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MODULE-I
D.C NETWORKS
1.1 Kirchoff’s Laws:-
1.1.1. Kirchoff’s current law or point law (KCL)
Statement:- In any electrical network, the algebraic sum of the currents meeting at a point is zero. Σ I = 0 ……………………at a junction or node
Assumption:- Incoming current = positive
Outgoing current = negative 1.1.2. Kirchoff’s voltage law or mesh law (KVL)
Statement:- The algebraic sum of the products of currents and resistances in each of the conductors in any closed path (or mesh) in a network plus the algebraic sum of the emfs in that path is zero. Σ IR +Σemf = 0 …………………………..round the mesh Assumption:- i) Rise in voltage (If we go from negative terminal of the battery to positive terminal) = positive ii) Fall in voltage (If we go from positive terminal of the battery to negative terminal) = negative iii) If we go through the resistor in the same direction as current then there is a fall in potential. Hence this voltage is taken as negative. iv)If we go through the resistor against the direction of current then there is a rise in potential. Hence this voltage drop is taken as positive.
Example :- Write the loop equation for the given circuit below ( Supplementary exam 2004 )
i
r 1 E 1
r 2
E 2
E^ r^3 3 Solution: Apply KVL to the loop,
1 1 2 2 3 3
1 2 3 1 2 3
1 2 3 1 2 3
ir E ir E ir E 0
E E E ir ir ir
E E E i r r r
1.2. MAXWELL’S LOOP CURRENT METHOD (MESH ANALYSIS)
Statement:- This method determines branch currents and voltages across the elements of a network. The following process is followed in this method:-
- Here, instead of taking branch currents (as in Kirchoff’s law) loop currents are taken which are assumed to flow in the clockwise direction.
- Branch currents can be found in terms of loop currents
- Sign conventions for the IR drops and battery emfs are the same as for Kirchoff’s law.
- This method is easier if all the sources are given as voltage sources. If there is a current source present in a network then convert it into equivalent voltage source.
1 (^1 2 )
1 2
-I × 1 0 - I -I × 2 0 -1 0 = 0
⇒ 3 I -2 I = -1 ( 1 )
For mesh BEFCB,
2 (^2 3 )^ (^2 1 ) 1 2 3
4 0 -I × 2 0 + 1 0 - I -I × 1 0 - I -I × 2 0 = 0
⇒ 2 I -5 I + I = -5 (2 )
For mesh EGHFE,
3 (^3 2 ) 2 3
-1 0 I 5 0 1 0 1 0 0 I -2 I = -4 ( 3 )
Equation (2) x 2 + Equation (3)
4 I 1 - 9 I 2 = - 1 4 ( 4 )
Solving eqn^ (1) & eqn^ (4)
I 1 = 1 A, I 2 = 2 A, I 3 = 3 A
1.3. NODAL ANALYSIS
Statement :- This method determines branch currents in the circuit and also voltages at individual nodes.
The following steps are adopted in this method:-
Identify all the nodes in the network.
One of these nodes is taken as reference node in at zero potential
The node voltages are measured w.r.t the reference node
KCL to find current expression for each node
This method is easier if all the current sources are present. If any voltage source is present, convert it to current source
The number of simultaneous equations to be solved becomes (n-1) where ‘n’ is the number of independent nodes.
Explanation :-
At node ‘a’ I 1 +I 2 = I 3
By ohms law,
Therefore,
or,
or,
Hence,
Node voltage multiplied by sum of all the conductance connected to this node. This term is positive
The node voltage at the other end of each branch (connected to this node multiplied by conductance of this branch). This term is negative.
1 2 1 2 3 1 2 3
a (^) , a (^) , a E V E V V I I I R R R
− − = = =
1 2 1 2 3 1 2
a^0
E E
V
R R R R R
+^ +^ −^ −^ =
1 2 1 2 3 1 2
a^0
E E
V
R R R R R
+^ +^ −^ −^ =
1 2 1 2 3 1 2
1 1 1 a^0
E E V R R R R R
+^ +^ −^ −^ =
1 2 2 2
1 2 1 2
V V V V
V V
V V
Solving Eq (1) and (2) we get V 1 = 3.924 Volt and V 2 = 3.584 volt
Current through 2 Ω resistance =
12-V 1 = 12-3.924=4.038 A
Current through 1 Ω resistance =
0-V 1 =-3.924 A
Current through 3 Ω resistance =
V -V 1 2 =0.1133 A
Current through 5 Ω resistance =
0-V 2 =-0.7168 A
Current through 4 Ω resistance =
6-V 2 =0.604 A
As currents through 1Ω and 5Ω are negative, so actually their directions are opposite to the assumptions.
1.4. STAR-DELTA CONVERSION
Need :- Complicated networks can be simplified by successively replacing delta mesh to star equivalent system and vice-versa.
In delta network, three resistors are connected in delta fashion (∆) and in star network three resistors are connected in wye (Y) fashion.
Fig. 1.4.1.
a) Delta connection b) Star connection
1.4.1. Delta to Star Conversion :- From Fig. 1.4.1 (a), ∆ : Between A & B, there are two parallel path. Resistance between terminal A & B =
From Fig. 1.4.1 (b), STAR: Between A & B two series resistances are there RA + RB. So, terminal resistances have to be the same.
Eq {(1)-(2)}+(3) & Solving,-
AB^ (^ BC CA ) AB BC CA
R R R
R R R
A B^ (^ B C C A )^ .................(1) A B A B B C C A
R R R R R R R R
( ) B C B C^ C A^ A B .................( 2 ) A B B C C A
R R R
R R
R R R
C A^ (^ A B B C )^ .................( 3 ) C A A B B C C A
R R R R R R R R
A AB^ CA^ .................(4) AB BC CA
R R R R R R
×
B AB^ BC^ .................(5) AB BC CA
R R R R R R
×
Example(delta to star ):- Convert the following Delta Resistive Network into an Star Network.
NETWORK THEOREMS
• SUPERPOSITION THEOREM
• THEVENIN’S THEOREM
• NORTON’S THEOREM
• MAXIMUM POWER TRANSFER THEOREM
1. Superposition theorem Statement:- In a network of linear resistances containing more than one generator (or source of of all the currents which would flow at that point if each generator were considered separately and all the other generators replaced for the time being by resistances equal to their internal resistance
Convert the following Delta Resistive Network into an
NETWORK THEOREMS
SUPERPOSITION THEOREM
THEVENIN’S THEOREM
NORTON’S THEOREM
MAXIMUM POWER TRANSFER THEOREM
Superposition theorem In a network of linear resistances containing more than one generator (or source of emf), the current which flows at any point is the sum of all the currents which would flow at that point if each generator were considered separately and all the other generators replaced for the time being by resistances equal to their internal resistance.
Convert the following Delta Resistive Network into an equivalent
In a network of linear resistances containing more than one emf), the current which flows at any point is the sum of all the currents which would flow at that point if each generator were considered separately and all the other generators replaced for the time being
Example :- By means of superposition theorem, calculate the currents in the network shown.
Step 1. Considering 10 V battery
e q
1 b
2 b
3 b 1 b 2 b
R 1 2. 8
I 3. 5 7 A
I 3. 5 7 3. 2 1 A
I I I 0. 3 6 A
×
= × =
Proof :-
From Eq. (1) & (2)
- STATEMENT : The two circuits are said to be electrically equivalent if they supply equal load currents with the same resistance connected across their terminals.
- voltage source having a voltage V and source resistance Rx can be replaced by I(= V/Rx) and a source resistance Rx in parallel with current source.
- Current source I and source resistance Rx can be replaced by a voltage source V (=IRx) and a source resistance Rx in series with V.
- THEVENIN’S THEOREM :- Statement :- Any pair of terminals AB of a linear active network may be replaced by an equivalent voltage source in series with an equivalent
L X L
V I (1) R R
=
X L X L
R
I I (2)
R R
V = IR (^) X (3)
resistance Rth. The value of Vth (called the Thevenin’s voltage) is equal to potential difference between the terminals AB when they are open circuited, and Rth is the equivalent resistance looking into the network at AB with the independent active sources set to zero i.e with all the independent voltage sources short-circuited and all the independent current sources open- circuited. Example:- Find VX by first finding VTH and RTH to the left of A-B
Solution:- step1. First remove everything to the right of A-B.
Notice that there is no current flowing in the 4 Ω resistor (A-B) is open. Thus there can be no voltage across the resistor.
Step 2. To find Rth
We now deactivate the sources to the left of A-B and find the resistance seen looking in these terminals.
30 V +_ 6 ΩΩΩΩ 2 ΩΩΩΩ VX
+
_
A
• ••• B
12 ΩΩΩΩ^4 ΩΩΩΩ
30 V +_ 6 ΩΩΩΩ
A
• ••• B
A B 6 1 2
V ==== ==== V
Example : Find the Norton equivalent circuit to the left of terminals A-B for the network shown below. Connect the Norton equivalent circuit to the load and find the current in the 50 Ω resistor.
Solution:
Fig. Circuit to find INORTON
It can also be shown that by deactivating the sources,We find the resistance looking into terminals A-B is RN and RTH will always be the same value for a given circuit. The Norton equivalent circuit tied to the load is shown below.
+ _
10 A
50 V
A
B
+ _
10 A
50 V
ISS
I SS ==== 10.7 A
RN ==== 55 ΩΩΩΩ
10.7 A (^55) ΩΩΩΩ 50 ΩΩΩΩ
5. MAXIMUM POWER TRANSFER THEOREM :
- Statement : For any power source, the maximum power transferred from the power source to the load is when the resistance of the load RL is equal to the equivalent or input resistance of the power source (Rin = RTh or RN).
- The process used to make RL = Rin is called impedance matching.
Explanation :
( )
TH TH L 2 (^2) TH L L L (^2) TH L L L L L TH 2 2 (^2) TH L TH L L 2 L^ L
I=^ V
R R
V R
P I R
(R R )
dP For P to be maximum, 0 dR Or, R R V R V So, Maximum power drawn by R =I R = 2R 4R
( )
2 TH TH L
Power supplied by the source= V R +R
s s t r
s s
d i V = R i + L d t i = i i V i R
=
t r
t r
i R i + L d i 0 d t d i R (^) i = 0 d t L d i R (^) i d t L d i R i L l n R t ; i L
R (^) t L
d t
i K e
−
= =
= −
= −
= − =
s s
i V R
i K e
t R (^) t i V K e^ L R
−
=
=
= +
At t=0, i=0 So,
0 V + K R K = - V R
1
R (^) t i V e^ L R
−
=
= (^) −
-t R
t
L R λ= is called time constant and is called damping coefficient of the circuit R L
V iR=V 1-e
Emf of self inductance is -L di i R dt
= ^ λ
=
If t=λ, then it = I 0 e-1^ = I 0 e-1=I 0 /e =I 0 /2.718 = 0.37 I 0
Hence, time period of a circuit is the time during which the transient current decrease to 0.37 of its initial value.
Transient in R-C Series Circuit :
Consider an ac circuit containing a resistor of resistance R ohms and a capacitor of capacitance C farad across an a.c source of rms voltage V volts as shown in Fig. below:-
