An RC circuit contains a resistor and capacitor in series. When power is applied, maximum current (I0) flows which charges the capacitor. The charge on the capacitor (Q) is equal to the capacitance (C) multiplied by the voltage (Ɛ). The expressions for the charge (q(t)), voltage across the capacitor (VC), and current (I) during the charging phase are given. The time constant (RC) represents the time for the current to decrease to 37% of its initial value. For the discharging phase, the expression for the remaining charge is given.

1. ‫الرح‬ ‫الرحمن‬ ‫هللا‬ ‫بسم‬‫يم‬
RL + RC CIRCUIT

2. A circuit containing a series combination of a resistor and a
capacitor is called an RC circuit.
Maximum current of the circuit, I0 =
Ɛ
𝐑
[When, t = 0, Maximum current flows]
Maximum Charge on Capacitor, Q = CƐ
RC circuit. Charging case

3. Expression of Charge q(t), voltage VC and
current I during charging phase of an RC
circuit: The voltage across a capacitor cannot
change instantaneously.By applying KVL, We get,
I =
𝐝𝐪
𝐝𝐭Putting this value of I and after rearranging, we get,

4. Equation of instantaneous current can be obtained by differentiating
the equation of charge,
Voltage across the capacitor is, VC = q(t) / C

5. Graph: Time vs Charge (or voltage) Graph: Current vs Time
time constant =RC
represents the time interval during which the current decreases
to 1/e of its initial value; that is, after a time interval t, the current
decreases
i = 0.368 Ii

6. By applying KVL in opposite direction, we get,
Now, I =
𝐝𝐪
𝐝𝐭
. Again, when t=0 then q = Q
RC circuit Discharging case
This is the equation of charge
remaining in the capacitor. The
equation of current can be
obtained by differentiating this
equation.

8. VL = – L
diL
dt
and VR = iL R
By applying KVL we get .. E – VR – VL = 0
or E – iL R – L
diL
dt
= 0
Let,
E
R
– iL = x then,
diL
dt
= –
dx
dt
x +
L
R
dx
dt
= 0 ,,
dx
x
= –
R
L
dt
RL circuit
By integrating within the limit (x0 to x) and (0 to t),
ln
x
x𝟎
= –
R
L
t , x = x0 e –Rt/L
When t = 0, current iL = 0 thus, x = x0 =
E
R
When t = t, current = iL thus, x =
E
R
– iL
Now,
E
R
– iL =
E
R
e –Rt/L
, iL =
E
R
( 1 – e –Rt/L )

9. 𝐕𝐨𝐥𝐭𝐚𝐠𝐞 𝐚𝐜𝐫𝐨𝐬𝐬 𝐢𝐧𝐝𝐮𝐜𝐭𝐨𝐫, VL = E e –t/τ
𝐕𝐨𝐥𝐭𝐚𝐠𝐞 𝐚𝐜𝐫𝐨𝐬𝐬 𝐫𝐞𝐬𝐢𝐬𝐭𝐨𝐫, VR = E (1 – e –t/τ )
The current in the circuit , iL =
E
R
( 1 – e –Rt/L )
time constant = τ =
L
R
Physically, τ is the time it takes the current
in the circuit to reach ( 1 – e –1 ) = 0.637 or
63.7% of its final value
E
R
.

10. By applying KVL we get .. VR – VL = 0
or iL R – L
diL
dt
= 0
At no E
d𝒊
i
= –
R
L
dt
Ln i= –
R
L
dt + const At , t=0, i=
E
R
const=ln
E
R Ln
𝒊𝑹
𝑬
=
−𝑹
𝑳
t
𝐕𝐨𝐥𝐭𝐚𝐠𝐞 𝐚𝐜𝐫𝐨𝐬𝐬 𝐢𝐧𝐝𝐮𝐜𝐭𝐨𝐫, VL = -E e –t/τ
𝐕𝐨𝐥𝐭𝐚𝐠𝐞 𝐚𝐜𝐫𝐨𝐬𝐬 𝐫𝐞𝐬𝐢𝐬𝐭𝐨𝐫, VR = E e –t/τ
The current in the circuit , iL =
E
R
e –Rt/L

11. Volt across L
Volt across R
E
E=0