Prepared by

Md. Amirul Islam

Lecturer

Department of Applied Physics & Electronics

Bangabandhu Sheikh Mujibur Rahman Science &

Technology University, Gopalganj – 8100

A circuit containing a series combination of a resistor and a

capacitor is called an RC circuit.

Maximum current of the circuit, I0 =Ɛ𝐑

[When, t = 0, Maximum

current flows]

Maximum Charge on Capacitor, Q = CƐ [When t = tf = 5τ]

Reference: Physics II by Robert Resnick and David Halliday, Topic – 28.4, Page – 883

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.

Reference: Physics II by Robert Resnick and David Halliday, Topic – 28.4, Page – 883

By applying KVL, We get,

Current through the circuit is the time rate of change of the

charge on the capacitor plates. So,I =

𝐝𝐪

𝐝𝐭

Putting this value of I and after rearranging, we get,

Reference: Physics II by Robert Resnick and David Halliday, Topic – 28.4, Page – 883

This is the equation of charge stored in a capacitor. Voltage

across the capacitor is, VC = q(t) / C

Equation of instantaneous current can be obtained by

differentiating the equation of charge,

Reference: Physics II by Robert Resnick and David Halliday, Topic – 28.4, Page – 883

Graph: Time vs Charge (or voltage)

The charge is zero at t = 0 and approaches the maximum value

CƐ as t → ∞ . The current I has its maximum value at t = 0 and

decays exponentially to zero as t → ∞. From the graph, we

observe that after 5

Graph: Current vs Time

Expression of Charge q(t), voltage VC and current I during

discharging phase of an RC circuit:

Reference: Physics II by Robert Resnick and David Halliday, Topic – 28.4, Page – 885

By applying KVL in clockwise direction, we get,

Now, I =𝐝𝐪

𝐝𝐭. Again, when t=0 then q = Q and when t=t then q = q

Reference: Physics II by Robert Resnick and David Halliday, Topic – 28.4, Page – 885

This is the equation of charge remaining in the capacitor. The

equation of current can be obtained by differentiating this equation.

Reference: Circuit Analysis by Robert Boylestad, Figure– 10.24 & 10.39, Page – 395

Figure: Charging – discharging network

Here,

VC – Voltage across capacitor

iC – Circuit current

VR – Voltage across resistor

Figure: Charging – discharging cycles

Math. Problem: Find the mathematical expressions for the

transient behavior of vC, iC, and vR for the circuit of Figure

when the switch is moved to position 1.

b. How much time must pass before it can be assumed, for all

practical purposes, that iC ≈ 0 A and vC ≈ E volts?

c. When the switch is placed to position 2, find the mathematical

expressions for the transient behavior of vC, iC, and vR

Reference: Circuit Analysis by Robert Boylestad, , Example – 10.5, Page – 393

Math. Problem: An uncharged capacitor and a resistor are

connected in series to a battery, as shown in Figure. If C = 5μF

and Ɛ = 12V, R = 0.8 MΩ, find the time constant of the circuit,

the maximum charge on the capacitor, the maximum current in

the circuit, and the charge and current as functions of time.

Reference: Physics II by Robert Resnick and David Halliday, Example– 28.11, Page – 887

Time constant, τ = RC = 4s

Maximum charge, Q = CƐ = 60 μC

Maximum Current, I0 =ƐR = 15 μA

A circuit containing a series combination of a resistor and an

inductor is called an RL circuit.

an inductor in a circuit opposes changes in the current through that

circuit.

Reference: Physics II by Robert Resnick and David Halliday, Quick Quiz– 32.1, Page – 1018

When switch S is closed at t = 0.

The current in the circuit begins

to increase, and a back emf that

opposes the increasing current is

induced in the inductor. The back

emf is,VL = – L

diLdt

In R-L circuits, the energy is stored in the form of a magnetic field

established by the current through the coil.

Expression of transient current iL for the storage cycle of an

RL circuit:

Reference: Physics II by Robert Resnick and David Halliday, Quick Quiz– 32.1, Page – 1018

Here, VL = – LdiLdt

and VR = iL R

By applying KVL we get,

E – VR – VL = 0

or, E – iL R – LdiLdt

= 0

Let,ER

– iL = x then,diLdt

= –dxdt

Now, x +LR

dxdt

= 0

or,dxx

= –RL

dt

Reference: Physics II by Robert Resnick and David Halliday, Quick Quiz– 32.1, Page – 1018

By integrating within the limit (x0 to x) and (0 to t),

lnxx𝟎

= –RL

t

or, x = x0 e –Rt/L

When t = 0, current iL = 0 thus, x = x0 =ER

When t = t, current = iL thus, x =ER

– iL

Now,ER

– iL =ER

e –Rt/L

or, iL =ER

( 1 – e –Rt/L )

For RL circuit time constant, τ =LR

Reference: Physics II by Robert Resnick and David Halliday, Quick Quiz– 32.1, Page – 1019

Thus the equation of current, iL =ER

( 1 – e –t/τ )

When t → ∞ current reaches

final value or steady state

value and then, iL =ER

. That is,

the inductor acts as a short

circuit.

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 ER

.

Figure: Storage Cycle

T𝐡𝐞 𝐞𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐨𝐟 𝐕𝐨𝐥𝐭𝐚𝐠𝐞 𝐚𝐜𝐫𝐨𝐬𝐬 𝐢𝐧𝐝𝐮𝐜𝐭𝐨𝐫, VL =ER

e –t/τ

T𝐡𝐞 𝐞𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐨𝐟 𝐕𝐨𝐥𝐭𝐚𝐠𝐞 𝐚𝐜𝐫𝐨𝐬𝐬 𝐫𝐞𝐬𝐢𝐬𝐭𝐨𝐫, VR =ER

(1 – e –t/τ )

Find the mathematical expressions for the transient behavior

of iL and vL for the circuit of figure after closing of the switch.

Reference: Circuit Analysis by Robert Boylestad, Example– 12.4, Page – 484

The switch in figure is thrown closed at t = 0

(a) Find the time constant of the circuit.

(b) Calculate the current in the circuit at t = 2 ms

Reference: Physics II by Robert Resnick and David Halliday, Example– 32.3, Page – 1021

RL Circuit – decay Phase:

Reference: Physics II by Robert Resnick and David Halliday, Problem – 18, Page – 1036, 1020

At the instant S2 is closed, S1 is

opened, the battery is no longer part

of the circuit. Stored magnetic energy

inside the inductor starts to decay. By

applying KVL now,

– iL R – LdiLdt

= 0

or,diLiL

= –RLdt

After integrating within the limit (i0 to iL) and (0 to t) we get,

ln iL – ln i0 = –RL

t

or, iL = i0 e – Rt/L = i0 e – t/τ =ER

e – t/τ

So the current through the inductor decrease according to this equation.

A circuit containing a series combination of a resistor, an

inductor and a capacitor is called an RLC circuit.

Let, the capacitor has an initial

charge Qmax. Thus energy is stored

in the capacitor and inductor.

When the switch is closed, the

resistor causes transformation to

internal energy. The rate of energy

transformation is:

Reference: Physics II by Robert Resnick and David Halliday, Topic – 32.6, Page – 1031

The negative sign indicates that the energy of the circuit is

decreasing in time.

Reference: Physics II by Robert Resnick and David Halliday, Topic – 32.6, Page – 1031

This energy is equal to the energy of capacitor and inductor. Thus,

Solution of this equation is,

ωd is the angular frequency at which the circuit oscillates.

Reference: Physics II by Robert Resnick and David Halliday, Topic – 32.6, Page – 1031

A plot of the charge versus time for the damped oscillator is shown

in Figure. the value of the charge on the capacitor undergoes a

damped harmonic oscillation.

When R << 4L/C , second term of ωd can be neglected. Then ωd =

1/ LC and we get un-damped oscillation i.e. pure sinusoidal wave.

The circuit where no resistance is present, the LC circuit is called

tank circuit and we can get pure sinusoidal wave form this circuit.