ELECTRIC CIRCUITS
1.2 CIRCUIT ANALYSIS TECHNIQUES
Analysis is important in electric domain because it is 1. easy to analyse.
2. easy to measure the electrical quantities.
3. easy to process the electrical signals.
4. easy to store.
5. easy to visualise, etc.
Analysis typically involves the calculation of the response of a known circuit or system to a given excitation. Synthesis
1.2 CIRCUIT aNaLYSIS TEChNIQUES 15
+
+ I
− −
+− 5Ω
10V 10V 10V
The voltage at 5 Ω resistor is V
5Ω = 10 V, therefore current is
I5
10 5
W = =2A
Ans. (2) Problem 1.10: Determine the power delivered by the 10 V source in the following circuit.
+−
+
− 5Ω 10V
i
2A 2A
10V
Solution: The voltage at 5 Ω resistor is V5Ω = 10 V
I5
10 5
W = =2A Then applying KCL, we have
−i - 2 + 2 = 0 ⇒i = 0 Thus, the power delivered by the 10 V source
= 10.
(0) = 0 Watts
P2A = 2 × 10 = 20 Watts (delivered) P5Ω = 22× 5 = 20 Watts (absorbed)
Ans. (20) V1
V2
i1 i
i
= i1 + i2
V = V1 + V2
≅ ≅
+− +−
+−
(c)
10V
10A 10A
10V Open circuit
Short circuit
≅ ≅
+−
+ +
+ +
−
−
−
−
−
− (d)
Figure 1.26
|
Equivalent circuits.Problem 1.9: Determine the current through the 5 Ω resistor in the following figure.
+
+− +
− − 5Ω
5V 10V
Solution: Consider the following equivalent circuit
+− +
− 5Ω
5V
10V
Since there is violation of the KVL in the network, the circuit connection is physically not possible.
Applying KVL, we get
5 - 10 = 0 ⇒ −5 = 0 which is not possible.
Now consider the following equivalent circuit
+− +
− 5Ω
10V 10V
Here applying KVL, we get
10 + 10 = 0 ⇒ 20 = 0
It is also violation of KVL, so physically not possible.
The value of current can be determined using the following circuit.
1.2.2 Series and Parallel Networks for Resistors, Capacitors and Inductors
In a series circuit, the current through the elements is the same. In a parallel circuit, the voltage across the ele- ments is the same.
1.2.2.1 Parallel Impedance and Current Division
Two or more circuit elements are said to be in parallel if the same voltage appears across each of the elements (Fig. 1.27).
I1
I2 I
I Z1
Z2
Figure 1.27
|
Parallel circuit elements.Z
j C
= 1
w ⇒ V VC
C C
1
2
1 2
= + V
VC
C C
2
1
1 2
= + I = I
1 + I
2
Z R I
IR
R R
= =
+
⇒ 1 2
1 2
I
IR
R R
2
1
1 2
= +
Z
j C I
I j C j C j C
I C
C C
= =
+
= + 1
1
1 1
1
2
1 2
1
1 2
w
w
w w
⇒
⋅ ⋅
I
I C
C C
2
2
1 2
= +
⋅
Z =jwL⇒ I
I j L j L j L
1
2
1 2
=
+ w
w w =
+ I L
L L
⋅ 2
1 2
I
I L
L L
2
1
1 2
= +
⋅
1.2.2.2 Series Impedance and Voltage Division
Two or more circuit elements are said to be in series if they exclusively share a single node and consequently carry the same current (Fig. 1.28).
V I
Z1 Z2
+ − + −
Figure 1.28
|
Series circuit elements.I
V
Z Z
=
1+ 2
Therefore, V I Z
V Z
Z Z
1 1
1
1 2
= =
+
⋅ ⋅
V
VZ
Z Z
2
2
1 2
= +
R Z V
VR
R R
=
+
⇒ 1= 1
1 2
V
VR
R R
2
2
1 2
= + Z =jwL⇒ V
VL
L L
1
1
1 2
= + V
VL
L L
2
2
1 2
= +
Problem 1.11: Determine equivalent capacitance in the circuit given below.
CC
CC CC CC CS
B
(0V)
(0V) Y
Solution: The equivalent circuit is B
B
Y
Y CC
CC CC
(CC + CS) (CS + CC)
CC CS
CS
C C
C C C C
BY = +
+
= +
C
C S C S
2
3 2
Problem 1.12: Twelve 1 Ω resistors are used as edges to form a cube. The R
eq seen between the two diago- nally opposite corners of cube is
R
I
V R
R R R
R
R R
R
R
R R
− + (a)
6 5
W (b) 5 6
W (c) 6 Ω (d) 5 12
W
1.2 CIRCUIT aNaLYSIS TEChNIQUES 17
1.2.3 Source Transformation
Source transformation states that an independent (or dependent) voltage source V
S in series with a resistance R is equivalent to an independent current source I
S = V R
S/ S
in parallel with a resistance R
S. or
An independent (dependent) current source I
S in paral- lel with a resistance R
S is equivalent to an independent voltage source V
S = I
S R in series with a resistance R
S. 1. It is a simplification technique and is applicable
only for practical sources.
2. It is impossible to convert an ideal voltage into an equivalent ideal current source and vice versa.
The two circuits shown in Fig. 1.29 are equal only with respect to the performance point of view, but from the ele- ments and connections point of view, they are not equal.
The source transformation is applicable even for the depen- dent sources, provided the controlled variables are outside the branches where the source transformation is applied.
RS
RS VS
IS ISR
+−
+− RS
VS
Figure 1.29
|
Source transformation.Problem 1.13: If in the figure shown in Problem 1.12, if all resistors are replaced by L Henry inductors, determine the equivalent inductor between to two diagonally opposite corners of cube.
Solution: From the corresponding equivalent circuit, we have
V I
Z I
Z I
− − − Z
3 6 3
L L L =0
V I Z
V I
Z
=
=
⋅ ⋅
⋅ 5 6 5 6
L
L
Note: If capacitors are used in the circuit, we have
Z Z
eq = 6 5 C
Note: For DC input, the inductor acts as short circuit. The capacitor will act as an open circuit in steady state.
Solution: The equivalent circuits is
R
I
V R
R I
R R R
R I/3
I/3 I/3
I/6 I/3 I/3 I/3
I/6 I/6
I/6 I/6 I/6
R R
R
R R
− + Applying KVL, we get
V I
R I
R I
− − − R
3 6 3
0
= V I
R R R
I R
= + + =
3 6 3
5 6
⇒V I
R
= =
eq
5 6
Ω
Ans. (b)
Problem 1.14: Simplfy the network between the ter- minals a and b in the following circuit.
3Ω 6V
1A
3V 4V a
b 2Ω
2Ω
1Ω +−
+− +
− Note: Whenever the inductor and capacitors are pres-
ent in the network, an AC source of frequency (w) is always used.
For an ideal diode V
f = 0 and R
f = 0. So, +
+ −
−
1Ω 2Ω
2V 5V
id +
+
−
− + −
+−
5 - 2i
d - 1i
d - 2 = 0 ⇒ i
d = 1 A
Ans. (1) Solution: The simplified circuits are:
⇒
2Ω 1Ω
2Ω
3Ω 1Ω 1A
a
b
2A 3A
2A 3A
Ω 1A
3 4
⇒ ⇒
⇒
15 A a
a a
b b
b 1Ω
3V
7 Ω
7 4 15 4
+− +−
+− Ω 3 4
Ω 7 4 V
3 4
Problem 1.15: Determine the current through the ideal diode (D) in the following circuit.
4Ω
4Ω 2A
D
10V
id 1Ω +
−
Solution: Since the diode is a non-linear element, the network is non-linear and hence only source trans- formation is applicable (superposition theorem is not applicable).
2Ω 1Ω 2A
D
D id
1Ω 2Ω
2V 5V
id 5
2 A
+− +
−
2Ω 1Ω 2A
D
D id
1Ω 2Ω
2V 5V
id 5
2 A
+− +
−
1.2.4 Nodal and Mesh Analysis
The Nodal and Mesh analyses are two network solving tech- niques that are applicable only for lumped electric circuits.
Kirchhoff’s current law (KCL) + Ohm’s law = Nodal analysis (98% used).
Kirchhoff’s voltage law (KVL) + Ohm’s law = Mesh analysis (2% used).
1.2.4.1 Mesh Analysis
Mesh analysis provides another general procedure for analyzing circuits using mesh currents as the circuit vari- ables. A mesh is a loop that does not contain any other loop within it. Mesh analysis is applicable for only planar circuits. A planar circuit is one that can be drawn in a plane with no branches crossing one another; otherwise, it is non-planar. The steps involved in mesh analysis are:
1. Identifying the meshes.
2. Assigning the mesh currents.
3. Using KVL + Ohm’s laws to write the mesh equations.
Applying mesh analysis at mesh 2 in Fig. 1.30 and assuming i
2>> i
1, we get
−V −V t −V −V
R R =
3 ( ) 3 2 0
−i R −V t −i R −L − d dt
i i
2 2 ( ) 2 3 (2 1) =0
V2
V(t) i(t)
V1 C R2
R1 LL
iL iC
i1 i2
iR2
R3 +−
Figure 1.30
|
Circuit for mesh analysis.1.2 CIRCUIT aNaLYSIS TEChNIQUES 19
Solving, we get i
3 = 2A, i
1 = 9A, i
2 = 2.5A Inside the super mesh, applying KCL, we get
i1 = 7 + i
3- 7 - i
3 + i
1 = 0 i1- i
3 = 7 (iii)
So, the power dissipation is P = (i
2- i
3)2× 3 = (0.5)2.
3 ⇒ P = 0.75 W Ans. (0.75) Note: Since the voltage across an ideal current source can be of any value, it is not possible to write the mesh equations for the meshes (1) and (2) indepen- dently. Hence, the supemesh procedure is followed.
Problem 1.16: A segment of a circuit is shown in the following figure. At V
R = 5 V, V
C = 4 sin 2t. Find V
L.
2A
1F iR
iC
iC VR
VL 2H 5Ω
− +
−
− +
+
(a) 16 sin2t V (b) 32 sin2t V (c) 16 cos2t V (d) 32 cos2t V Solution: Applying KCL, we get
−2A - i
R + i
L + i
C = 0
− −2 ×
5 5
1 4 2 0
+i + =
d dt
L ( sin t)
−2 − 1+i
L + 4 cos 2t(2) = 0 iL = 3 - 8 cos 2t Now, V L
di dt
L t
L ( ( sin ) )
= ⋅ =2× − −0 8 2 ×2 = 32 sin 2t VL = 32 sin 2t V
Ans. (b) Supermesh
If a current source (independent or dependent) is common between two meshes, we can create a super- mesh by avoiding the current source and any elements connected in series with it.
Problem 1.17: Find the power dissipation in 3 Ω resistor.
3Ω 1Ω
1Ω 7A i1
i2
i3
2Ω
1Ω 7V
+−
− +
− +
Solution: Applying KCL at mesh 2, we get
−2i
2- 3(i
2- i
3) - 1(i
2- i
1) = 0 (i) The combined mesh equations for meshes (1) and (3) are
(7) (i
1- i
2) - 3(i
3- i
2) - 1.i
3 = 0 (ii)
1.2.4.2 Nodal Analysis
Nodal analysis provides a general procedure for analys- ing circuits using node voltages as the circuit variables.
1. Identify the nodes.
2. Assign the node voltages w.r.t. ground node.
3. By using KCL + Ohm’s laws, write the equations.
Consider the circuit shown in Fig. 1.30 above. Applying nodal analysis at node 2, we get
iC + i
L + i
R = 0 C
d dt
V V
t
V dt
V V t
R R
( )
( )
2 1 2
2
2 3
1
− − 0
−
+ +
+
=
∞
∞
∫
Note: Whenever the dependent sources are present in the network always nodal analysis is used to calculate the responses.
Supernode
If the voltage source (dependent or independent) is connected between two non-reference nodes, the two non-reference nodes form a generalised node called the supernode. Both KCL and KVL are used to determine node voltages.
Problem 1.18: Determine V
1 and V
2 in the following circuit.
3Ω Super
node
4A 5V 9A
V1 V2
+ +
+
−
−
−
+ Ω −
1
2 1Ω
6
Problem 1.20: Voltage V
x and current I
x in the given circuit are
––––.
VA 10A
4V Ix
2Ω 2Ω 1
4Ω
− +
Solution: Applying KVL at node 1 and using nodal analysis, we get
−10 4
4 2
0
+ +
+
=
V V
x x
−40 2 8
4
0
+ + +
=
V V
x x
3V 32 V 10 66
x = ⇒ x = . V
The current through concerned branch is Ix =
+
= 10 66 4
2
7 33 .
. A
Ans. (10.66, 7.33) Solution: Applying KVL, we get
− − −
− 4
1 2
1 3
1 3
1 6
9 0
1 1 2 2 1 2
+ + + + =
V V V V V V
−4 + 2V
1 + 6V
2- 9 = 0 2V1 + 6V
2 = 13 (i) Since the current i through an ideal voltage source can be of any value; it is not possible to write the nodal equations at nodes (1) and (2) independently.
Hence, a supernode procedure is followed.
Inside the supernode, applying KVL, we get V1- 5 - V
2 = 0 V1- V
2 = 5 (ii)
Solving Eqs. (i) and (ii), we get V1 = 5.375 V and V
2 = 0.375 V
Ans. (5.375, 0.375)
Problem 1.19: Determine the current i in the circuit.
4A 20V
2Ω
2Ω
2Ω V
i
+ +
+
− −
−
Solution: Applying nodal analysis
− −
4
20 2
0
+i+ =
V
− −
4 2
20 2
0
+ + =
V V
− − −
4
2 20 2
0 V
=
So, we have V = 14 V ⇒ i V
= =
2 7 A.
Ans. (7) Note: A resistor in series with an ideal current source is always neglected from the nodal analysis. But in the power calculation, it cannot be neglected.
Problem 1.21: Determine voltage V
2 in the following circuit.
10Ω
60V 20Ω 20Ω 20Ω 5V
10Ω V2 V2
+− +
−
Solution: Applying KVL, we have
V V V
2 60 2 2
10 10
5 10
0
− −
+ + =
3V2 = 65 ⇒ V
2 = 21.6 V
Ans. (21.6) Note: A register parallel with ideal voltage source is always negligible, but not in power calculations.
Problem 1.22: If R of 10 Ω connected across terminal A and B as in the following circuit what is the value of current?