Initial & Final Condition | Network Theory (Electric Circuits) - Electrical Engineering (EE) PDF Download

 Page 1


                                  
 
 
UNIT-1  
D.C Transient Analysis 
 
? Transient response of R-L, R-C, R-L-C circuits (Series and parallel 
combinations) for D.C. excitations  
? Initial conditions 
? Solution using differential equation and Laplace transform 
method. 
? Summary of Important formulae and Equations 
? Illustrative examples 
 
 
 
 
 
 
 
 
 
  
Page 2


                                  
 
 
UNIT-1  
D.C Transient Analysis 
 
? Transient response of R-L, R-C, R-L-C circuits (Series and parallel 
combinations) for D.C. excitations  
? Initial conditions 
? Solution using differential equation and Laplace transform 
method. 
? Summary of Important formulae and Equations 
? Illustrative examples 
 
 
 
 
 
 
 
 
 
  
                                  
 
Introduction:  
In this chapter we shall study transient response of the RL, RC series and RLC circuits with 
external DC excitations. Transients are generated in Electrical circuits due to abrupt changes in 
the operating conditions when energy storage elements like Inductors or capacitors are 
present. Transient response is the dynamic response during the initial phase before the steady 
state response is achieved when such abrupt changes are applied. To obtain the transient 
response of such circuits we have to solve the differential equations which are the governing 
equations representing the electrical behavior of the circuit. A circuit having a single energy 
storage element i.e. either a capacitor or an Inductor is called a Single order circuit and it’s 
governing equation is called a First order Differential Equation. A circuit having both Inductor 
and a Capacitor is called a Second order Circuit and it’s governing equation is called a Second 
order Differential Equation. The variables in these Differential Equations are currents and 
voltages in the circuit as a function of time.   
A solution is said to be obtained to these equations when we have found an expression for the 
dependent variable that satisfies both the differential equation and the prescribed initial 
conditions. The solution of the differential equation represents the Response of the circuit. 
Now we will find out the response of the basic RL and RC circuits with DC Excitation. 
RL CIRCUIT with external DC excitation: 
Let us take a simple RL network subjected to external DC excitation as shown in the figure. The 
circuit consists of a battery whose voltage is V in series with a switch, a resistor R, and an 
inductor L. The switch is closed at t = 0. 
 
 
 
 
Fig 1.1: RL Circuit with external DC excitation 
 
When the switch is closed current tries to change in the inductor and hence a voltage V
L
(t) is 
induced across the terminals of the Inductor in opposition to the applied voltage. The rate of 
change of current decreases with time which allows current to build up to it’s maximum value. 
 
Page 3


                                  
 
 
UNIT-1  
D.C Transient Analysis 
 
? Transient response of R-L, R-C, R-L-C circuits (Series and parallel 
combinations) for D.C. excitations  
? Initial conditions 
? Solution using differential equation and Laplace transform 
method. 
? Summary of Important formulae and Equations 
? Illustrative examples 
 
 
 
 
 
 
 
 
 
  
                                  
 
Introduction:  
In this chapter we shall study transient response of the RL, RC series and RLC circuits with 
external DC excitations. Transients are generated in Electrical circuits due to abrupt changes in 
the operating conditions when energy storage elements like Inductors or capacitors are 
present. Transient response is the dynamic response during the initial phase before the steady 
state response is achieved when such abrupt changes are applied. To obtain the transient 
response of such circuits we have to solve the differential equations which are the governing 
equations representing the electrical behavior of the circuit. A circuit having a single energy 
storage element i.e. either a capacitor or an Inductor is called a Single order circuit and it’s 
governing equation is called a First order Differential Equation. A circuit having both Inductor 
and a Capacitor is called a Second order Circuit and it’s governing equation is called a Second 
order Differential Equation. The variables in these Differential Equations are currents and 
voltages in the circuit as a function of time.   
A solution is said to be obtained to these equations when we have found an expression for the 
dependent variable that satisfies both the differential equation and the prescribed initial 
conditions. The solution of the differential equation represents the Response of the circuit. 
Now we will find out the response of the basic RL and RC circuits with DC Excitation. 
RL CIRCUIT with external DC excitation: 
Let us take a simple RL network subjected to external DC excitation as shown in the figure. The 
circuit consists of a battery whose voltage is V in series with a switch, a resistor R, and an 
inductor L. The switch is closed at t = 0. 
 
 
 
 
Fig 1.1: RL Circuit with external DC excitation 
 
When the switch is closed current tries to change in the inductor and hence a voltage V
L
(t) is 
induced across the terminals of the Inductor in opposition to the applied voltage. The rate of 
change of current decreases with time which allows current to build up to it’s maximum value. 
 
                                  
 
It is evident that the current i(t) is zero before t = 0.and we have to find out current i(t) for time 
t >0. We will find i(t)for time t >0 by writing the appropriate circuit equation and then solving it 
by separation of the variables and integration.  
Applying Kirchhoff’s voltage law to the above circuit we get: 
V  =  v
R
(t)+ v
L
(t) 
i (t) = 0 fort <0and 
Using the standard relationships of Voltage and Current for the Resistors and Inductors we can 
rewrite the above equations as  
    V  =    Ri + Ldi/dt for t >0 
 
One direct method of solving such a differential equation consists of writing the equation in 
such a way that the variables are separated, and then integrating each side of the equation. The 
variables in the above equation are I and t. This equation is multiplied by dt and arranged with 
the variables separated as shown below: 
 
Ri. dt + Ldi = V. dt 
 
i.e    Ldi= (V – Ri)dt 
 
i.e    Ldi / (V – Ri)     =   dt 
 
Next each side is integrated directly to get : 
 
- (L/R ) ln(V- Ri) =t + k 
 
Where k is the integration constant. In order to evaluate k, an initial condition must be invoked. 
Prior to t = 0, i (t) is zero, and thus i (0-) = 0. Since the current in an inductor cannot change by 
a finite amount in zero time without being associated with an infinite voltage, we have i (0+) = 
0. Setting i = 0 at t = 0, in the above equation we obtain 
- (L/R ) ln(V)            = k 
and, hence, 
- L/R[ln(V- Ri) - ln V]= t 
Rearranging we get  
 ln[ (V- Ri) /V] =  - (R/L)t 
Taking antilogarithm on both sides we get  
(V–Ri)/V= e
-Rt/L
 
From which we can see that   
i(t) = (V/R)–(V/R)e
-Rt/L
 for  t >0 
 
Thus, an expression for the response valid for all time t would be 
 
i(t) = V/R [1- e
-Rt/L
 ] 
 
This is normally written as: 
Page 4


                                  
 
 
UNIT-1  
D.C Transient Analysis 
 
? Transient response of R-L, R-C, R-L-C circuits (Series and parallel 
combinations) for D.C. excitations  
? Initial conditions 
? Solution using differential equation and Laplace transform 
method. 
? Summary of Important formulae and Equations 
? Illustrative examples 
 
 
 
 
 
 
 
 
 
  
                                  
 
Introduction:  
In this chapter we shall study transient response of the RL, RC series and RLC circuits with 
external DC excitations. Transients are generated in Electrical circuits due to abrupt changes in 
the operating conditions when energy storage elements like Inductors or capacitors are 
present. Transient response is the dynamic response during the initial phase before the steady 
state response is achieved when such abrupt changes are applied. To obtain the transient 
response of such circuits we have to solve the differential equations which are the governing 
equations representing the electrical behavior of the circuit. A circuit having a single energy 
storage element i.e. either a capacitor or an Inductor is called a Single order circuit and it’s 
governing equation is called a First order Differential Equation. A circuit having both Inductor 
and a Capacitor is called a Second order Circuit and it’s governing equation is called a Second 
order Differential Equation. The variables in these Differential Equations are currents and 
voltages in the circuit as a function of time.   
A solution is said to be obtained to these equations when we have found an expression for the 
dependent variable that satisfies both the differential equation and the prescribed initial 
conditions. The solution of the differential equation represents the Response of the circuit. 
Now we will find out the response of the basic RL and RC circuits with DC Excitation. 
RL CIRCUIT with external DC excitation: 
Let us take a simple RL network subjected to external DC excitation as shown in the figure. The 
circuit consists of a battery whose voltage is V in series with a switch, a resistor R, and an 
inductor L. The switch is closed at t = 0. 
 
 
 
 
Fig 1.1: RL Circuit with external DC excitation 
 
When the switch is closed current tries to change in the inductor and hence a voltage V
L
(t) is 
induced across the terminals of the Inductor in opposition to the applied voltage. The rate of 
change of current decreases with time which allows current to build up to it’s maximum value. 
 
                                  
 
It is evident that the current i(t) is zero before t = 0.and we have to find out current i(t) for time 
t >0. We will find i(t)for time t >0 by writing the appropriate circuit equation and then solving it 
by separation of the variables and integration.  
Applying Kirchhoff’s voltage law to the above circuit we get: 
V  =  v
R
(t)+ v
L
(t) 
i (t) = 0 fort <0and 
Using the standard relationships of Voltage and Current for the Resistors and Inductors we can 
rewrite the above equations as  
    V  =    Ri + Ldi/dt for t >0 
 
One direct method of solving such a differential equation consists of writing the equation in 
such a way that the variables are separated, and then integrating each side of the equation. The 
variables in the above equation are I and t. This equation is multiplied by dt and arranged with 
the variables separated as shown below: 
 
Ri. dt + Ldi = V. dt 
 
i.e    Ldi= (V – Ri)dt 
 
i.e    Ldi / (V – Ri)     =   dt 
 
Next each side is integrated directly to get : 
 
- (L/R ) ln(V- Ri) =t + k 
 
Where k is the integration constant. In order to evaluate k, an initial condition must be invoked. 
Prior to t = 0, i (t) is zero, and thus i (0-) = 0. Since the current in an inductor cannot change by 
a finite amount in zero time without being associated with an infinite voltage, we have i (0+) = 
0. Setting i = 0 at t = 0, in the above equation we obtain 
- (L/R ) ln(V)            = k 
and, hence, 
- L/R[ln(V- Ri) - ln V]= t 
Rearranging we get  
 ln[ (V- Ri) /V] =  - (R/L)t 
Taking antilogarithm on both sides we get  
(V–Ri)/V= e
-Rt/L
 
From which we can see that   
i(t) = (V/R)–(V/R)e
-Rt/L
 for  t >0 
 
Thus, an expression for the response valid for all time t would be 
 
i(t) = V/R [1- e
-Rt/L
 ] 
 
This is normally written as: 
                                  
 
i(t) = V/R [1- e
-t./t
 ]  
 
where ‘t’ is called  the time constant of the circuit and it’s unit is seconds.  
 
The voltage across the resistance and the Inductor for t >0can be written as : 
  
v
R
(t)    =i(t).R   = V [1- e
-t./t
 ] 
 
v
L
(t)   =      V -v
R
(t)   =  V -V [1- e
-t./t
 ]  = V (e
-t./t
) 
 
A plot of the current i(t)  and the voltages v
R
(t) &  v
L
(t)  is shown in the figure below.  
 
 
 
 
Fig 1.2: Transient current and voltages in the Series RL circuit.  
 
At   t =   ‘t’ the voltage across the inductor will be 
 
                                                          v
L
(t)     =    V (e
-t /t
)  =  V/e    =   0.36788 V 
 
and the voltage across the Resistor  will be       v
R
(t)    =     V [1- e
-t./t
 ]  =  0.63212 V  
 
The plots of current i(t)  and the voltage across the Resistor v
R
(t) are called exponential growth 
curves and the voltage across the inductor v
L
(t)is  called exponential decay curve. 
 
RCCIRCUIT with external DC excitation: 
 
A series RC circuit with external DC excitation V volts connected through a switch is shown in 
the figure below. If the capacitor is not charged initially i.e. it’s voltage is zero ,then after the 
switch S is closed at time t=0, the capacitor voltage builds up gradually and reaches it’s steady 
state value of V volts after a finite time. The charging current will be maximum initially (since 
initially capacitor voltage is zero and voltage across a capacitor cannot change instantaneously) 
and then it will gradually comedown as the capacitor voltage starts building up. The current and 
the voltage during such charging periods are called Transient Current and Transient Voltage.  
Page 5


                                  
 
 
UNIT-1  
D.C Transient Analysis 
 
? Transient response of R-L, R-C, R-L-C circuits (Series and parallel 
combinations) for D.C. excitations  
? Initial conditions 
? Solution using differential equation and Laplace transform 
method. 
? Summary of Important formulae and Equations 
? Illustrative examples 
 
 
 
 
 
 
 
 
 
  
                                  
 
Introduction:  
In this chapter we shall study transient response of the RL, RC series and RLC circuits with 
external DC excitations. Transients are generated in Electrical circuits due to abrupt changes in 
the operating conditions when energy storage elements like Inductors or capacitors are 
present. Transient response is the dynamic response during the initial phase before the steady 
state response is achieved when such abrupt changes are applied. To obtain the transient 
response of such circuits we have to solve the differential equations which are the governing 
equations representing the electrical behavior of the circuit. A circuit having a single energy 
storage element i.e. either a capacitor or an Inductor is called a Single order circuit and it’s 
governing equation is called a First order Differential Equation. A circuit having both Inductor 
and a Capacitor is called a Second order Circuit and it’s governing equation is called a Second 
order Differential Equation. The variables in these Differential Equations are currents and 
voltages in the circuit as a function of time.   
A solution is said to be obtained to these equations when we have found an expression for the 
dependent variable that satisfies both the differential equation and the prescribed initial 
conditions. The solution of the differential equation represents the Response of the circuit. 
Now we will find out the response of the basic RL and RC circuits with DC Excitation. 
RL CIRCUIT with external DC excitation: 
Let us take a simple RL network subjected to external DC excitation as shown in the figure. The 
circuit consists of a battery whose voltage is V in series with a switch, a resistor R, and an 
inductor L. The switch is closed at t = 0. 
 
 
 
 
Fig 1.1: RL Circuit with external DC excitation 
 
When the switch is closed current tries to change in the inductor and hence a voltage V
L
(t) is 
induced across the terminals of the Inductor in opposition to the applied voltage. The rate of 
change of current decreases with time which allows current to build up to it’s maximum value. 
 
                                  
 
It is evident that the current i(t) is zero before t = 0.and we have to find out current i(t) for time 
t >0. We will find i(t)for time t >0 by writing the appropriate circuit equation and then solving it 
by separation of the variables and integration.  
Applying Kirchhoff’s voltage law to the above circuit we get: 
V  =  v
R
(t)+ v
L
(t) 
i (t) = 0 fort <0and 
Using the standard relationships of Voltage and Current for the Resistors and Inductors we can 
rewrite the above equations as  
    V  =    Ri + Ldi/dt for t >0 
 
One direct method of solving such a differential equation consists of writing the equation in 
such a way that the variables are separated, and then integrating each side of the equation. The 
variables in the above equation are I and t. This equation is multiplied by dt and arranged with 
the variables separated as shown below: 
 
Ri. dt + Ldi = V. dt 
 
i.e    Ldi= (V – Ri)dt 
 
i.e    Ldi / (V – Ri)     =   dt 
 
Next each side is integrated directly to get : 
 
- (L/R ) ln(V- Ri) =t + k 
 
Where k is the integration constant. In order to evaluate k, an initial condition must be invoked. 
Prior to t = 0, i (t) is zero, and thus i (0-) = 0. Since the current in an inductor cannot change by 
a finite amount in zero time without being associated with an infinite voltage, we have i (0+) = 
0. Setting i = 0 at t = 0, in the above equation we obtain 
- (L/R ) ln(V)            = k 
and, hence, 
- L/R[ln(V- Ri) - ln V]= t 
Rearranging we get  
 ln[ (V- Ri) /V] =  - (R/L)t 
Taking antilogarithm on both sides we get  
(V–Ri)/V= e
-Rt/L
 
From which we can see that   
i(t) = (V/R)–(V/R)e
-Rt/L
 for  t >0 
 
Thus, an expression for the response valid for all time t would be 
 
i(t) = V/R [1- e
-Rt/L
 ] 
 
This is normally written as: 
                                  
 
i(t) = V/R [1- e
-t./t
 ]  
 
where ‘t’ is called  the time constant of the circuit and it’s unit is seconds.  
 
The voltage across the resistance and the Inductor for t >0can be written as : 
  
v
R
(t)    =i(t).R   = V [1- e
-t./t
 ] 
 
v
L
(t)   =      V -v
R
(t)   =  V -V [1- e
-t./t
 ]  = V (e
-t./t
) 
 
A plot of the current i(t)  and the voltages v
R
(t) &  v
L
(t)  is shown in the figure below.  
 
 
 
 
Fig 1.2: Transient current and voltages in the Series RL circuit.  
 
At   t =   ‘t’ the voltage across the inductor will be 
 
                                                          v
L
(t)     =    V (e
-t /t
)  =  V/e    =   0.36788 V 
 
and the voltage across the Resistor  will be       v
R
(t)    =     V [1- e
-t./t
 ]  =  0.63212 V  
 
The plots of current i(t)  and the voltage across the Resistor v
R
(t) are called exponential growth 
curves and the voltage across the inductor v
L
(t)is  called exponential decay curve. 
 
RCCIRCUIT with external DC excitation: 
 
A series RC circuit with external DC excitation V volts connected through a switch is shown in 
the figure below. If the capacitor is not charged initially i.e. it’s voltage is zero ,then after the 
switch S is closed at time t=0, the capacitor voltage builds up gradually and reaches it’s steady 
state value of V volts after a finite time. The charging current will be maximum initially (since 
initially capacitor voltage is zero and voltage across a capacitor cannot change instantaneously) 
and then it will gradually comedown as the capacitor voltage starts building up. The current and 
the voltage during such charging periods are called Transient Current and Transient Voltage.  
                                  
 
 
 
 
 
Fig 1.3: RC Circuit with external DC excitation 
 
Applying KVL around the loop in the above circuit we can write  
 
V  =  v
R
(t) + v
C
(t) 
 
Using the standard relationships of voltage and current for an Ideal Capacitor we get  
 
v
C
(t)  = (1/C ) ?? ?? ?? ??   or i(t)  =  C.[dv
C
(t)/dt]   
 
and using this relation,  v
R
(t)  can  be written asv
R
(t)  =   Ri(t)   = R. C.[dv
C
(t)/dt]      
 
Using the above two expressions for v
R
(t) and  v
C
(t)the above expression for V can be rewritten 
as : 
 
V = R. C.[dv
C
(t)/dt] + v
C
(t) 
 
Or finally dv
C
(t)/dt + (1/RC). v
C
(t)  =  V/RC 
 
The inverse coefficient of v
C
(t)  is known as the time constant of the circuit t and is given by t = 
RC and it’s units are seconds.  
 
The above equation is a first order differential equation and can be solved by using the same 
method of separation of variables as we adopted for the LC circuit.  
 
Multiplying the above equation dv
C
(t)/dt + (1/RC). v
C
(t)  =  V/RC 
 
both sides by ‘dt’   and rearranging the terms so as to separate the variables v
C
(t) and t we get: 
 
dv
C
(t)+ (1/RC). v
C
(t) . dt =  (V/RC).dt