• By Admin Koushi
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• April 18, 2025

Capacitor Part – 9

Charging and discharging of a capacitor:

Charging: Let us consider a capacitor C is connected to a battery of emf E through a resistance R. During charging of the capacitor the potential difference across the plates becomes equal to the emf of the battery. Let at any time t, I is the current through the resistance R and q is the charge of capacitor. Using KVL, V_C + V_R = E  

Or,  + IR = E  

Or,  + R = E  

Or, R = E –  

Or, CRdq = (EC – q)dt  

Or,  =  

Or, =  

Or, – log_e =

Or, 1 –  = e^{– t/RC}  

Or, q = EC(1 – e^{– t/RC})

Or, q = q₀(1 – e^{– t/RC}) at t = , q = q₀ = EC.

= CR is called capacitive time constant, i.e. the time in which charge of the capacitor is 0.632 times of its maximum value during charging.

The charging current is I = =  = I₀e^{– t/RC} where I₀ =  = .

At t = 0, I = I₀ i.e. initially it acts as short circuit or as a simple conducting wire.

At t = , I = 0 it acts as an open circuit.

Discharging: If a capacitor C with charge q₀ is discharged through a resistance R then at any time t, the PD of the resistance is V = IR.

Using KVL, V_C + V_R = 0

Or, R  +  = 0

Or,  = –  

Or, q = q₀e^{– t/RC} [This is the charge of the capacitor during discharging.]

The capacitive time constant = CR is the time in which charge becomes    i.e. 0.368 times the initial value of charge q₀.

The discharging current is I = – = –  = I₀e^{– t/RC} where I₀ =  = .

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Admin Koushi