- Consider a circuit containing a capacitor of capacitance C and a resistor R connected to a constant source of emf (battery) through a key (K) as shown below in the figure

- Source of EMF E can be included or excluded from circuit using this two way key

- when the battery is included in the circuit by throwing the switch to a, the capacitor gradually begins to charge and because of this capacitor current in the circuit will vary with time

- There are two factors which contributes to voltage drop V across the circuit i.e. if current I flows through resistor R,voltage drop across the resistor is IR and if there is a charge Q on the capacitor then voltage drop across it would be Q/C

- At any instant, instantaneous potential difference across capacitor and resistor are

V_{R}=IR and V_{C}=q/C

Therefore total potential difference drop across circuit is

V=V_{R}+V_{C}=IR+q/C

Where V is a constant

- Now current is circuit

Initially at time t=0 ,when the connection was made, charge on the capacitor q=0 and initial current in the circuit would be I_{max}=V/R ,which would be the steady current in the circuit in the absence of the capacitor

- As the charge q on the capacitor increase ,the term q/RC becomes larger and current decreases until it becomes zero .Hence for I=0

V/R=q/RC

or q=CV=Q_{f}

where Q_{f}is the final charge on the capacitor

- Again consider equation (14)

we know that I=dq/dt

So,

rearranging this equation we get

Integrating this we get

Where A is the constant of integration

Now at t=0,q=0 So

A=CRlnCV

From this we have

As CV=Q_{f}so,

Where Q_{f}=CV as defined earlier is the final charge on the capacitor when potential difference across it becomes equal to applied to EMF

- Equation (15) represents the growth of charge on the capacitor and shows that it grows exponentially as shown below in the figure

- Now since

Again I_{max}=V/R,so we have,

Thus from equation (16) we see that current decreases exponentially from its maximum value I_{max}=V/R to zero

- Quantity RC in equation (15) and (16) is called capacitive time constant of the circuit

τ_{C}=CR

- Smaller is the value of τ
_{C},charge will grow on the capacitor more rapidly.

- Putting t= τ
_{C}=CR in equation (15)

q=Q_{f}(1-e^{-1})

=6.32Q_{f}

Thus τ_{C}of CR circuit is the time which the charge on capacitor grows from 0 to .632 of its maximum value

- Again consider figure (8),by throwing the switch(S) to b,we can now exclude the battery from the circuit

- After the removal of external emf the charged capacitor now begins to discharge through the resistance R

- Putting V=0 in equation (14) we have

I=-q/RC

or dq/dt=-q/RC

On integrating this equation we get

Where A_{1}is the constant of integration .Initially at t=0,q=Q_{f},since capacitor is fully charged thus

A_{1}=CRlnQ_{f}

Hence

This is the equation governing discharge of capacitor C through resistance R

- From equation (17) we see that charge on a capacitor decays exponentianally with time as shown below in the figure

- Current during discharge is obtained by differentiating the equation (17) so,

Thus smaller the capacitive time constant ,the quicker is the discharge of the capacitor

- Putting t= τ
_{C}=CR in equation (17)

We get

q=Q_{f}e^{-1}=Q_{f}(.368)

Thus the capacitive time constant can also be defined as the time in which the charge on the capacitor decays from maximum to .368 of the maximum value

Class 12 Maths Class 12 Physics

- NCERT Solutions: Physics 12th
- NCERT Exemplar Problems: Solutions Physics Class 12
- Concepts of Physics - Vol. 2 HC Verma
- Dinesh New Millennium Physics Class -XII (Set of 2 Vols) (Free Complete Solutions to NCERT Textbook Problems & NCERT Exemplar Problems in Physics-XII)
- Principles of Physics Extended (International Student Version) (WSE)
- university physics with modern physics by Hugh D. Young 13th edition
- CBSE Chapterwise Questions-Solutions Physics, Class 12
- CBSE 15 Sample Question Paper: Physics for Class 12th

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