First Order Differential Circuits

MADE BY-Aadish Bhagat (130110111003)Arjun Gupta (130110111021)Shubham Kurup (130110111027)Pragnesh Patel (130110111042)Denish Shah (140113111012)

GUIDED BY – Prof. Shankar K. Parmar

G. H. Patel College of Engineering & Technology

Subject :- Circuits and Networks (2130901)

First Order Differential Circuits• Consider a differential equation of nth order• • Where,• a0,a1,a2…. Are positive constants

• i=dependent variable• t=independent variable• v(t)=forcing function

• solution of a non homogeneous differential equation using integrating factor

• consider a first order non homogeneous equation as

• +pi=q, Where

• P is constant and q may be function of independent variable or constant

• Multiplying the equation by ept

• ept + ept pi = qept

• (i ept)=q ept

• now integrating the above equation

• ept= + k-------------(1)• equation 1 is generalized solution of the equation

• consider q is constant• i(t)=e-pt ept + k e-pt

• i(t) = + k e-pt

• at t=0 switch is getting closed

FIG

• after switching actionv = iR + Where p =and q = • i(t) = + K

• ABOUT INDUCTOR(L): 1)L behaves as open circuit if it is initially not charged

• 2)L behaves as current source if it is initially charged

• 3)L behaves as short circuit at steady state

Seven important steps to solve first order differential equation

• 1)Draw the circuit after switching action takes place.

• 2)Apply KVL or KCL to get first order differential equation.

• 3)Get generalized solution.• 4)Draw the circuit at t=0-.

• 5)Identify whether L or C is charged or not.• 6)Replace it at t=o+ and get the quantity for which

we have found generalized solution.• 7)Get particular solution.

• At t=0 switch is getting closed.• Steady state is attained previously.

QUESTION 30Ω 20Ω

𝟏𝟐𝑯10V

20V

𝒊 (𝒕 )

Answer• After switching

30Ω 20Ω

10V

20V

𝟏𝟐𝑯

𝒊 (𝒕 )

• ←first order differential equation

P Q

𝒊𝒍(𝟎−)

30Ω

20Ω

20V

10V

𝒕=𝟎−

• Putting the value of K in general solution𝐾=0.1

𝒕=𝟎+¿¿ 30Ω

20Ω

20V

10V

𝟑𝟓𝑨

Equation For Capacitor

After Switching

𝑖(𝑡)

𝑖(𝑡)

• V = iR + .•

Differentiate•

+ I = 0.•

Q = 0 , P = •

i(t) = k e

Note : Voltage across capacitor can not change instantly

T =

Vc(0-) = 0 V

i(0) = i(0) = =K i(t) =

T =

Vc (0-)=0v

• ABOUT Capacitor (C): 1)C behaves as short circuit if it is initially not charged

2)C behaves as voltage source if it is initially charged 3)C behaves as open circuit at steady state

Question:- find

5V

1Ω

2Ω

𝟏𝟐Ω

1Ω𝟏𝟐𝐅

𝑖𝑐

• After switching:-

0 2V -V32

1

02

1 )V -V(2

1

5-V

b a

b a a

dt

dVdt

dV

a

aNode A:-

5V

1Ω

2Ω

𝟏𝟐Ω

1Ω

𝟏𝟐𝐅

𝑖𝑐

ANSWER

Node B :-

7

5 4V V

5 4V V72

5 2V V]21

2

1[

0 )V -V(21

V

2

5 -V

a b

a b

a b

a bb b

tK

dt

dVdt

dVdt

dVdt

dV

a

a

a

a

e 726

26

90 (t)V

7

90 V

7

267

45 V

7

13

2

17

10 5 V]

7

83[

2

1

5]7

5 4V[ 2 -V3

2

1

a

a

a

a

a a

At t=

v312

3

5*23

)0-

( V c

5V

1Ω

𝟑𝟐Ω

-

+

𝒗𝒄𝟎−

At

e

e

e

t

t

dt

dV

t

K

a

7

26*

14

12

]7

26*7

26*

26

12[

2

1

2

1(t)i

7

26

26

12

26

90(t)V

26

903(0)V

3V (0)V

c

a

a

a

0t

3V

1Ω2Ω

1Ω

𝟏𝟐Ω

5V

THANK YOU