When a capacitor and resistor are connected in series to a voltage source via a switch, the capacitor will charge exponentially when the switch is closed and discharge exponentially when the switch is opened. The time constant (τ) of the circuit, which is equal to the product of the resistance (R) and capacitance (C), determines the rate of charge and discharge. Specifically, it represents the time required for the capacitor to reach 63.2% of its maximum charge during charging or 36.8% of its maximum charge during discharging. The current through the circuit also follows an exponential decay/growth formula determined by the time constant τ.

1. GROWTH AND DECAY OF CHARGE THROUGH RC CIRCUIT
Consider a circuit which consisting a capacitor C and a resistance R, are connected in series. The
combination is connected across a source of EMF E0 through a pressing key K as shown.
a) Growth of charge :
Before the key K is pressed no current flows through the circuit,hence Q=0 when t=0. After the key K
is pressed, the capacitor begins to receive charge. The chargingprocess continues till the steady state value,
proportional to the e m f E0 of the sourceis reached.Let the instantaneous chargeon the capacitor beQ. The
instantaneous PD across C is E0 −
Q
C
which is equal to RI drop across the resistance R, where I be the
instantaneous current. Then from Ohms law, we can have
E0 −
Q
C
= RI … … … … … … … … … … … … … … … … 1
By definition time rate of flow of charge is current i.e
dQ
dt
= I, hence we can have
E0 −
Q
C
= R
dQ
dt
… … … … … … … … … … … … … … … … 2
On rewriting the above equation we get
dt =
R
E0 −
Q
C
dQ … … … … … … … … … … … … .. . . 3
Multiplying and dividing the numerator of RHS by –C, we get
dt =
−CR (−
1
C
)
E0 −
Q
C
dQ … … … … … … … … … … … … 4
Integrating equation 4 yield,
t = −CR loge [E0 −
Q
C
] + A … … … … … … … … … … … … 5
Where A is the constant of integration, its value can be evaluated by initial conditions. When t=0, Q=0,
Therefore
0 = −CR loge
[E0
] + A
Or A = CR loge
[E0
]
On substituting for A in equation (6) we get,
t = −CR loge [E0 −
Q
C
] + CR loge E … … … 6
or
t = −CR loge [
E0 −
Q
C
E
] … … … … … … … … 7
In exponential form we can put above as,
E0 −
Q
C
E
= e
−
t
CR
or Q = CE0 (1 − e
−
t
CR) … … … … … … … … 8
We have CE0 = Q0 is the maximum steady state charge on the capacitor,
Hence
Q = Q0 (1 − e
−
t
CR) … … … … … … … … … . . 9
A graph of Q against t is plotted, as shown.
Q0
From the graph it is clear that the charging of
capacitor follows an exponential law. In the
equation (9) CR is known as tome constant.
Q
Q =0.632 Q0

2. Time constant:
The symbol used for time constant is 𝜆; i.e 𝜆 = 𝐶𝑅
Q = Q0 (1 − e
−
t
𝜆) ………………………….. 10
Now put t= 𝜆 in the relation (10) we get,
Q = Q0 (1 − e
−
t
t )
On simplification, we get
Q = Q0 (1 −
1
e
)
Q = CE0 (1 −
1
2.718
)
Q = 0.632CE0
Q = 0.632Q0 … … … … … … … … . . 11
Definition: From the relation (11) we define time constant of RC circuitas,itis the amount of time taken by the
capacitor to charge from its minimum value to 0.632 times the maximum steady state value Q0.
Rate of growth of charge of capacitor:
Consider the relation,
Q = Q0 (1 − e
−
t
CR) … … … … … … … … … . . 12
Differentiating the equation w.r.t to t, we get,
dQ
dt
=
d
dt
[Q0 (1 − e
−
t
CR )]
Or
dQ
dt
=
Q0
CR
e
−
t
CR
Or
dQ
dt
=
Q0
CR
(
Q0 − Q
Q0
)
Or
dQ
dt
=
1
CR
(Q0 − Q) … … … … … … 13
From relation (13) itis learntthatas thetime constantCRbecome larger,the growth of chargedoes more slowly
and vice-versa.
Current in the circuit:
To obtain the current equation, consider the relation,
Q = Q0 (1 − e
−
t
CR) … … … … … … … … … . . 14
We know that
dQ
dt
= I , hence on differentiating and simplifying the relation (15) we get,
dQ
dt
=
Q0
CR
e
−
t
CR
I = I0e
−
t
CR …………………………. 15
b) Decay of charge :
Before the key K is released,the chargeon capacitor isQ=Q0 when t=0. After the key K is released,the capacitor
begins to discharge through resistance R.. Let the instantaneous charge on the capacitor be Q. The
instantaneous PD across C is
Q
C
which is equal to RI drop across the resistanceR, where I is the instantaneous
current. Then from Ohms law, we can have
Q
C
= RI … … … … … … … … … … … … … … … … 1
By definition time rate of charge is current i.e −
dQ
dt
= I, here negative sign indicate the reverse current in the
circuit, hence we can have
Q
C
= −R
dQ
dt
… … … … … … … … … … … … … … … … 2
On rewriting the above equation we get
dt =
−CR
Q
dQ … … … … … … … … … … … … .. . . 3
Integrating equation 3 yield,

3. t = −CR loge
[Q] + A … … … … … … … … … … 4
Where A is the constant of integration, its value can be evaluated by initial conditions. When t=0, Q=Q0,
Therefore
0 = −CR loge
[Q0
] + A
Or A = CR loge
[Q0
]
Hence on substituting for A in equation (6) we get,
t = −CR loge
[Q] + CR loge
[Q0
] … … … … … … . . 5
or
t = −CR loge [
Q
Q0
] … … … … … … … … 6
In exponential form we can put above as,
[
Q
Q0
] = e
−
t
CR
or Q = Q0e
−
t
CR … … … … … … … … 7
A graph of Q against t is plotted, as shown. From the graph itis clear that the chargingof
capacitor follows an exponential law. In the
equation (7) CR is known as tome constant.
Q0
Q
Q = 0.368 Q0
𝜆 t
Time constant:
The symbol used for time constant is 𝜆; i.e 𝜆 = 𝐶𝑅
Q = Q0e
−
t
CR ……………… 8
Now put t= 𝜆 in the relation (8) we get,
Q = Q0e
−
t
t
On simplification, we get
Q = Q0
1
e
Q = Q0
1
2.718
Q = 0.368Q0 … … … … … 9
Definition:
From the relation (9) we can define time constantof RC circuitas,itisthe amount of time taken by the capacitor
to discharge from its maximum value to 0.368 times the maximum steady state value Q0.
Rate of decay of charge of capacitor:
Consider the relation,
Q = Q0e
−
t
CR … … … … … … … … … … . 10
Differentiating the equation w.r.t to t, we get,
dQ
dt
=
d
dt
[Q0e
−
t
CR ]
Or
dQ
dt
=
Q0
CR
e
−
t
CR
Or
dQ
dt
=
Q0
CR
[
Q
Q0
]
Or
dQ
dt
=
1
CR
Q … … … … … … 11
From relation (11) itis learntthat,as thetimeconstantCR become larger,the growth of chargedoes more slowly
and vice-versa.

4. Current in the circuit:
To obtain the current equation, consider the relation,
Q = Q0e
−
t
CR … … … … … … … … … . . 12
We know that -
dQ
dt
= I , hence on differentiating and simplifying the relation (12) we get,
dQ
dt
=
Q0
CR
e
−
t
CR
Hence
I = I0e
−
t
CR …………………………. 13
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