Unit – 1

D.C. Circuit Analysis

1.1           Basic concepts of electric circuits

Electric circuit: A circuit is an interconnection of electrical components.

    Figure 1. Circuit

Electric charge: is one of the fundamental quantities and exists in every atom.

 Symbol: Q or q

Unit: Coulomb (C).

Electric Current: The time rate of change of charge.

i(t ) = dq/dt

Symbol: i t( ) or i or I depending on whether the current is constant or time varying quantity.

Unit: Ampere (A); A = 1.C / 1.s

Voltage (Potential Difference):

Voltage => Work done on q to move it from point A to point B per unit charge

 => Difference in potential energy per unit charge

Symbol: V or v(t ) v t

Unit: Volt (V); 1.V = 1.J/1.C

Power: The time rate of change of energy is defined as power.

Symbol: P or p.

Unit: Watts (W). 1 W = 1 J / 1 s.

In general i, v and p are used to represent time varying parameters.

 ∴ Change in energy from time t1 to t2

 ∆w =   =

1.2 Ohm’s Law

Statement - Ohm’s Law states that the current flowing through a conductor is directly proportional to the potential difference applied across its ends, provided the temperature and other physical conditions remain unchanged. Mathematically it can be represented as,

Figure 2. Ohms law

Potential difference ∝ Current

V ∝ I

(When the value of V increases the value of I increases simultaneously)

V = IR  ………………………. (eq.1)

Where,

Calculating Various Parameters Using Ohm’s Law

As an equation (eq.1), the Ohm’s Law serves as master for calculating the current (I), when the resistance (R) and the potential difference (V) are known. Similarly, if any two parameters in the equation (eq.1) are known, then the unknown third one can be easily calculated as follows:

V = IR …………………………… (eq.2)

I=VR ……………………………… (eq.3)

R=VI  ……………………………… (eq.4)

1.3 Independent energy sources

An independent voltage source  maintains a voltage fixed or varying with time not affected by any other quantity.

Similarly an independent current source  maintains a current fixed or time-varying which is unaffected by any other quantity. The usual symbols are shown in figure.

Figure 1.3: Symbols for independent sources 

 
1.4 Dependent energy sources

Some voltage or current sources have their voltage or current values varying with some other variables. They are called dependent   voltage or current sources or controlled voltage or current sources, and their usual symbols are shown in figure 

Figure 1.4: Symbols for dependent sources. Variables in brackets are the controlling variables whose values affect the value of the source.  

1.5 Passive elements

A passive element is an electrical component that does not generate power, but instead dissipates, stores, and/or releases it. Passive elements include resistances, capacitors, and coils also called inductors. These components are labelled in circuit diagrams as R, C and L, respectively. In most circuits, they are connected to active elements, typically semiconductor devices such as amplifiers and digital logic chips.

1.6 Circuit properties

The components in electrical circuits can be connected in series or in parallel.

Series connections

Components that are connected one after another on the same loop of the circuit are connected in series. The current that flows through each component connected in series is the same.

Two lamps connected in series

The circuit diagram shows a circuit with two lamps connected in series. If one lamp breaks, the other lamp will not light.

      Figure 1.5 Two lamps connected in series with open switch and cell                                         

The sum of all the potential differences across the components in a series circuit is equal to the total potential difference across the power supply.

Parallel connections

Components that are connected on separate loops are connected in parallel. The current is shared between each component connected in parallel. The total amount of current flowing into the junction, or split, is equal to the total current flowing out. The current is described as being conserved.

Two lamps connected in parallel

The circuit diagram shows a circuit with two lamps connected in parallel. If one lamp breaks, the other lamp will still light.

Figure 1.6 Two lamps connected in parallel

The lights in most houses are connected in parallel. This means that they all receive the full voltage and if one bulb breaks the others remain on.

For a parallel circuit the current from the electrical supply is greater than the current in each branch. The sum of all the current in every branch is equal to the current from the electrical supply.

1.7 Kirchoff’s laws

Kirchhoff’s Current  Law (KCL)

Statement - The algebraic sum of currents meeting at a junction or node in a electric circuit is zero  or the summation of all incoming current is always equal to summation of all outgoing current in an electrical network.

Explanation

    Figure 1.7 KCL

Assuming the incoming current to be positive and outgoing current negative we have

I e incoming current = ∑ outgoing current thus, the above Law can also be stated as the sum of current flowing towards any junction in an electric circuit is equal to the sum of currents flowing away from that junction.

Kirchhoff’s Voltage Law (KVL)

The second law of Kirchhoff’s laws is Kirchhoff’s Voltage Law. It states that in any closed loop network, the sum of the emf values in any closed loop is equal to the sum of the potential drops in that loop.

Fig. 1.8  – Kirchhoff’s Voltage Law

In another words, it can also be said as “The total voltage around the loop is equal to the sum of all the voltage drops within the same loop”, which is equal to zero.

Example :

Consider the following circuit :

-12 + V1 + V2 -6 +V3 + V4 =0

V1+V2+V3+V4 = 12 +6

V1 + V2 + V3 + V4 = 18

1.8 Applications of Kirchoff’s laws

The applications include:

1.9 Nodal and Loop methods of Analysis

Nodal analysis

In the nodal method we are finding the node voltages with the following steps:

Consider the following circuit.

   Figure 1.9 Nodal Analysis

Let us consider a reference node – 3.

In a real circuit a reference node is ground is assumed to have potential.

By the Kirchhoff’s Law having the following current relation:

I1 = i1 + i2

I2 = i2 -i3

Let us represent currents like ratio of voltage and resistance by the Ohm’s Law:

i 1 = v1 /R1 ;i 2 = v1 -v2 /R2 ; i3 = v2/R3;

or

i 1= v1 * G1 ; i2 =(v1 -v2)*G2 ; i3 =v2 * G3

i1 = v1 * G1 + (v1-v2) * G2 ;

i2 = (v1-v2) * G2 – v2 * G3

i1= v1 * (G1 +G2) – v2 * G2

i2 = v1 *G2 – v2 *(G3 +G2)

or

And we achieve the following matrix equation, which can be resolved by the Cramer’s rule:

[ G1 + G2 -G3 -G3 ] [v1] = [I1]

[     G2                -G2][v2] =[I2]

Consider the following figure :

  Figure 10. Mesh circuit

There are two possibilities that exist:

Applying Kirchhoff’s Law we achieve the following equations:

i  1 + i4 = i2 + i3

-v2 +5+v3 =0

v 1 =10

By Ohm’s Law

These equations will help determine node voltages.

Loop Analysis

Loop analysis is a special application of KVL on a circuit. We use a special kind of loop called a 'mesh' which is a loop that does not have any other loops inside of it. A mesh starts at a node and traces a path around a circuit, returning to the original node without hitting any nodes more than once. We can only apply mesh analysis to planar circuits, that is circuits without crossover connections. If a circuit cannot be redrawn without the intersecting disconnected lines then we cannot use mesh analysis.

1.10 Superposition Theorem

Superposition Theorem

Question 1. Find the current through resistance.

Solution:

1= 0

   2=

   1 + 2 

Special Case:

Since two voltage sources with different magnitude in parallel which cannot be connected as in single branch two different current is not possible (if 5V than I = zero).

Question:

1 =

2 =

=

1 + 2

=

1.11 Thevenin’s and Norton’s Theorem

Thevenin’s and Norton’s Theorem

 Thevenin’s equivalent of A

Norton’s equivalent of A

  sc = Vth/Rth

CONDITIONS FOR APPLICATION

     Method for finding Rth :

Firstly, open circuit terminal A and B.

2.     If network A is operating with independent and dependent sources:

3.     If network is operating with only dependent sources:

Connect generation between A and B

     Method for Vth:

First open circuit terminal A and B.

Find out the voltage between A and B this is Vth

     Method for Isc:

Isc =

Question:

Answer:

Finding Isc from circuit directly:

By KCL,

Question:

Answer

Also, clear from circuit that Vth = 1V.

By applying KVL we get,

1-3Isc=0

Isc=A

Que:

Ans;

Rth=3k+2k=5k

By applying KVL we get

  Therefore,

Question:

Solution: For Rth

By KCL,

     But,

By KVL,

Question:

Solution: Since, no independent source is present so,

  Isc = 0

  And we know that,

  Since Rth cannot be zero

         But

Question: Find out the Norton’s equivalent

Solution:

Since, there is no significance of current source

1.12 Reciprocity Theorem

The reciprocity theorem states that in a linear passive bilateral network by changing the voltage source from branch 1 to branch2, the current I in the branch 2 appears in branch 1.

   Figure 12. Reciprocity theorem

The resistances R1, R2 and R3 is connected in the circuit diagram. It is clearly shown the voltage and current sources are interchanged for solving the network.

Step 1 – Firstly, select the branches between which reciprocity is established.

Step 2 – The current in the branch is obtained using any conventional network analysis method.

Step 3 – The voltage source is interchanged between the branch which is selected.

Step 4 – The current in the branch where the voltage source was existing earlier is calculated.

Step 5 – Hence it is seen that the current obtained in the previous connection, that is  step 2 and the current which is calculated when the source is interchanged, that is step 4 are identical to each other.

Problem:

Verify the reciprocity theorem for the given network.

Find the current in branch AB

RT = 2+[3||(2 + 2 || 2)

RT = 3.5 Ω

IT = V / RT = 20/3.5 = 5.71 A

Apply current division technique for the circuit to find the current through branch AB

= 5.71 x 3 / 6 = 2.855 A

Current in branch AB = 2.855 x 2/4 = 1.43A

Interchanging the source to branch AB

Total resistance RT = 3.23Ω IT = 20/3.23 = 6.19A

Apply current division technique for the circuit to find current through branch CD = 6.19 x 2/5.2 = 2.38A

 Current in branch CD = 2.38 x 3/5 = 1.43A

Hence reciprocity theorem verified.

Verify the reciprocity theorem for the given circuit:

Apply current division technique for the circuit to find current through branch 3Ω

= 10 x 2 / 5 = 4 A.

Voltage across 3Ω = 4 x 3 = 12 V

Replace the current source and find the open circuit voltage.

Apply current division technique and find the current through 2 Ω resistor.

= 10 x 3 / 5 = 6

Voltage across 2Ω resistor = voltage across AB = 6 x 2 = 12 V

1.13 Maximum Power Transfer Theorem

   Figure 13. Maximum Power Transfer

  where Rth is Thevenin’s equivalent resistance across a and b.

Maximum power is absorbed by ZL  when

 Condition:

       Comparing real and imaginary parts

   OR

       Maximum power absorbed by ZL is

Question: Find out the value of load resistance if power absorbed is maximum.

Solution: find Thevenin’s equation

Question: Find maximum power delivered is RLif its value is

Ω

Solution

  Therefore,

1.14 Millman’s Theorem

Millman’s Theorem

The Millman’s Theorem states that  when a number of voltage sources (V1, V2, V3……… Vn) are in parallel having internal resistance (R1, R2, R3………….Rn) respectively, the arrangement can replace by a single equivalent voltage source V in series with an equivalent series resistance R.  

In other words; it determines the voltage across the parallel branches of the circuit have more than one voltage sources which reduces the complexity of the electrical circuit.

   Figure 14. Millmans theorem

V = V1 G1 V2 G2   ……………………… Vn Gn / G1 + G2 ……………+Gn 

And

R = 1/G = 1/ G1 + G2 + …………………..+Gn

Assuming a DC network of numerous parallel voltage sources with internal resistances supplying power to a load resistance RL as shown in the figure below

   Figure 15. DC network
Let I represent the resultant current of the parallel current sources while G the equivalent conductance as shown in the figure below

   Figure 16. Equivalent

I = I1 + I2 + I3 + …………………….. and

G = G1 + G2 + G3 + …………………

Next, the resulting current source is converted to an equivalent voltage source as shown in the figure below
 

   Figure 17.Final circuit 

Thus,

V = 1/G = I1 I2 …………………… In / G1 + G2 + ………………..+ Gn

Positive (+) and negative (-) sign appeared to include the cases where the sources may not be supplying current in the same direction.

Also,

R = 1/G = 1/ G1 + G2 +………………….+ Gn

And as we know,

I = V/R, and we can also write R = I/G as G = I/R

So, the equation can be written as

V = V1/R1 V2 / R2 ……………………Vn/Rn / 1/R1 + 1/R2 +……………..+1/Rn

Where R is the equivalent resistance connected to the equivalent voltage source in series.

Thus, the final equation becomes

V = V1G1 V2 G2 ……………………VnGn  / G1 + G2 +……………..+Gn or

V = /   and Gk = 1/Rk

Steps for Solving Millman’s Theorem

Following steps are used to solve the network by Millman’s Theorem

Step 1 – Obtain the conductance (G1, G2,….) of each voltage source (V1, V2,….).

Step 2 – Find the value of equivalent conductance G by removing the load from the network.

Step 3 – Now, apply Millman’s Theorem to find the equivalent voltage source V by the equation shown below

V = V1G1 V2 G2 ……………………VnGn  / G1 + G2 +……………..+Gn

Step 4 – Determine the equivalent series resistance (R) with the equivalent voltage sources (V) by the equation

R = 1/G

Step 5 – Find the current IL flowing in the circuit across the load resistance RL by the equation

IL = V / R + RL

1.15 Star-Delta or delta-star transformation

Star to delta conversion to final equivalent resistance

    Figure 18. Delta to star

We know that (from delta to star conversion)

R1 =    …….①

R2 = …..②

R1 = ……③

Multiply ① X ② L.H.S and R.H.S

R1 R2  = …….④   where  

Similarly multiply  ② X ③

R2 R3  = …….⑤

And ③  X①

R1 R3  = …….⑥

Now add equation ④, ⑤, and ⑥ L.H.S and R.H.S

……refer eq. ②

= + +

=  +

(Delta)      star             star

Similarly R23 = R2+R3 +

R23 = R1+R2 +

Delta       Star

= R12// (R23 + R13)         =R1 + R2

=       = R1 + R2

=    

Here let R = R12 + R23 + R13

=    R2 + R3

   R1 + R3

Now the 3 equations after equating L.H.S. and R.H.S

R1 + R2 = …….①

R2 + R3 = ……②

R1 + R3 = …..③

Now subtract ② and ① on L.H.S. and R.H.S

R2+ R3 – R1 – R2 =

R3 – R1 = …..④

Now add equation ④ and ③

R3 – R1 + R1 + R3 =

2R3 =

Similarly R1 =

And R2 = R23 R12/R       where R = R12 + R23 + R13

ie  star equivalent from delta network is ratio of product of adjacent branches in delta to the addition of all branches in delta.

1.16 Applications of network theorems

Maximum power transfer theorem:

The circuit as shown in figure provides power to the load 27. This can be viewed as simulating a power amplifier delivering power to a complex load. It is the Thevenin equivalent of a more complex circuit. Calculate and plot a graph of the power delivered to the load for each of the following frequencies: 10 kHz, 30 kHz, 50 kHz, 80 kHz, and 100 kHz.

   Figure 20. Maximum Power Transfer

For f= 10 kHz,

Xc = 1/ 2 f C = 1/ 2 (10kHz) (0.01 μ F) = 1.59 kΩ

XL = 2 π f L = 2 (10KHz) (1mH) = 62.8 Ω

The magnitude of the total impedance is

Ztot = (Rs + RL) 2 + (XL – Xc) 2 = (20 Ω) 2 + (1.53 k Ω) 2 = 1.53 k Ω

The current is

I = Vs / Ztot = 10/1.53k = 6.54 m A

The load power

PL = I 2 RL = (6.54 m A) 2 x 10 Ω = 428 μ W

For f=30KHz

Xc = 1/2∏(30kHz)(0.01μF)

XL = 2 ∏ (30kHz) (1mH) = 189 Ω

Zout = (20 Ω) 2   + (342Ω) 2 = 343 Ω

I = Vs / Zout  = 10V / 343 Ω = 29.2 m A

PL = I 2 RL (29.2 mA) 2 (10 Ω) = 8.53 m W

F=50kHz

Xc = 1/2∏(50kHz)(0.01μF)=318Ω

XL = 2 ∏ (50kHz) (1mH) = 314 Ω

Zout = (20 Ω) 2   + (4Ω) 2 = 20.4 Ω

I = Vs / Zout  = 10V / 20.4 Ω = 490 m A

PL = I 2 RL (490 mA) 2 (10 Ω) = 2.40 m W

For f =80kHz

Xc = 1/2∏(80kHz)(0.01μF)=199Ω

XL = 2 ∏ (80kHz) (1mH) = 503 Ω

Zout = (20 Ω) 2   + (304Ω) 2 = 305 Ω

I = Vs / Zout  = 10V / 305 Ω = 32.8 m A

PL = I 2 RL (490 mA) 2 (10 Ω) = 10.8 m W

F=100kHz

Xc = 1/2∏(100kHz)(0.01μF)=159Ω

XL = 2 ∏ (100kHz) (1mH) = 628 Ω

Zout = (20 Ω) 2   + (304Ω) 2 = 469 Ω

I = Vs / Zout  = 10V / 305 Ω = 21.3 m A

PL = I 2 RL (490 mA) 2 (10 Ω) = 4.54 m W

1.17 P-spice for DC circuit analysis

Figure 21: Open new schematic

After opening the new schematic before jumping into designing first save the schematic by clicking on the file button at the top left corner and then selecting save as so that we can access it anytime in the future. Refer to the figure below,

Figure 22: Saving schematic

Click on the get new part icon at the top bar of the schematic window in order to search for the components that are needed for circuit designing.

Figure23 : Getting new part

the figure below,

Figure 24: Place resistor

with each other. the resistors.

Figure : Resistors placed

Again open the get new part window and in the pat name block type Vdc, select the supply from the list given and then click on place & close as shown in the figure below

Figure 25. Input dc placed

Next step is to place a ground, do the same again and in the part name type Gnd and select the ground and then click on place & close as shown in the figure below,

Figure 26: Ground placed

The placed components in the schematic window are shown in the figure below,

Figure 27: Placed components

Click on the draw wire icon at the top bar of the schematic window in order to connect the already placed components for circuit designing.

Figure 28: Drawing wire

Connect all the resistors in parallel with the wire as shown in the figure below,

Figure 29: Parallel resistor

Connect the remaining components to complete the circuit diagram as shown in the figure below,

Figure 30: Complete circuit diagram

Next step is to set the attributes of the input dc supply. Double click on the dc supply you connected in the circuit previously and set the dc voltage of the supply to 10V as shown in the figure below,

Figure 31: DC attributes

On the top of the schematic window, click on the Voltage/Level Marker button as shown in the figure below,

Figure 32: Voltage marker

Place it at each resistor node as shown in the figure below,

Figure 33: Voltage marker placed

Next step is to adjust the properties of the simulations in order to produce the graph of the voltage at the marker. Click on analysis and then click on Setup as shown in the figure below,

Figure 34: Analysis setup

A widow will appear, click on the transient block on the window and adjust the properties of the window according to your requirement, refer to the figure below

Figure  35.Transient properties

If we are interested in checking the voltage on a specific wire in spite of checking it at a node, double click on the wire and inn the window that appear as a result, type the name of the wire you want to label it with, as shown in the figure below,

Figure  36: Labeling the wire

Now comes the simulation part, click on the analysis at the top bar of the schematic window and then click on simulate as shown in the figure below,

Figure 37: simulation

As all the resistors are in parallel and we know that the voltage remains the same in parallel regardless of the value of the resistances connected in parallel. If two or more resistors are connected in parallel, then the voltage across each of the parallel connected resistances remains same no matter what value we assign to the resistance. The voltage supplies to the parallel resistor appears across every resistor connected parallel to it. This concept is illustrated in the output of the circuit we simulated above.

Figure38 : Output Voltages

Figure39 : Adding Y axis

Figure40 : Updating resistance value

Also change the marker from the voltage marker to the current marker as shown in the figure below,

Figure 41: Current marker

Simulate the circuit as we have done before, now the output of the circuit will show three different current depending upon the value of the resistor as shown in the figure below,

Figure 42: Current output

References