### Capacitors: Calculations & Simulations on Charging, Discharging Time & Voltage

Assume charging a completely discharged capacitor, under resistance, R with an applied potential, E having the capacitance, C.

Potential across the capacitor is Ec.

At time, t = 0s or to:   EC < E; EC = 0 V.

If it is fully charged then EC = E and t = tc.

‘E’ is the maximum potential attained by the capacitor.

Time to reach maximum potential is: (tc – to).

At any given time, ‘t’ on charging in between to and tc the potential of the capacitor, EC is related to the maximum potential, E by the relation,

where ‘t’ is the elapsed time during charging

‘R’ is the resistance in Ω

‘C’ is the capacitance in F.

The product of R and C is called the time constant and is represented as, τ. So, τ = R × C.

Unit for τ

τ = R × C = Ω × {F} = Ω × {C × V–1}

C = Charge in Coulombs = A × s

= Ω × {A × s × V–1} = = Ω × {V × Ω–1 × s × V–1} = s.

At one time constant, capacitor is 63.2% charged.

In other words, if (t/τ)= 1, then EC = E × 0.632.

If, τ = 2s, the capacitor is 86.5% charged.

If, τ = 3s, the capacitor is 95% charged.

If, τ = 5s, the capacitor is 99.3% charged.

Time required, ‘t’ to reach EC, under the applied potential, E is given as:

In terms of current flow, it is given as:

Discharging

On discharging the relevant equations are:

But during discharging at (initial) time, t = 0 s, the potential across the capacitor is E (maximum starting potential).

Potential across the capacitor at any given time, ‘t’ during discharging is EC and EC < E, and EC decreases with time, t.

The time required, ‘t’ to discharge to the given potential, EC from E (initial potential) is given as:

In terms of current flow, it is given as:

Hence, 1 τ about 36.8 % is discharged from the initial potential, E.

At about 5 τ, the capacitor is almost discharged and the charge remaining will be about 0.7 % only.

Example

Estimate the time required for the fully discharged capacitor to charge 90 %. Also, estimate the charge at 5 minutes.

Formula:

‘t’ is the required time in seconds

‘E’ is applied potential or maximum potential = 60 V.

‘EC’ is the required potential of the capacitor to charge = 90 % of E.

EC = 0.9 × 60 V = 54 V

τ is one time constant; τ = R × C

R is 4000 Ω and Cap is 3000 × 10–3 F = 3 F

So one-time constant, τ = 4000 × 3 = 12000 s

Substituting these values, time required to charge 90 % (54 V) is 27631 s.

In the second part, EC is required at t = 5 minutes or after 300 seconds.

Required equation is:

t = 300 s and τ = 12000 s, substituting these values, EC = 1.48 V.

To understand more about τ, check EC at τ.

One time constant, τ = 4000 × 3 = 12000 s.

So, EC at 12000 s, from the above equation:

EC = 37.93 V, in terms of percentage,

(EC/E) × 100 = (37.93/60) × 100 = 63.22%.

In terms of current flow, at one time constant value of IC is given as:

Current flow at time, t = 0 is given as:

With the increase in EC, current flow decreases, as R is constant (as per Ohm’s law: V = I × R).

Example

Estimate the time required for a fully charged capacitor at 30 V with 600 mF capacitance to discharge 30 % of its charge at a load of 3000 Ω. Also, estimate the percentage of charge left after 10 minutes of discharge.

Required equation is:

where E = 30 V (at t = 0): EC = (100 – 30) = 70 % remaining.

So, EC = 30 × 0.7 = 21 V.

τ = R × Cap = 3000 Ω × 600 × 10–3 F = 1800 s.

Hence, one time constant is 1800 s. Substituting these values, then t = 642 s.

So, at 642 s, the capacitor is discharged from 30 V to 21 V at 3000 Ω.

In terms of time constant, 642/1800 = 0.36 τ it is discharged to 30 %.

In the second part, required equation is:

E = 30 V; t = 10 min. = 600 s and τ = 1800 s (one-time constant).

Substituting these values into the equation then, EC = 21.5 V.

In terms of percentage, (21.5/30) × 100 = 71.7 % charge is remaining.

So, after 600 s, (100 – 71.7) = 28.3 % or (30 – 21.5) = 8.5 V is discharged.

This is explained through circuit simulations in the following videos.

Part I

Part II

Part III