## Charging and discharging of capacitors

### Charging

When a capacitor C is connected to a battery through R then charging of capacitor takes place.The eqn. of emf at any time t is

$R I+\frac{q}{C}=E$

$\frac{R d q}{d t}+\frac{q}{C}=E$

$\int_{0}^{Q} \frac{d q}{(C E-q)}$

$=-\int_{0}^{t} \frac{d t}{R C}$

This on solving gives $Q=Q_{0}\left(1-e^{-t / R C}\right)$

where $Q_{0}=C E$

#### Important Points

- The charge on a capacitor increases exponentially with time
- The current during charging process is$I=\frac{d Q}{d t}$$=-\frac{Q_{0}}{R C} e^{-t / R C}$$=-I_{0} e^{-t / R C}$

The current decreases exponentially with time.

- $\tau=R C$ is the capacitive
**time constant**. It is the time in which the charge on the capacitor reaches 0.632 times of its maximum value during charging. **The time constant**is the time in which current reduces to $\frac{1}{\mathrm{e}}$ times or 0.368 times of its initial value- At initial time $\mathrm{t}=0$ and $I=I_{\max }$ so circuit acts as short circuit or conducting wire.At $t=\infty \quad I=0$ so circuit acts as an open circuit or as a broken wire.
- The voltage increases exponentially with time as $V=V_{0}\left(1-e^{-t / R C}\right)$ during charging.

### Discharging

If a completely charged capacitor C having charge $\mathrm{Q}_{0}$ is discharged through a resistance R then equation of emf at any time t is$\mathrm{RI}+\frac{\mathrm{q}}{\mathrm{C}}=0$

Or

$R \frac{d q}{d t}+\frac{q}{C}=0$

Or

$\int_{Q_{0}}^{Q} \frac{d q}{q}=-\int_{0}^{t} \frac{d t}{R C}$

Or

$Q=Q_{0} e^{-t / R C}$

#### Important Points

- During discharging charge on a capacitor decreases exponentially with time.
- The current during discharging process$I=\frac{d Q}{d t}=-\frac{Q_{0}}{R C} e^{-t / R C}$$=-I_{0} e^{-t / R C}$ The current decreases exponentially with time.
**Time constant**is the time in which charge on capacitor become $\frac{1}{\mathrm{e}}$ or 0.368 times of its initial value $\mathrm{Q}_{0}$- Time constant is the time in which current reduces to $\frac{1}{\mathrm{e}}$ or 0.368 times of its initial value $\mathrm{I}_{0}$.The direction discharging current is opposite to that of charging.
- During discharging voltage decreases with time as $V=V_{0} e^{-t / R C}$

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