Question

Explain charging and discharging of a capacitor.​

Answer 1

Explanation:

A Capacitor is a passive device that stores energy in its Electric Field and returns energy to the circuit whenever required. ... When a Capacitor is connected to a circuit with Direct Current (DC) source, two processes, which are called "charging" and "discharging" the Capacitor, will happen in specific conditions.

Answer 2

Hi mate

Here is ur answer ⤵

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Charging of a Capacitor

When the key is pressed, the capacitor begins to store charge. If at any time during charging, I is the current through the circuit and Q is the charge on the capacitor, then

Potential difference across resistor = IR, and

Potential difference between the plates of the capacitor = Q/C

Since the sum of both these potentials is equal to ε,

RI + Q/C = ε … (1)

As the current stops flowing when the capacitor is fully charged,

When Q = Q0 (the maximum value of the charge on the capacitor), I = 0

From equation. (1),

Q0 / C = ε … (2)

From equations. (1) and (2),

$RI + \frac{Q }{C} = \frac{Q0 }{C}$

$\frac{Q0-Q}{CR} ..........(3)$

When t = 0, Q = 0 and when t = t, Q = Q.

Integrating both sides within proper limits, we get

$\int\limits_{0}^{Q}{\frac{dQ}{\left( {{Q}_{0}}-Q \right)}}=\int\limits_{0}^{t}{\frac{dt}{CR}}=\frac{1}{CR}\int\limits_{0}^{t}{dt} </p><p>$

or

$∣ - ln( Q0 - Q )∣ \binom{Q}{0} = \frac{1}{cr} ∣ t∣ \binom{t}{0}$

or

$Q = Q0(1 - {e}^{ - t /τ</p><p>) } .......(4)$

Where

$\tau =CRτ=CR$

Discharging of a capacitor

When the key K is released , the circuit is broken without introducing any additional resistance. The battery is now out of the circuit and the capacitor will discharge itself through R. If I is the current at any time during discharge, then putting ε = 0 in RI + Q/C = ε, we get

$RI+\frac{Q}{C}=0\,\,\,or\,\,\,R\frac{dQ}{dt}+\frac{Q}{C}=0$

$R \frac{dt}{dq} = - \frac{q}{c} \: \: or \: \frac{dq}{q} = - \frac{dt}{cr}$

When t = 0, Q = Q0 and when t = t, Q = Q.

Integrating both sides within proper limits, we get

$\int\limits_{{{Q}_{0}}}^{Q}{\frac{dQ}{Q}}=-\int\limits_{0}^{t}{\frac{dt}{CR}}=-\frac{1}{CR}\int\limits_{0}^{t}{dt}$

$\left| \ln Q \right|_{{{Q}_{0}}}^{Q}=-\frac{1}{CR}\left| t \right|_{0}^{t}$

or

$lnQ−lnQ0=− \frac{t}{cr} </p><p>$

Or

$Q=Q0e−t/CR=Q0e−t/τ … (5)</p><p></p><p>$

where \tau =CRτ=CR

Eqn. (5) gives the value of the charge on the capacitor at any time during discharging.

If t = CR, then from eqn. (5).

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Hope it helps to you ⤴⤴