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1.1. Define following terms
(a) Current
Free electron Copper wire
Conventional current V^ Electron current
Figure 1 .1Concept of electric current Flow of electron in closed circuit is called current. Amount of charge passing through the conductor in unit time also called current. Unit of current is charge/second or Ampere (A).
, Current Charge Time
I Q
t Where I Q t
(b) Potential or Voltage The capacity of a charged body to do work is called potential. Unit of potential is joule/coulomb or Volt (V).
= Potential or Voltage = Workdone
V W
Q
Where, V W
(c) Potential difference
+ 12 V + 7 V
A B
Conventional current Figure 1. 1 Potential differences The difference of electrical potential between two charged bodies is called potential difference. Unit of Potential Difference is Volt (V). If potential of body A is +12V and potential of body B is +7V then potential difference is +5V. i.e. (+12V) - (+7V) = +5V
(d) Electro Motive Force (emf) The force is required to move electron from negative terminal to positive terminal of electrical source in electrical circuit is called emf. Unit of emf is volt (V).
Emf is denoted as ε.
(e) Energy Ability to do work is called energy. Unit of energy is Joule or Watt-sec or Kilowatt-hour (KWh). 1KWh is equal to 1 Unit.
W P t VIt I Rt V t R Where W P t
2 2
, =Energy =Power =Time
(f) Power Energy per unit in time is called power. Unit of Power is Joule/Second or Watt (W).
P W
t
(g) Resistance Property of a material that opposes the flow of electron is called resistance. Unit of resistance is Ohm (Ω). R V I Where R
, Resistance (h) Conductance Property of a material that allows flow of electron. It is reciprocal of resistance. Unit of conductance is (Ω-^1 ) or mho or Siemens(S).
G R Where G
, Conductance
(i) Resistivity or Specific Resistance Amount of resistance offered by 1m length of wire of 1 m^2 cross-sectional area. Resistivity is denoted as a ρ. Unit of Resistivity is Ohm-meter (Ωm).
R l
a
(b) Dependent sources
Voltage controlled voltage source (VCVS) Voltage controlled current source (VCCS)
+
-
a
b
c
d
+
-
+
-
Vab μ Vab Vcd
Figure 1.5VCVS Voltage controlled voltage source is four terminal network components that established a voltage Vcdbetween two- point c and d. Vcd μVab The voltage Vcddepends upon the control voltage Vab and μis constant so it is dimensionless. μis known as a voltage gain.
a
b
+
-
gm Vab
Icd c
d
+
-
Vab Vcd
Figure 1.6VCCS Voltage controlled current source is four terminal network components that established a current icdin the branch of circuit. icd g Vm ab icddepends only on the control voltageVab and constant gm,is called trans conductance or mutual conductance. Unit of transconductance is Ampere/Volt or Siemens(S).
Current controlled voltage source (CCVS) Current controlled current source (CCCS)
a
b
+
-
r iab
c
d
+
-
Vcd +-
iab
Figure 1.7CCVS Current controlled voltage source is four terminal network components that established a voltage Vcdbetween two- point c and d. Vcd riab Vcd depends on only on the control current iaband constant r and r is called trans resistance or mutual resistance. Unit of transresistance is Volt/Ampere or Ohm (Ω).
a
b
+
-
β iab
c
d
+
-
iab icd
Figure 1.8CCCS Current controlled current source is four terminal network components that established a current Icd in the branch of circuit. icd βiab icddepends on only on the control current iab and constant β and β is called current gain. Current gain is constant. Current gain is dimensionless.
1.3. Explain source conversion
A voltage source with a series resistor can be converted into an equivalent current source with a parallel resistor. Conversely, a current source with a parallel resistor can be converted into a voltage source with a series resistor. Open circuit voltages in both the circuits are equal and short circuit currents in both the circuit are equal.Source transformation can be applied to dependent source as well.
I V
R
I=V/R R
Figure 1. 9 Source conversion Network simplification techniques
V 1 +
V 2
V1 + V 2
(V1 > V 2 )
(a)
(b)
V 1 +
V 2
V1 - V 2
Vs
(g)
is
(h)
is
R
Vs + R^
Figure 1.10Rules under which source may be combined and separated
1.4. Explain ideal electrical circuit element.
There are major three electrical circuit elements which are discussed below. (a) Resistor Resistor is element which opposes the flow of current.
a
l
Figure 1.11Resistor Figure 1.12Conductor Resistance is property of material which opposes the flow current. It is measured in Ohms (Ω). Value of resistance of conductor is Proportional to its length. Inversely proportional to the area of cross section. Depends on nature of material. Depends on temperature of conductor. R l a R ρl a
(b) Inductor An inductor is element which store energy in form of magnetic field. The property of the coil of inducing emf due to the changing flux linked with it is known as inductance of the coil. Inductance is denoted by L and it is measured in Henry (H).
1. 13 Inductor Value of inductance of coil is Directly proportional to the square of number of turns. Directly proportional to the area of cross section. Inversely proportional to the length. Depends on absolute permeability of magnetic material. 0
0 (^02) 0
, =Inductance of coil = Number of turns of coil Φ = Flux link in coil = Magneto motive force(MM
r
r r r
F NI NI NIμ μ A S S l l μ μ A N NIμ μ A Now L N^ l N^ μ μ A I I l Where L N
F
0
F)
= Current in the coil = Mean length of coil = Permiability of free space = Relative permiability of magnetic material = Cross sectio
r
I
l μ μ A nal area of magnetic material (c) Capacitor Capacitor is an element which stored energy in form of charge. Capacitance is the capacity of capacitor to store electric charge. It is denoted by C and measured in Farad (F).
Figure 1.14Capacitor Value of capacitance is Directly proportional to the area of plate. Inversely proportional to distance between two plates. Depends on absolute permittivity of medium between the plates.
Branches are meeting at a junction ‘J’ Incoming current are denoted with (+ve) sign Outgoing currents are denoted with (-ve) sign
+
-
+
- R 1
R 2
R 3
R 4
R 5
E 1
E 2
I 1 I 2
A (^) J I 3 B
E D C Figure 1.16Kirchhoff’s law diagram Then,
1 2 3 1 2 3 1 2 3
I
I I I
I I I
I I
Incoming current
I
Outgoing current
(b) Kirchhoff’s voltage law (KVL)
Statement : “Algebraic sum of all voltage drops and all emf sources in any closed path is zero” Let, Suppose Loop current in clockwise or anticlockwise direction Circuit current and loop current are in same direction than voltage drop is denoted by (-ve) sign. Circuit current and loop current are in opposite direction than voltage drop is denoted by (+ve) sign. Loop current move through (+ve) to (-ve) terminal of source than direction of emf is (-ve). If Loop current move through (-ve) to (+ve) terminal of source than direction of emf is (+ve).
R V= -IR V= +IR
+ +
E= -E
R
E= +E
I I
Figure 1.17Sign convention for Kirchhoff’s voltage law
1 2 2 3 2 1 1 1
3 4 3 5 2 2 3
IR E
KVL to loop AJDEA I R I R E I R E KVL to loop JBCDJ I R I R E I R
1.7. Explain series and parallel combination of resistor
Series combination of resistor Parallel combination of resistor R 1 R 2
V
V 1 V 2 +
-
I
I 1 I 2
Figure 1.18Series combination of resistors
1 2
1 2 1 2 1 2 1 2 1 2
1 2 3
eq
eq n
As per KVL
For n resistor are connected in ser
Here I I I
V V V V IR IR V I R R V (^) R R I
i
R R R
R R R R
s R
e
R 2
R 1
V
I 1
I 2
I
+
-
V 1
V 2
Figure 1.19Parallel combinations of resistors
1 2
1 2
1 2
1 2
1 2
1 2
eq
Here V V V
I I I I V^ V R R
I V
As p
R R
I
V R R
R R R
er KCL
^
^
^
1 2 3
eq n
For n resistor are connected in Paral l
R
l
R R R
e
R
Value of equivalent resistance of series circuit is bigger than the biggest value of individual resistance of circuit.
Value of equivalent resistance of parallel circuit is smaller than the smallest value of individual resistance of circuit.
1.9. Derive the equation of delta to star and star to delta transformation
R 1
R 2 R 3
R 12
R 23
1
2
3
R 31
Figure 1.22Delta connected network
12 23 31 12 23 31 12 23 31
23 12 31 23
( ) ( )
( ) (
1 & 2
2 & 3
Resistance between terminal
Resistance between termi
R R R R R R R R R
R R
a R R
n l
12 31 12 23 31
31 12 23 31 12 23 12 23 31
)
( ) ( )
Resistance bet 3 & 1
R R R R R
R R R R R R R R R
ween terminal
R 1
R 2 R 3
R 12
R 23
1
2
3
R 31
Figure 1.23Star connected^ network
1 2
2 3
3 1
1 & 2
2 & 3
3 & 1
Resistance between terminal
Resistance between terminal
Resi
R R
R R stance between terminal R R
1 2 12 23 31 12 23 31
( )
1 & 2 1 & 2 R^ R^ R^ R
Resistance between terminals in delta equal to resistance between terminals i R R R R
n star
2 3 23 12 31 12 23 31
( )
( )
,
i
R R
Similarl
R
y R R R R R (^ ii )
3 1 31 12 23 12 23 31
R R R^^ (^ R^ R )^ ( iii ) R R R
(a) Delta to star conversion 1 2 2 3 3 1 12 23 31 23 12 31 31 12 23 12 23 31 12 23 31 12 23 31
Simplify i ii iii on both the side of equatio R R R R R R R^ R^ R^ R^ R^ R^ R^ R^ R R R R R R R R
n
R R
s ^ ^
12 23 12 31 23 12 23 31 31 12 31 23 12 23 31 12 23 31 12 23 31 12 23 12 31 23 12 23 31
( ) (^) + ( ) (- )
(
R R R R R R R R R R R R R R R R R R R R R R R R R R R R R
^ ^ ^ ^ ^31 12 31 12 23 31 2 12 23 12 23 31 2 12 23 12 23 31 1 12 31 12 23 31
S ,
R R R R R R R R R^ R R R R R R^ R R R R imilarly R R^ R R R R
(^)
(^)
3 23 31 12 23 31
R R^ R (^) R R R
(b) Star to delta conversion 1 2 2 3 2 3 3 1 3 1 1 2 12 23 31 23 12 31 23 12 31 31 1 12 23 31 12 23 31 12 23 31
( )( ) ( )( ) ( )( ) ( ) ( ) ( ) (
Simplify i ii ii iii iii i on bot R R R R R R R R R R R R R R R R R R R R R R R R R R R R R
h th
R R R
e side of equation
^ ^ ^ ^ ^ ^ ^
2 23 31 12 23 12 23 31 2 2 12 223 31 12 23 31 12 23 31 1 2 1 3 2 2 3 2 3 2 1 3 3 1 3 1 3 2 1 1 2 12 23 12 31 23 12 23 31 12 23 31 12 23 3
R ) R ( R R ) R ( R R ) R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R
^ 23 12 23 31 31 12 31 23 31 12 31 23 12 23 12 31 1 12 23 31 12 23 31 12 23 31 12 23 31 1 2 2 3 3 1 12 22 32 232 122 12 232 31 122 23 31 1
3 3 3
R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R
^ ^ ^ ^ ^
2 23 312 122 23 31 12 232 31 12 23 312 232 312 122 23 31 122 312 12 232 31 12 23 312 12 23 31 2 12 23 31 2 12 23 312 232 122 12 232 31 122 23 31 12 23 3
R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R
(^) (^)
12 122 23 31 12 232 31 12 23 312 232 312 122 23 31 122 312 12 232 31 12 23 312 12 23 312 ( 12 232 31 122 23 31 12 23 312 122 23 31 12 232 31 12 23 312 122 23 31 12 232 31 12 2
R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R
^ ^ ^ ^ ^ ^
3 312 232 122 232 312 122 312 12 23 312 12 23 31 23 12 31 12 23 31 12 23 31 232 122 232 312 122 312 12 23 31 2 12 23 312 12 23 31 12 23 31 12 23 31
) ( )
( ) ( )
(3 3 3 )
R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R
232 122 232 312 122 312 2 12 23 31 2 12 23 31 2 12 23 312 12 23 31 12 23 31 232 122 232 312 122 312 12 23 31 2 12 23 31 2 12 23 31 2 12 23 312 3 12
3 ( )
3
R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R
^ ^
(^)
^ 22 32 12
1 2 2 3 3 1 12 22 32 3 12 22 32 12 1 2 2 3 3 1 3 12
3 3 3 3 3 3 3 3
R R R Now equation become R R R R R R R R R R R R R R R R R R R R R R
12 1 2 1 2 3
23 2 3 2 3 1 31 3 1 3 1 2
R R R R R R Similarly R R R R R R R R R R R R
1 1 2 3 3 1
(^1 1 1 1) ( ) A B V V V i R R R R R
(^) (^) (^) Node B R 1 R 3
R 2 R 4
R 5
V 1 V 2
I 3 I 4
VA VB I 5
+
-
+
-
VC
Figure 1.26Node analysis network for node B
3 4 5 3 4 5 2 3 4 5 2 3 3 4 5 5
0 0
1 1 1 1
,
B A B C B
A B
I I I I I I V V V V
Apply the KC
V V R R R V V V R R R
L at nod
R
e B
R
(^) (^) (^)
1 1 2 3 3 1 2 3 3 4 5 5
( ) From, equation (i) & (ii) 1 1 1 1
1 1 1 1
A B
ii
V R R R R V R V V R R R R R
(^) (^) (^) ^ (^) (^) ^
One can easily find branch current of this network by solving equation (i) and (ii),if V 1 , V 2 and all resistance value are given.
1.11. Explain Mesh analysis
R 1 R 3
R 2 R 4
R 5
V 1 V^2
+
-
+ I 1 I^2 I^3 -
Figure 1.27Mesh analysis network Mesh: It is defined as a loop which does not contain any other loops within it.
The current in different meshes are assigned continues path that they do not split at a junction into a branch currents. Basically, this analysis consists of writing mesh equation by Kirchhoff’s voltage law in terms of unknown mesh current.
Steps to be followed in mesh analysis: Identify the mesh, assign a direction to it and assign an unknown current in it. Assigned polarity for voltage across the branches. Apply the KVL around the mesh and use ohm’s law to express the branch voltage in term of unknown mesh current and resistance. Solve the equations for unknown mesh current. Loop 1
R 1 R 3
R 2 R 4
R 5
V 1 V^2
+
-
+
-
I 1
I 1 I^2
Figure 1. 28 Mesh analysis network for loop- 1
1 1 1 2 2 1 1 1 1 2 2 2 1 1 2 1 2 2 1
1, I R I I R V I R I R I R V R R
Now apply the KVL in
I R I
loop
V
(i)
Loop 2
R 1 R 3
R 2 R 4
R 5
V 2
+
-
+ V (^1) -
I 2
I (^1) I 2 I 2 I 3
Figure 1.29Mesh analysis network for loop- 2
2 3 2 3 4 2 1 2 2 3 2 4 3 4 2 2 1 2 1 2 2 3 4 2 3 4 2 1 3 4 2 2 4 3
I R I I R I I R
I R I R I R I R I R
I R I R R R I R
Now Apply th
R I
e
R R R I R I
KVL loop
(ii)
According to the application of the superposition theorem. It may be noted that each independent source is considered at a time while all other sources are turned off or killed. To kill a voltage source means the voltage source is replaced by its internal resistance whereas to kill a current source means to replace the current source by its internal resistance. To consider the effects of each source independently requires that sources be removed and replaced without affecting the final result. To remove a voltage source when applying this theorem, the difference in potential between the terminals of the voltage source must be set to zero (short circuit) removing a current source requires that its terminals be opened (open circuit). Any internal resistance or conductance associated with the displaced sources is not eliminated but must still be considered. The total current through any portion of the network is equal to the algebraic sum of the currents produced independently by each source. That is, for a two-source network, if the current produced by one source is in one direction, while that produced by the other is in the opposite direction through the same resistor, the resulting current is the difference of the two and has the direction of the larger. If the individual currents are in the same direction, the resulting current is the sum of two in the direction of either current. This rule holds true for the voltage across a portion of a network as determined by polarities, and it can be extended to networks with any number of sources. The superposition principle is not applicable to power effects since the power loss in a resistor varies as the square (nonlinear) of the current or voltage. Steps to be followed to apply the superposition theorem: Select any one energy source. Replace all the other energy sources by their internal series resistances for voltage sources. Their internal shunt resistances for current sources. With only one energy source calculate the voltage drops or branch currents paying attention to the voltage polarities and current directions. Repeat steps 1, 2 and 3 for each source individually. Add algebraically the voltage drops or branch currents obtained due to the individual source to obtain the combined effect of all the sources. Example network:
R 1 R 2
V 1
+
-
+ R (^3) -
A
B
r V 2
Figure 1.31Superposition theorem network
Step- 1 R 1 R 2
V (^1) +
- R 3 r
I (^1) I 2
A
B Figure 1.32Superposition theorem network for step- 1 Now apply Mesh analysis in loop
Now apply Mesh analysis in l
I R I R I R I r V
I R I R I R Now current flow from R branch is a ebric
oop
sum of I and I
1 1 1 3 2 3 1 1
2 2 2 3 1 3 3 1 2
1,
2,
Step- 2 R 1 R 2
I 3 R (^3) I 4 +- r V^2
A
B Figure 1.33Superposition theorem network for step- 2
I R I R I R I r
I R V I
Now apply Mesh analysis in loop
Now apply Mesh analysis R I R Now current flow from R branch is a ebric sum of I and I Finally current flow from
in
R
loop
3 1 3 3 4 3 3
4 2 2 4 3 3 3 3 3 4 3
1,
2
,
is a lg ebric sum of step 1 and step -
1.13. Explain Thevenin’s theorem
Thevenin theorem is an analytical method used to change a complex circuit into a simple equivalent circuit consisting of a single resistance in series with a source voltage. Thevenin’s can calculate the currents and voltages at any point in a circuit. Thevenin’s Theorem states that “Any linear circuit containing several voltages and resistances can be replaced by just one single voltage in series with a single resistance
