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‫الرح‬ ‫الرحمن‬ ‫هللا‬ ‫بسم‬‫يم‬
RL + RC CIRCUIT
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
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,
Equation of instantaneous current can be obtained by differentiating
the equation of charge,
Voltage across the capacitor is, VC = q(t) / C
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
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.
RL + RC CIRCUIT  ( بسم الله الرحمن الرحيم )
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 )
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
.
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
Volt across L
Volt across R
E
E=0

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RL + RC CIRCUIT ( بسم الله الرحمن الرحيم )

  • 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