Charging and Discharging of Capacitor

A capacitor is charged when we supply a voltage with current running through the capacitor, in the process the capacitor accumulates charges and is called charged.

Charging and Discharging of Capacitor

A charged capacitor can then be discharged by draining the current through it’s two terminals or connecting some load through it’s terminals. Consider a circuit consisting of Resistor (R Ohms) , Capacitor (C Farads) , a Voltage Source (V voltage) and a switch as shown below:

If we put the switch to the point B , then the circuit of capacitor is completed with resistor and voltage source, the voltage source will charge the capacitor by running the current through capacitor and if we put the switch to point A then the circuit of capacitor is completed with resistor which will drain current from capacitor and discharge the capacitor.
Charging and Discharging of capacitorFigure 1: Charging and Discharging of capacitor

Charging of Capacitor:

In the above circuit let the Capacitor is fully discharged or does not contains any charge and the switch is connected to point B as shown below:
Charging of capacitorFigure 2: Charging of capacitor

The Capacitor starts getting charged or it slowly starts accumulating charges on it’s plates. If Ic is charging current through capacitor then  Ic is maximum at the beginning and it slows starts getting smaller until the capacitor is fully charged or the Potential difference built across capacitor is equal to the supply voltage V.
Let, At any instance when the capacitor is getting charged;

Vc = Potential difference across the capacitor.
Ic = The instantaneous charging current.
q = total charge on capacitor plate.

Then, We know,
At any instance the Total voltage V is equal to the sum of Voltage drop across resistor R and voltage across Capacitor. Or, V = Vr + Vc
Or, V = I_c R + V_c

We also know, I_c = \frac{dq}{dt} = \frac{d}{dt} C.V_c=C.\frac{dV_c}{dt}

Thus,  V = I_c R+ V_c = CR.\frac{dV_c}{dt} + V_c

Or, -\dfrac{dV_c}{V-V_c} = -\dfrac{dt}{CR}

Or,  \displaystyle \int \dfrac{-dV_c}{V-V_c} = -\dfrac{t}{CR} \int dt

Or, \log_c (V-V_c) = -\frac{1}{CR}+K

Where, K is the constant of integration.
We know at the start of the process of charging t=0 and Vc=0
If we substitute these values in above equation we get: \log_C V=K
And the equation becomes:
\log_e (V-v_c) = \dfrac{-t}{CR} + \log_e V
If we denote the constant CR by \lambda
\log_e (V-v_c) = - \dfrac{t}{\lambda} + \log_e V
Or, \dfrac{V-v_c}{V} = e^{\frac{-t}{\lambda}}

Or, v_c = V (1 - e^{\frac{-t}{\lambda}}) – – – (i)
This gives the variation of voltage across the terminals of capacitors as time varies, where, v_c = Voltage across the capacitor, V = the voltage source supplied, t = time in seconds and \lambda = CR or the multiple of the Capacitance and Resistance in series which is also known as time constant.

The general graph of voltage across a capacitor as it is charged is shown in the figure below:
Voltage Across Capacitor while chargingFigure 3: Voltage Across Capacitor while charging

We can also convert the above equation in  terms of the charge accumulated in the plates of the capacitor ( q ):
We know,  v_c = \frac{q}{C} and  V = \frac{Q}{C}
Or, the equation (i) above becomes: q = Q(1-e^{\frac{-t}{\lambda}}) – – – (ii)

This gives the variation of charge across the terminals of capacitors as time varies, where, q_c = Charge across the capacitor, Q = The total charge that the capacitor can accumulate or the multiple of C & V, t = time in seconds and \lambda = time constant.

The general graph of charge across a capacitor as it is charged is shown in the figure below:
charge across capacitor while charging

Figure 4: charge across capacitor while charging

If we differentiate both sides of the equation (ii) above we get:
\frac{dq}{dt} = Q \frac{d}{dt} (1-e^{\frac{-t}{\lambda}})

Or, i_c = Q(\frac{1}{\lambda}e^{\frac{-t}{\lambda}})

Or, i_c = \frac{VC}{RC}e^{\frac{-t}{\lambda}}

Or, i_c = I e^{\frac{-t}{\lambda}}

This gives the variation of current through the terminals of capacitor as time varies, where, i_c = Current across the capacitor at any time, I = The maximum current that can flow through Capacitor or the initial current, t = time in seconds and \lambda = time constant.

The general graph of current through a capacitor as it gets charged is shown in the figure below:
Current across capacitor while charging

Figure 5: Current across capacitor while charging

Time Constant:

As discussed above the Time Constant is the product of C ( Capacitance) & R (Resistance) in a circuit consisting of capacitor and resistor. But with further analysis the time constant is not only a purely mathematical product of two constants, It also have various practical meanings in the circuit.

Significance of CR as time required to fully charge capacitor if initial voltage rise is maintained:

If we differentiate the equation (i) above with respect to “t” or time, At , t=0
\dfrac{dv_c}{dt} = \dfrac{V}{CR} . e^{-0} = \dfrac{V}{CR} volts / seconds
The equation above means the initial rate of change of voltage of capacitor is V/CR volts per seconds , which means if we maintain the initial rise of voltage between the terminals of capacitors in the circuit then the Capacitor will get fully charged  up to voltage V in time CR.

Significance of CR as time required to attain 0.632/1 part of voltage of V by Capacitor:

Now if we again put CR = t in the equation (i) above we get:
v_c = V (1 - e^{\frac{-CR}{CR}}) = V (1 - \frac{1}{e})

Or, v_c = V . (1 - \frac{1}{2.718}) = V . 0.632
The equation above signifies that the the time taken for which the voltage of capacitor arises to 0.632 / 1 part of it’s final V voltage is equal to time constant CR.

Discharging of Capacitor:

When a capacitor is charged we can discharge it or use the electrical power/energy stored in it by joining the two terminals of the capacitor by a load as shown in the figure below:

Discharging of capacitorFigure 6: Discharging of capacitor

It is obvious that the discharging current will flow in the opposite direction of the charging current. So, if we take the direction of charging current as positive then the discharging current is taken negative as it flows in opposite direction.
The discharging current is maximum at the starting of the process of discharge but the discharging current slowly decreases until it becomes zero and the capacitor becomes fully discharged.

We know,

Or, V = I_c R + V_c
Where,
V = the voltage supplied to the capacitor which is zero in the case of discharging.
Ic = the positive ( charging current or negative of discharging current) current through capacitor.
Vc = Voltage between terminals of the capacitor.

We also know, I_c = \frac{dq}{dt} = \frac{d}{dt} C.V_c=C.\frac{dV_c}{dt}

Thus,  V = I_c R+ V_c = CR.\frac{dV_c}{dt} + V_c

Now, because Vc = 0
0 = CR.\frac{dV_c}{dt} + V_c

Or, V_c = -CR.\frac{dV_c}{dt}

Or, \dfrac{dV_c}{V_c} = \dfrac{-dt}{CR}
Integrating both sides:
\displaystyle \int \dfrac{dV_c}{V_c} = \dfrac{-1}{CR} \int dt

Thus, \log_e V_c = - \dfrac{t}{CR} +K

To Solve for the value of K ; we k now at the start of discharging, t = 0 and Vc = V
Putting these values in above equation to solve for K we get: K = \log _e V

So, substituting the value of K in above equation it becomes:
\log_e V_c = - \dfrac{t}{\lambda} + \log_e V

Or,  \log_e \dfrac{V_c}{V} = - \dfrac{t}{\lambda}

Or, V_c = V. e^{- \frac{t}{\lambda}}

This gives the variation of voltage across the capacitor (Vc) while discharging the capacitor as time varies. Where, Vc = Voltage across the capacitor ,  V = Original voltage of the capacitor , t = time in seconds and \lambda = CR = Time constant.

The general graph of the Voltage across the terminals of a capacitor as it is discharged is shown below:
Voltage Across Capacitor while discharging

Figure 7: Voltage Across Capacitor while discharging

One Thought to “Charging and Discharging of Capacitor”

  1. leo

    how can i attach it in my existing circuit as a controller of beeping sound?

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