Chapter 1 D.

C Circuits

1) Ohm’s law: Statement: The potential difference (p.d) across any two
points on a conductor is directly proportional to current flowing through it,
provided physical conditions of the conductor remains same
V I
V=RI
The constant R is called Resistance of the given conductor.
If p.d and currents are expressed in Volts and ampere respectively then R
is expressed in Ohms
For DC circuits ----- V= R* I
For AC circuits ----- V = Z * I, The constant Z is called impedance (i.e.
total resistance) of the circuits, while V and I represents instantaneous
values of p.d and current. In case of Ac Circuits Z also depends on
frequency of an supply

Where X is net reactance (a.c resistance)

2) Power dissipation in Resistance: It can be defined as rate of doing

work.
P = V* I = I2 * R = V2/ R

Where V is applied voltage and R is resistance of the conductor. Power P

is expressed in Watts (W).

3) Energy: It is defined as capacity to do the work.

E = Power * time
It is expressed in Joules.

4) Resistance: It is property of substance due to which it opposes flow

of current through it. Substances having very less resistance to flow of
current are called Conductors e.g. copper. Substances having very high
resistance to flow of current are called as insulators e.g. mica, glass etc.

The resistance of the conductor depends on following factors

1) Directly proportional to its length (L)
2) Inversely area
3) Depends on material

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4) Depends on Temperature of the conductor.

We can write

Where ρ is receptivity or specific resistance of conductor in

ohm meter

Specific Resistance: It is defined as resistance of unit length of material

having unit cross sectional area.

Effect of temperature on resistance:

a) In case of conductors as temp.increases resistance also increases
b) In case of insulators resistance decreases as temp.increases.
The change in resistance of a material with rise in temperature can be
expressed by means of temperature coefficient of resistance hence
conductors have positive and insulators have negative temperature
coefficient.

Consider a conductor having resistance of R 0 at 0oc and Rt ay toC .Then

increase in resistance (Rt – R0) is proportional to
1) initial temperature
2) rise in temp.

hence

Where is a constant called as temperature

coefficient of resistance at 0o C

Temperature Coefficient: It is defined as increase in resistance per ohm

original resistance per 0o C rise in temperature.

Effect of temperature on temperature coefficient

Consider a conductor is heated from initial temp. t 1oC to a final temp.t2oC.
Let R1 and R2 be the resistance of the material at t1oC and t2oC resp. Let
α1and α2 be corresponding temperature coefficient. Then

------------ (1)

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We can also write and

Substitute in equation 1

Solving further we get

Similarly

Illustrating Numerical
A coil has 25 ohm resistance at 40 oC and 45 ohm resistance at 100 oC.
Find its resistance and resistance temperature coefficient at 0 oC.
Consider initial temperature as 0oC
Writing equations
R 40 = R 0 ( 1+ α 0 ( 40 ) ) 25 = R 0 ( 1+ α 0 ( 40 ) )
i.e. ----- (1)
R 100 = R 0 ( 1+ α 0 (100 ) ) 45 = R 0 ( 1+ α 0 ( 100 ) )
i.e. ----- (2)

Dividing two equations we get

α 0 = 0 . 0285 /0C
Substitute this value in any one equation we get
R 0 =11. 68 Ω

Resistances in series connection:

Consider resistors R1and R2 connected in series.

We note following points:

1) Current flowing through each resistor is same.
2) Potential difference across each resistor can be calculated by ohm’s law
(V=IR)

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3) V =V1+V2
4) Equivalent resistance of the circuit is Rs= R 1+R2 i.e. Total resistance is
sum of individual resistance connected in series.
5) The combined resistance in series will always be greater than any of
the individual resistance.

Voltage division Rule:

V1 = V * R1
(R1+R2)
Similarly, V2 = V * R2
(R1+R2)
Example: Find V1,V2 and V3 by voltage divider.

Resistance connected in parallel:

Consider resistors R1 and R2 connected in parallel.

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We note following points:
1) The voltage across each resistor is same and equal to applied voltage.
2) Current through each resistor can calculated by ohm’s law (I=V/R)
3) I=I1+I2

4) Equivalent resistance of the circuit is

5) The combined resistance in parallel will always be less than any of the
individual resistance.

Current division Rule:

In the diagram

Current through one resistor = Total current * Neighboring resistor i.e.

Sum of Resistances

I1= I and I2= I

Example:In the figure consider current I=10A, R1=2Ω and R2=3Ω

Calculate I1 and I2

I1=10*(2/5) = 4A
I2= 10* (3/5) =6A

In some cases resistances are neither in series nor in parallel. Such

network can be simplified by star-delta transformations.

Open and short circuit

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When two points are connected together, they are said to be short
circuited.
1) Short has practically zero resistance
2) V = IR= 0
3) Current through short is very large, practically infinity.
Two points are said to be open when there is no connection between them
1) Resistance between two points is infinity
2) Current flowing is zero

Star-Delta Conversion
Consider following diagrams

Case 1 Delta to Star conversion (Δ to Ύ)

The two arrangements will be electrically equivalent if resistance between
any two terminals in Δ is equal to resistance between corresponding
terminals in Ύ.
Resistance between 1 and 2 in Δ =

Resistance between 1 and 2 in Ύ = R1+ R2 (ii)

Equating i and ii we get

R1+ R2 = similarly (iii)

R2 + R 3 = (IV)

R3 + R 1 = (v)
Subtracting equation (iv) from (iii) and adding the result in (v) we get

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 (vi) Similarly

(vii)

(viii)

Case 2 Star to Delta Conversion (Ύ to Δ)

(vi) ÷ (vii) gives hence

(vii) ÷ (viii) gives hence

Substituting values of R31 and R12 we get,

Solving we get

And

Illustrating Numerical
Find RAB

Converting outer delta into star we get,

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We have (40\\40), (100\\100), and (60\\60)
Solving we get

Thus we get, (50\\50)

RAB =25 Ω

Independent Sources: The internal values of these sources (either

voltage source or current source) – that is, the generated voltage Vs or
the generated currents I are not affected by the load connected across
the source terminals or across any other element that exists elsewhere in
the circuit or external to the source.
Conversion From voltage to current source:
A practical voltage source with a voltage Vs and an internal source
resistance Rs can be replaced by an equivalent practical current source
with a current Is = Vs/Rs and a source internal resistance Rs in parallel.

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Conversion From current to voltage source

A current source with a magnitude of currents I and a source internal

resistance Rs can be replaced by an equivalent voltage source of
magnitude Vs= Is*Rs and an internal source resistance Rs in series.

Kirchhoff’s laws:
Kirchhoff’s first law (Current law or KCL)
Statement:” The sum of all the current meeting at a junction of a network
is equal to sum of currents going away from the junction”.
I.e. The Algebraic sum of currents meeting at any junction is zero.
In the figure
i1 - i2 - i3 =0 or i 1 = i2 + i3

Kirchhoff’s second law (voltage law or KVL)

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Statement:” In any closed circuit, the Algebraic sum of all voltages in a
closed path is zero”.
Sign convention for battery emf: Traveling from –ve terminal to +ve
terminal of the battery is rise in voltage hence given +sign and via versa.

Sign convention for voltage drops: Current flows from higher potential to
lower potential hence voltage drop across resistor in the direction of
current is given –ve sign.

In the figure

Definition:
Loop: It is any closed part of the circuit.
Mesh: It is a closed circuit which cannot be further divided into other
loops.

Kirchhoff’s branch current method:

Steps to be followed:
1) Identify no of independent loops equal to branch currents.
2) By applying KCL assign current directions
3) Apply KVL in each independent loop.

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4) Solve equations simultaneously and find each branch current.

Illustrating Numerical
Find current through 60 ohm by Kirchhoff's law

Assigning currents according to KCL we get

Write and solve KVL equations for three loops to get x, y and z.
Current through 60 Ω is Z

Maxwell’s Mesh current method :

Steps to be followed:
1) Identify no. of independent Meshes.
2) Consider current in each mesh in any direction(For convenience
assume clockwise direction)
3) Apply KVL in each mesh and get equations.
4) Solve equations simultaneously and get all mesh currents
5) If value of any mesh current comes –ve, it means the actual
direction of mesh current is opposite to what you have assumed.
While calculating the current through circuit element, special attention
is to be given to those elements which are common to two meshes.
Current in such element is algebraic sum of two mesh currents.

Illustrating Numerical
Find current through 2 Ω resistor.

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Here there are three meshes
Assign three mesh currents in clockwise direction

Write KVL equations for three meshes

10 - 6 i1 - 6 (i1-i2) = 0
7i1 – i2 = 0 --------- (I)
- (i2-i1) – 2i2 – 3(i2-i3) = 0
- i1+ 6 i2 -3 i3 = 0----- (II)
Similarly -3i2 + 13i3 = -20----- (III)
Solving three equations we get i1= 1.34 A
i2= -0.62 A
i3= -1.68 A
Current through 2 ohm resistor is i2=-0.62 A.
–ve Sign indicates that the actual direction of mesh current is opposite to
what you have assumed

-9–
Nodal Analysis : In this method one of the node is taken as reference
node and potential of all remaining nodes are w.r.t reference node.
Node is defined as junction where two or more circuit elements meet each
other.
Steps to be followed:
1) Identify no of nodes.

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 2) Assume voltages at all nodes w.r.t reference node( zero voltage
node)
3) Mark currents in all branches.
4) Apply KCL at all node except reference node.
5) Write current as Potential difference/Resistance between two nodes.
6) Solve equations simultaneously.

Illustrating Numerical
Find voltages at nodes 1 and 2. (Jan 2003)

Here there are three nodes.

Voltage at node 1and 2 are V1 and V2 resp.
Assign current in every branch as shown in the diagram. Branch
containing current source has same current

KCL at node 1 gives

1= i1 + i2
i1= (V1-0)/2
i2 = (V1-V2)/2
V 1 V 1−V 2
+ = 1 −−−−− (1)
Hence 2 2

KCL at node 2 gives

2 + i2 = i3
i3 = (V2-0)/1
V 1−V 2 V 2
2+ = −−−−−( 2 )
therefore 2 1

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solving 1 and 2 we get
V1= V2=

Superposition Theorem:
Statement: In a linear network containing more than one active sources
(i.e. constant voltage and constant current source), the resultant current
in any element is the algebraic sum of current that would flow when each
source acting alone, all other sources replaced by their internal
resistances.
Internal resistance of voltage source is zero hence represented by short
circuit. Internal resistance of current source is infinite hence represented
by open circuit. A linear network is one which obeys ohm’s law i.e. the
relation between voltage and current is linear

Steps to be followed:
1) Consider any source and draw diagram with other sources replaced
by short or open circuit.( If internal resistance is given consider that
value)
2) Solve by any method explained earlier and find current in resistor.
3) Write your answer with direction assumed.
4) Repeat these steps for all other sources.
5) Add all current algebraically to get final answer.

Illustrating Numerical

1. Determine current in 10 Ω resistor.

Consider only 10V acting. Draw the diagram with 10V acting and 4A
replaced by open circuit.

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We have two meshes. Apply KVL and write two equations.
10-i 1-10(i1-12) =0------------- (1)
-2 i2- 5 i2- 10(i1-12) =0------------- (2)
Solving we get i1= and i2=
Current through 10 ohm resistance is (i1-i2)=
Consider only 4A acting. Draw the diagram with 4A acting and 10V
replaced by Short circuit.

12-i1= 4 and -12 i1-5 i2 = 0

Solving we get 1=
Current through 10 ohm resistance is i 1=
By superposition theorem I 10 ohm =

Thevenin’s Theorem :
Statement: Any network between terminals A and B can be replaced by
single source of emf (Vth) in series with single resistance (Rth).

A network is connected between A and B. It consists of resistances and

sources connected in any manner. According to theorem network
connected between A and B can be replaced by emf source V th and
resistance Rth as shown in figure. This circuit is called Thevenin’s
equivalent circuit.

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Steps to be followed:
1) To find Vth: Remove load resistance between A and B. calculate
voltage between open terminals (Vth =VAB) by applying KVL to the
path connecting A and B with A as +ve w.r.t B.
2) To find Rth: With load resistance removed replace sources by their
respective internal resistances and Calculate resistance between A
and b (Rth =RAB).
3) Draw Thevenin’s equivalent circuit. Connect load resistance
between terminals A and B and find current flowing through it by
simply applying ohm’s law.

Example: Using Thevenin theorem, calculate the current flowing through

the 4 ohm resistor

Step 1. If we remove the 4-Q resistor, the circuit becomes as shown

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The two resistors of 3 Ω and 6 Ω are connected in series across the 12 V
battery. Hence, drop across 6 Ω resistor =12 x 6/(3 + 6) =8 V.
Since full 10 A current passes through 2Ω resistor, drop across it is 10 x 2
=20 V.
Appling KVL From A(assume as + terminal) to B(-ve terminal) we get
-8 + 20 + Vth =0
Vth = -12V (means B is +ve wrt A)

Step 2. Finding Rth

Replaced Voltage source by a short circuit (zero resistance) and replaced
by an 'open' (infinite resistance) and draw the circuit

Rth= 6 // 3 + 2 = 4 Ω
Step 3: to draw Thevenins Equivalent and connect Load resistance RL

I L = (12)/ (4+ 4)
= 1.5 A From B to A

Norton’ Theorem :
Statement: Any network between terminals A and B can be replaced by
current source of output IN in parallel with Resistance RN.

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In diagram above a network is connected between A and B. It consists of
resistances and sources connected in any manner. According to theorem
network connected between A and B can be replaced by Current source I N
and resistance RN as shown in figure. This circuit is called Norton’s
equivalent circuit

Steps to be followed:
1) To find IN: Remove load resistance between A and B place short
between them.
2) Find current flowing in short circuited path i.e. IN.
3) To find RN: With load resistance removed replace sources by their
respective internal resistances and Calculate resistance between A
and b (RN =RAB).
4) Draw Norton’s equivalent circuit. Connect load resistance between
terminals A and B and find current flowing through it by simply
applying current division rule.

Example: Using Norton's theorem, find the current which would flow in a
25 Ω Resistor connected between points A and B. (May 10)

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Circuit can be redrawn as

Step 1 Converting voltage sources into current sources we will found that
total current will flow in short circuited path. i.e

Step 2 To find RN we see that all three resistances 5,10and 20 are

connected cross A and B hence
RN = 5// 10//20 = 2.86 Ω
Step 3 Draw Norton’s equivalent and connect load resistance cross it.

By current divider formula we get

IL =5.5 * 2.86
(2.86 +25)
IL = 0.56 A (From A to B)

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Maximum Power Transfer Theorem : (MPTT)
Statement: A resistive load will abstract maximum power from the
network when the load resistance is equal to resistance of network
calculated between output terminals with all sources replaced by their
internal resistance.
Proof : Consider a network consist of emf source E with internal
resistance Rs, series resistance R and load resistance R L connected series
as shown in diagram.

Let Ri be internal resistance of the network. Ri = Rs + R

When RL = Ri maximum power will be transferred.

Circuit current

Power consumed by load is

For PL to be maximum,
Differentiating above equation we get

hence proof.

Maximum power Pmax =

Steps to be followed:

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 1) To find Vth (E): Remove load resistance between A and B. calculate
voltage between open terminals (Vth =VAB) by applying KVL to the
path connecting A and B with A as +ve w.r.t B.
2) To find Rth: With load resistance removed replace sources by their
respective internal resistances and Calculate resistance between A
and b (Rth =RAB).
3) For maximum power to transfer RL =Rth

4) Calculate Pmax =
Note following points :
1) Under condition of maximum power transfer efficiency of network is
η = output power
Input power

2) Due to low efficiency & greater voltage drop, electric power supply
never apply MPTT.Here emphasis is more on high efficiency than
maximum power transfer.
4) In case of Electronics & communication emphasis is more on
maximum power transfer though at reduced efficiency.
Example: Find the maximum power in 'RL' which is variable in the
circuit shown below.

Solve the above problem as per steps of Thevenin’s Theorem to get Vth
= 8 Volts and Rth = 49 Ω
As per Maximum power transfer RL = Rth = 49 Ω
Then draw thevenins equivalent and find IL
IL = 0.0816 A
Pmax = IL2* RL
= 0.3265 W

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