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Charging and Discharging of Capacitor

Charging and Discharging of Capacitor :- Let a capacitor of capacitance C be connected in series with a resistance R to a battery of emf ε with the help of a Morse key K as shown in the figure below :

(a) Charging of a capacitor

(Charging and Discharging of Capacitor)

As soon as the Morse key K is pressed, the resistor R and the capacitor C get connected to the battery, and current flows in the R–C circuit, causing the capacitor to start charging. Let at some time t,

charge on the plates of the capacitor = q

The value of the current flowing in the circuit = I

∴ Potential difference across the capacitor (C) = q/C

potential difference across the resistance (R) = IR

Since the electromotive force of the battery is ε, therefore

$\displaystyle \varepsilon =\frac{q}{C}+IR$ …..(1)

When the capacitor becomes fully charged, the charge on its plates becomes maximum, and the current flow in the circuit stops, that is,

When q = q₀, then $\displaystyle I=\frac{dq}{dt}=0$

Hence, from equation (1),

$\displaystyle \varepsilon =\frac{{{q}_{\text{0}}}}{C}+0\times R=\frac{{{q}_{\text{0}}}}{C}$

Putting this value in equation (1), we get

$\displaystyle \frac{{{q}_{\text{0}}}}{C}=\frac{q}{C}+IR$

At some time t in the process of charging, $\displaystyle I=\frac{dq}{dt}$ , hence

$\displaystyle \frac{{{q}_{\text{0}}}}{C}=\frac{q}{C}+\frac{dq}{dt}R$

$\displaystyle \Rightarrow \frac{{{q}_{\text{0}}}}{C}-\frac{q}{C}=\frac{dq}{dt}R$

$\displaystyle \Rightarrow \frac{{{q}_{\text{0}}}-q}{C}=\frac{dq}{dt}R$

$\displaystyle \Rightarrow \frac{dq}{\left( {{q}_{\text{0}}}-q \right)}=\frac{dt}{RC}$

Integrating both sides within appropriate limits,

$\displaystyle \int\limits_{0}^{q}{\frac{dq}{\left( {{q}_{\text{0}}}-q \right)}}=\int\limits_{0}^{t}{\frac{dt}{RC}}$

$\displaystyle \left[ \frac{lo{{g}_{e}}\left( {{q}_{\text{0}}}-q \right)}{(-1)} \right]_{0}^{q}=\frac{1}{RC}\left[ t \right]_{0}^{t}$

$\displaystyle \Rightarrow \left[ lo{{g}_{e}}\left( {{q}_{\text{0}}}-q \right)-lo{{g}_{e}}\left( {{q}_{\text{0}}}-0 \right) \right]=-\frac{1}{RC}\left[ t-0 \right]$

$\displaystyle \Rightarrow lo{{g}_{e}}\frac{\left( {{q}_{\text{0}}}-q \right)}{{{q}_{\text{0}}}}=-\frac{1}{RC}t$

$\displaystyle \frac{{{q}_{\text{0}}}-q}{{{q}_{\text{0}}}}={{e}^{-{}^{t}\!\!\diagup\!\!{}_{RC}\;}}$

$\displaystyle 1-\frac{q}{{{q}_{\text{0}}}}={{e}^{-{}^{t}\!\!\diagup\!\!{}_{RC}\;}}$

$\displaystyle \Rightarrow \frac{q}{{{q}_{\text{0}}}}=1-{{e}^{-{}^{t}\!\!\diagup\!\!{}_{RC}\;}}$

$\displaystyle \therefore q={{q}_{\text{0}}}\left( 1-{{e}^{-{}^{t}\!\!\diagup\!\!{}_{RC}\;}} \right)={{q}_{\text{0}}}\left( 1-{{e}^{-{}^{t}\!\!\diagup\!\!{}_{\tau }\;}} \right)$ …..(2)

Where,

$\displaystyle \tau =RC$

Equation (2) represents the charging of the capacitor.

Time Constant

The quantity $\displaystyle \tau =RC$ is called the time constant of an RC circuit because $\displaystyle RC$ has the dimension of time, and for any given RC circuit, its value remains constant.

Substituting $\displaystyle \tau =RC$ in equation (2),

$\displaystyle q={{q}_{\text{0}}}\left( 1-{{e}^{-{}^{\tau }\!\!\diagup\!\!{}_{\tau }\;}} \right)={{q}_{\text{0}}}\left( 1-{{e}^{-1}} \right)={{q}_{\text{0}}}\left( 1-\frac{1}{e} \right)={{q}_{\text{0}}}\left( 1-\frac{1}{2.718} \right)$

$\displaystyle \Rightarrow q={{q}_{\text{0}}}\left( 1-0.368 \right)=0.632{{q}_{\text{0}}}$

∴ q = 63.2 % q₀

Thus, the time constant of an RC circuit is the time in which the charge on the capacitor reaches 63.2% of its maximum value.

Again in equation (2) for q = q₀ ,

$\displaystyle {{q}_{\text{0}}}={{q}_{\text{0}}}\left( 1-{{e}^{-{}^{t}\!\!\diagup\!\!{}_{\tau }\;}} \right)$

$\displaystyle \Rightarrow 1=1-{{e}^{-{}^{t}\!\!\diagup\!\!{}_{\tau }\;}}$

$\displaystyle \Rightarrow {{e}^{-{}^{t}\!\!\diagup\!\!{}_{\tau }\;}}=0$

$\displaystyle \Rightarrow \frac{1}{{{e}^{{}^{t}\!\!\diagup\!\!{}_{\tau }\;}}}=0$

$\displaystyle \Rightarrow {{e}^{{}^{t}\!\!\diagup\!\!{}_{\tau }\;}}=\infty$

$\displaystyle \Rightarrow t=\infty$

That is, in an RC circuit, it takes an infinite amount of time for the charge to reach its maximum value. However, in practice, within approximately five times the time constant, the charge becomes very close to its maximum value, meaning the capacitor becomes fully charged. The following graph illustrates the charging of a capacitor in an RC circuit :

(b) Discharging of a capacitor

(Charging and Discharging of Capacitor)

As soon as the Morse key K is released, the battery is disconnected from the circuit and the fully charged capacitor begins to discharge.

If at any time t, the charge on the plates of the capacitor is q, then since the battery has been disconnected from the circuit, by putting ε = 0 in equation (1),

$\displaystyle 0=\frac{q}{C}+IR$

$\displaystyle \Rightarrow IR=-\frac{q}{C}$

$\displaystyle \Rightarrow \frac{dq}{dt}R=-\frac{q}{C}$

$\displaystyle \Rightarrow \frac{dq}{q}=-\frac{dt}{RC}$

Integrating both sides within appropriate limits,

$\displaystyle \int\limits_{{{q}_{\text{0}}}}^{q}{\frac{dq}{q}}=-\int\limits_{0}^{t}{\frac{dt}{RC}}$

$\displaystyle \left[ lo{{g}_{e}}q \right]_{{{q}_{\text{0}}}}^{q}=-\frac{1}{RC}\left[ t \right]_{0}^{t}$

$\displaystyle \Rightarrow \left[ lo{{g}_{e}}q-lo{{g}_{e}}{{q}_{\text{0}}} \right]=-\frac{1}{RC}\left[ t-0 \right]$

$\displaystyle \Rightarrow lo{{g}_{e}}\frac{q}{{{q}_{\text{0}}}}=-\frac{1}{RC}t$

$\displaystyle \Rightarrow \frac{q}{{{q}_{\text{0}}}}={{e}^{-{}^{t}\!\!\diagup\!\!{}_{RC}\;}}$

$\displaystyle \therefore q={{q}_{\text{0}}}{{e}^{-{}^{t}\!\!\diagup\!\!{}_{RC}\;}}={{q}_{\text{0}}}{{e}^{-{}^{t}\!\!\diagup\!\!{}_{\tau }\;}}$ …..(3)

Where,

$\displaystyle \tau =RC$

Equation (3) represents the discharging of the capacitor.

Time Constant

As discussed earlier, the quantity $\displaystyle \tau =RC$ is called the time constant of an RC circuit. Substituting $\displaystyle t=\tau$ in equation (3),

$\displaystyle q={{q}_{\text{0}}}{{e}^{-{}^{\tau }\!\!\diagup\!\!{}_{\tau }\;}}={{q}_{\text{0}}}{{e}^{-1}}=\frac{{{q}_{\text{0}}}}{e}=\frac{{{q}_{\text{0}}}}{2.718}=0.368{{q}_{\text{0}}}$

∴ q = 36.8 % q₀

Thus, the time constant of an RC circuit is the time during the discharge of a capacitor in which the charge on the capacitor reduces to 36.8% of its maximum value.

Again, for q = 0 in equation (3)

$\displaystyle 0={{q}_{\text{0}}}{{e}^{-{}^{t}\!\!\diagup\!\!{}_{\tau }\;}}$

$\displaystyle \Rightarrow {{e}^{-{}^{t}\!\!\diagup\!\!{}_{\tau }\;}}=0$

$\displaystyle \Rightarrow \frac{1}{{{e}^{{}^{t}\!\!\diagup\!\!{}_{\tau }\;}}}=0$

$\displaystyle \Rightarrow {{e}^{{}^{t}\!\!\diagup\!\!{}_{\tau }\;}}=\infty$

$\displaystyle \Rightarrow t=\infty$

Thus, during the discharge of a capacitor in an RC circuit, it takes an infinite amount of time for the charge on the capacitor to become zero. The following graph shows the discharge of the capacitor in an RC circuit :-

Note :-

If the value of the time constant $\displaystyle \tau =RC$ is small, then from equation (2), the charge stored on the capacitor reaches its maximum value (q₀) in a shorter time. From equation (3), the charge also takes less time to fall to zero. Similarly, if the value of the time constant $\displaystyle \tau =RC$ is large, then both charging and discharging of the capacitor will take more time.

Thus, the time constant ( $\displaystyle \tau =RC$ ) of an RC circuit is a measure of the rate of charging and discharging of the capacitor.

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Complete List of Topics :-

Charging and Discharging of Capacitor

(a) Charging of a capacitor

(Charging and Discharging of Capacitor)

(b) Discharging of a capacitor

(Charging and Discharging of Capacitor)

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