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
(A) Growth of charge
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
(B) Decay of charge
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
Thanks for visiting our website. DISCLOSURE: THIS PAGE MAY CONTAIN AFFILIATE LINKS, MEANING I GET A COMMISSION IF YOU DECIDE TO MAKE A PURCHASE THROUGH MY LINKS, AT NO COST TO YOU. PLEASE READ MY DISCLOSURE FOR MORE INFO.