c(t)}=\frac{1}{C} \int_0^t I_c(t) \;dt$$ and $${I_c(t)}=C \frac{dV_c(t)}{dt}$$
If initial voltage of the capacitor is not equal to zero, then $${V_c(t)}=\frac{1}{C} \int_0^t I_c(t) \;dt+V_0$$ and $${I_c(t)}=C \frac{dV_c(t)}{dt}$$