Unit I

Basic Circuit Analysis and Simplification Techniques

Kirchhoff Current Law:

A node is a circuit connection that extends to all its neighboring elements but does not contain them.

Kirchhoff’s current law states that at a node:

Sum of the in currents = sum of out currents

Applying KCL at four nodes

0 = Ix + I1 + I2 KCL at node a

I2 = I3 KCL at node b

I3 + I4 = 0 KCL at node c

Ix + I1 = I4 KCL at node d

Adding equations 6-9 gives the following equation:

Ix + I1 + I2 + I3 + I4 = Ix + I1 + I2 + I3 + I4.

Kirchhoff Voltage Law:

A loop is a path around a circuit that starts and ends in the same place.

Kirchhoff's voltage law (KVL) states that around a loop:

sum of voltage rises=sum of voltage drops--------------------(1)

Applying KVL around the three loops we get

V1 = Vx KVL around loop 1

Vx + V4 = V2 + V3 KVL around loop2

V1 + V4 = V2 + V3 KVL around loop3

Find the voltage across the current source and the current passing through the voltage source.

R1 is in series with the current source; therefore, the same current passing through it as the current source:

IR1=I1=3A

and the voltage across R1 can be found by Ohm's law:

VR1=R1×IR1=2×3=6V

To find the voltage across the current source, KVL can be applied around the left- hand side loop:

The direction does not matter and would not change the result.

−VI1+VR1−V1=0A

VI1=VR1−V1=6−15=−9V

R2 and R3 are also in series and their equivalent is R23=R2+R3=3Ω+2Ω=5Ω

R23 is parallel with the voltage source. This means that its voltage is equal to the voltage of the voltage source.

VR23=V1=15V

Now, using the Ohm's law, the current passing through R23 can be calculated:

IR23=VR23/R23=15/5=3A

To find the current of the voltage source, we can apply KCL at one of the nodes:

IR23+I1+IV1=0

IV1=−IR23−I1=−3−3=−6A

Find out V1 and V2 using KVL.

-20 + v1 + v2 =0 ------------------------(1)

V1 = 2i from the circuit

V2 = 3i

Therefore

-20 + 2i + 3i =0

5i = 20

i=4A

v1 = 2i = 2(4) = 8V

v2 = 3i = 3(4) = 12V

Find the current io and voltage vo as shown in the figure using KCL .

At node ‘a’

0.5 io + 3 -io =0

3 – 0.5 io = 0

Io=6A

vo = 4io

vo = 4(6) = 24 V

The Independent and Dependent source means, whether the voltage or current sources are either depending upon some other source, or they are acting independently.

Energy Sources

There are two types of energy source direct sources and alternating sources.

Direct Source

The voltage and the current source are the direct sources. The direct source is further classified as independent voltage and current source and dependent voltage and the current source.

Independent Voltage and Current Source

Independent sources are that which does not depend on any other quantity in the circuit. They are two-terminal devices and has a constant value, i.e. the voltage across the two terminals remains constant irrespective of all circuit conditions.

The strength of voltage or current is not changed by any variation in the connected network the source is said to be either independent voltage or independent current source. In this, the value of voltage or current is fixed and is not adjustable

Dependent Voltage and Current Source

The sources whose output voltage or current is not fixed but depends on the voltage or current in another part of the circuit is called Dependent or Controlled source. They are four-terminal devices.

When the strength of voltage or current changes in the source for any change in the connected network, they are called dependent sources. The dependent sources are represented by a diamond shape.

The dependent sources are further categorised as:

Voltage Controlled Voltage Source (VCVS)

In voltage-controlled voltage source, the voltage source is dependent on any element of the circuit.

In the above figure, the voltage across the source terminal Vab is dependent on the voltage across the terminal Vcd,

Vab α Vcd or

Vab = k Vcd

Voltage Controlled Current Source (VCCS)

In the voltage controlled current source, the current of the source iab depends on the voltage across the terminal cd (Vcd) as shown in the figure below:

iab α Vcd

iab = Vcd

Thus,

Where ƞ is a constant known as transconductance and its unit is mho.

Current Controlled Voltage Source (CCVS)

In the current controlled voltage source voltage source of the network depends upon the current of the network as shown in the figure below

Here the voltage of source Vab depends on the current of the branch cd

Vab α icd

Vab = r icd

Where r is a constant.

Current Controlled Current Source (CCCS)

In the Current Controlled Current Source, the current source is dependent on the current of the branch another branch as shown in the figure below

iab α icd

iab = β icd

So,

Where β is a constant

Power Formula 1 – Electrical power equation: Power P = I × V = R × I2 = V2 ⁄ R

where power P is in watts, voltage V is in volts and current I is in amperes (DC).

If there is AC, look also at the power factor PF = cos φ and φ = power factor angle

(phase angle) between voltage and amperage.

Electric Energy is E = P × t − measured in watt-hours, or also in kWh. 1J = 1N×m = 1W×s

Power Formula 2 – Mechanical power equation: Power P = E ⁄ t where power P is in watts,

Power P = work / time (W ⁄ t). Energy E is in joules, and time t is in seconds. 1 W = 1 J/s.

Power = force multiplied by displacement divided by time P = F × s / t or

Power = force multiplied by speed (velocity) P = F × v.

Mesh Analysis:

Mesh analysis is applicable to the networks which are planar. Planar network is a network where branches are not passing over or under each other.

Nodal method uses Kirchhoff’s currents Law to consider nodal voltages, and Mesh method uses Kirchhoff’s voltages Law to consider mesh currents. Mesh is a loop, which does not contain any other loops.

There are only two meshes in abefa and bcdeb, abcdefa is not a mesh but a loop However, for Mesh method only meshes are in use.

Mesh analysis steps are the following:

The following equations correspond to the Kirchhoff’s Voltage Laws for the meshes:

V1 = i1 *(R1+R3) – i2 *R3

-V2 = -i1 *R3 + i2 * (R3 +R2);

[ R1 +R3 -R3 [i1]= [V1]

-R3 R2+R3] [i2] = [V2]

Using Cramer’s formula for resolving the matrix equations above we can find mesh currents.

I1 =i 1 I2 = i2 I1 -I2 =i3

Super Mesh Analysis

A super-mesh is a larger loop which has both meshes inside.

There are two methods.

1) Using

Mesh II:

R1 x (I2 – I1) + R2 x (I2) – V IS =0

Note that for R1, unlike the equation for Mesh I, the current is I2 – I1. This is because we are walking around the loop with the direction of I2, or briefly it is because we are writing the equation for mesh of I2.

Mesh III:

R3 x (I3 – I1) + VIS + R4 x (I3) =0

Let's add two equations:

R1 x (I2-I1) + R2 x (I2) – VIS + R3x (I3 -I1) + VIS + R4x(I3) = 0+0

Simplifying:

R1 x (I2-I1) + R2 x (I2) – VIS + R3x (I3 -I1) + VIS + R4x(I3) = 0+0

Method -II

Here is the supermesh:

Around the loop clockwise:

R1 x (I2-I1) + R2 x (I2) + R4 x (I3) + R3 x (I3 – I1 ) =0

As you can see, we were able to write the equation in one shot. That is why the supermesh method is preferred.

Nodal analysis

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

Consider the following circuit.

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 :

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.

Super node analysis:

Nodal Super node

The difference in both circuits is that there is an additional voltage source of 22V instead of 7Ω resistor between node 2 and node 3.

In Node or Nodal analysis, we use KCL at each non-reference nodes hence three nodes in fig (a).

There are two methods to simplify the circuit in the above fig (b).

Assign an unknown current value to the branch contains the voltage source. Then apply KCL three times on the 3 Nodes that is one KCL equation for each node. At last, apply KVL that is v3–v2 = 22V between Node2 and Node3.

Henc, we get four (4) equations for unknown values , which is complex to simplify.

The 2nd method is called Supernode analysis. In this apply KCL to both nodes that is node2 and Node3 at once. The supernode is indicated by the region enclosed by the dotted line. This is possible because, if the total current leaving Node2 is zero (0) and the total current leaving Node3 is zero (0), then total current leaving the combination is zero. This concept is shown in the following figure with the supernode indicated by the broken line.

Example:

Use Supernode analysis to find voltage across each current source i.e. v1 & v2 in the following fig 3 (a)?

Solution:

First, we redraw the circuit as shown in fig (b)

We begin by writing a KCL equation for Node1.

4 = 0 + 3v1 + 3v3 …. → Eq 1.

Now, consider the supernode (Combination of Node1 and Node2). Moreover, one current source and three resistors are connected. Thus,

Apply KCL at Supernode (Node1 & Node2)

9 = v2/1/2 – v3/1/6 + v3-v1/1/3 + v2-v1/0

9 = 2v2 + 6v3 + 3v3 – 3v1 + 0.

9 = – 3v1 + 2v2 + 9v3 …. → Eq 2.

Since we have three unknown values, therefore, we need one additional equation. Obliviously, we will go for the 5V voltage source between Nodes 2 and 3, which is;

v2 – v3 = 5 …. → Eq 3.

Solving equations 1, 2 and 3 we get

v3 = 0.575 V or 375mV.

v2 = 5.375 V.

v1 = 1.708 V.

Source Transformation: Conversion of Voltage Source into Current Source

When the voltage source is connected with the resistance in series and it has to be converted into the current source than the resistance is connected in parallel with the current source as shown in the above figure.

Where Is = Vs /R

Conversion of Current Source into Voltage Source

In the above circuit diagram a current source which is connected in parallel with the resistance is transformed into a voltage source by placing the resistance in series with the voltage source.

Where, Vs = Is / R

Source Shifting : We mean by either shifting current source or voltage source.

Shifting voltage source:

The figure shows node a at which the positive terminal of the voltage source V is connected to a couple of impedances Z1 to Z4. But we can't transform the voltage source, V, as it has no impedance in series with it.

However, we can push this voltage source through the node, towards the individual branches of the network. Care has to be taken to see that the current distribution through the circuit remains unaffected.

Figure b shows the resultant circuit obtained by the push through mechanism of the voltage source. At this instant, after V-shift, the voltage source is made to appear at every branch of the electrical network in series with the impedances present in each of them.

Consider the case where there are a couple of impedances joined at either ends of the voltage source. That is, Z1, Z2, and Z3 are connected to the negative terminal of the voltage source, V, while Z4 and Z5 are connected to its positive terminal.

The voltage source, V, in Figure a can either be (i) pushed through the node, a, towards the individual branches Z4 and Z5 or (ii) pulled through the node, b, towards the individual branches Z1, Z2, and Z3.

The networks resulting from such push-pull operation are shown by Figures b and c, respectively.

Shifting of current source:

In figure (a) the current source, I, connected between the nodes, a and c, such that current leaves node c and enters into node a.

The circuit can be redrawn as shown in Figure b wherein the identical current source is made to exit from node c, enter into node b, exit from node b, and enter into node a.

Note that even though the current is made to enter node b, it has no effect on the node. because an identical amount of current is made to exit from node b.

This means that the circuits shown in Figures a and b are equivalent to each other, except for the fact that one may apply current-to-voltage source transformation readily for the circuit shown in Figure b.

Superposition theorem states that in any linear, active, bilateral network having more than one source, the response across any element is the sum of the responses obtained from each source considered separately and all other sources are replaced by their internal resistance. The superposition theorem is used to solve the network where two or more sources are present and connected.

If a number of voltage or current sources are acting in a linear network, the resulting current in any branch is the algebraic sum of all the currents that would be produced in it when each source acts alone while all the other independent sources are replaced by their internal resistances.

Example:

The circuit diagram is shown below consists of two voltage sources V1 and V2.

First, take the source V1 alone and short circuit the V2 source as shown in the diagram below:

The value of current flowing in each branch, that is i1’, i2’ and i3’ is calculated by the following equations.

The difference between the two equations gives the value of the current i3’

Next, activate the voltage source V2 and deactivating the voltage source V1 by short-circuiting it, find the various currents, that is i1’’, i2’’, i3’’ flowing in the circuit diagram shown below:

Here,

And the value of the current i3’’ will be calculated by the equation shown below:

As per the superposition theorem, the value of current i1, i2, i3 is now calculated as:

The direction of the current should be taken care of while finding the current in the various branches.

The steps for solving Superposition Theorem

Considering the circuit diagramA, let us see the various steps to solve the superposition theorem:

Step 1 – Take only one independent source of voltage or current and deactivate the other sources.

Step 2 – In the circuit diagram B, consider source E1 and replace the other source E2 by its internal resistance. If its internal resistance is not given, then it is taken as zero and the source is short-circuited.

Step 3 – If there is a voltage source than short circuit it and if there is a current source then just open circuit it.

Step 4 – Thus, by activating one source and deactivating the other source find the current in each branch of the network. Taking the above example find the current I1’, I2’and I3’.

Step 5 – Now consider the other source E2 and replace the source E1 by its internal resistance r1 as shown in the circuit diagram C.

Step 6 – Determine the current in various sections, I1’’, I2’’ and I3’’.

Step 7 – Now to determine the net branch current utilizing the superposition theorem, add the currents obtained from each individual source for each branch.

Step 8 – If the current obtained by each branch is in the same direction then add them and if it is in the opposite direction, subtract them to obtain the net current in each branch.

The actual flow of current in the circuit C will be given by the equations shown below:

Problem:

Using Superposition theorem determine the voltage drop and current across the resistor 3.3K as shown in figure below

Step 1:

Remove 8V power supply from the original circuit such that the new circuit becomes as the following and measure the voltage across resistor.

Here 3.3 K and 2K are in parallel therefore the resultant resistance will be 1.245K

Using voltage divider rule across 1.245K will be

V1 =[1.245/(1.245+4.7)] * 5 = 1.047V

Step 2:

Remove the 5V power supply from the original circuit such that the new circuit becomes the following and then measure the voltage across resistor.

Here 3.3K and 4.7K are in parallel therefore the resultant resistance will be 1.938K. Using voltage divider rule voltage across 1.938K will be

V2 =[1.938(1.938+2)] *8 = 3.9377V

Therefore, voltage drop across 3.3K resistor is V1+V2 = 1.047 + 3.9377 = 4.9847V

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 connected across the load”.

To simplify any electrical circuit, to an equivalent two-terminal circuit with just a single constant voltage source in series with a resistance (or impedance) connected to a load as shown below.

Consider the circuit as shown in figure:

In order to analyse the circuit initially remove the 40Ω load resistor connected across the terminals AB and remove any internal resistance associated with the voltage source.

This is done by shorting out all the voltage sources connected to the circuit. Therefore v=0 or open circuit any connected current sources making i=0.

This because we want to make an ideal voltage source or ideal current source making for circuit analysis.

The value of the equivalent resistance, Rs is found by calculating the total resistance looking back from the terminals A and B with all the voltage sources shorted.

10 Ω resistor in parallel with 20Ω resistor

RT = R1 x R2/ R1 + R2 = 20 x 10 / 20 + 10 = 6.67 Ω

Find the equivalent voltage

To reconnect the two voltages back into the circuit Vs=VAB the current flowing around the loop is calculated as:

I = V/R = 20 V – 10V / 20 Ω + 10 Ω = 0.33 Amps.

The current is common to both resistors so the voltage drop across the 20Ω resistor or the 10Ω resistor is calculated as:

VAB = 20 –( 20 Ω x 0.33 amps) = 13.33 volts

VAB = 10 + (10 Ω x 0.33 amps) = 13.33 volts

The Thevenin’s Equivalent circuit is a series resistance of 6.67Ω and the voltage source of 13.33 V.

If 40Ω resistor is connected back to the circuit, we get

and from this the current flowing around the circuit is given as:

I = V/R = 13.33/6.67 + 40 = 0.286 amps.

Problem:

For the circuit as shown in figure find the current through RL = R2 = 1Ω resistor (Ia-b) branch using Thevenin’s theorem. Find the voltage across the current source.

Step 1: Disconnect the load resistance and reconnect the circuit.

Step 2: Apply any method to calculate Vth

At node C

2 + I1 + I2 =0

2 + (3 -Vc)/3 + (0-Vc/6) -- Vc = 6V

The currents I1 and I2 are computed by the following expressions:

I1 = Va -Vc/3 = 3-6/3 = -1A (I1 is flowing from c to a)

I2 = 0 – Vc/6 = -6/6 = -1A (I2 is flowing from c to a)

Step-3:

Redraw the circuit indicating the direction of currents in different branches. One can find the Thevenin’s voltage VTh using KVL around the closed path ‘gabg’

VTh = Vag − Vbg = 3 − 2 =1volt

Step 4:

Replace all sources by their internal resistances. In this problem, voltage source has an internal resistance zero (0) (ideal voltage source) and it is short-circuited with a wire.

On the other hand, the current source has an infinite internal resistance (ideal current source) and it is open-circuited (just remove the current source).

Thevenin’s resistance RTh of the fixed part of the circuit can be computed by looking at the load terminals ‘a’- ‘b’

RTh = ( R1 + R3 ) & R4 = ( 3 + 4 )&2 = 1.555Ω

Step-5: Place RTh in series with VTh to form the Thevenin’s equivalent circuit. Reconnect the original load resistance RL = R2 = 1 Ω to the Thevenin’s equivalent circuit .

Step-6: The circuit is redrawn to indicate different branch currents. Referring to one can calculate the voltage Vbg and voltage across the current source (Vcg ) using the following equations.

Vbg = Vag − Vab = 3 − 1 × 0.39 =2.61 volt.

Ibg = 2.61 2 = 1.305 A;

Icb = 1.305 − 0.39 = 0.915 A

Vcg = 4 × 0.915 + 2 ×1.305 =6.27 volt.

For the circuit shown in figure, find the current IL through 6 Ω resistor using Thevenin’s theorem.

Step-1:

Disconnect 6 Ω from the terminals ‘a’ and ‘b’ and the corresponding circuit diagram. Consider point ‘g’ as ground potential and other voltages are measured with respect to this point.

Step-2:

Apply any suitable method to find the Thevenin’s voltage (VTh )

KVL is applied around the closed path ‘gcag’ to compute Thevenin’s voltage.

42 – 8I – 4I -30 = 0 ; I = 1A

Vag = 30+4 = 34 volt

Vbg = 2 x 3 = 6 volt

Vth = Vab = Vag – Vbg = 34 -6 = 28 volt

Step-3:

Thevenin’s resistance RTh can be found by replacing all sources by their internal resistances all voltage sources are short-circuited and current sources are just removed or open circuited.

Rth = (8 x4) / 12 + 2 = 14/3 = 4.666Ω

Step-4:

Thevenin’s equivalent circuit is now equivalently represents the original circuit

IL = Vth / Rth + RL = 28 / 4.666 +6 = 2.625 A

Maximum power transfer theorem states that AC voltage source will deliver maximum power to the variable complex load only when the load impedance is equal to the complex conjugate of source impedance.

Proof:

Replace any two terminal linear network or circuit to the left side of variable load resistor having resistance of RL ohms with a Thevenin’s equivalent circuit. We know that Thevenin’s equivalent circuit resembles a practical voltage source.

The amount of power dissipated across the load resistor is

PL = I 2 R L ----------------------(1)

Substitute I = Vth / Rth + RL

PL = Vth / (Rth + RL ) 2 RL

PL = Vth 2 { RL / (RTH + RL) 2 ------------------------------(2)

Condition for Maximum Power Transfer

For maximum or minimum, first derivative will be zero. So, differentiate Equation 1 with respect to RL and make it equal to zero.

dPL / d RL = V Th 2 { (Rth + RL) 2 x 1 -RL x 2(Rth + RL) / (Rth + RL) 4 } =0

(Rth + RL) 2 – 2RL (Rth + RL) = 0

(Rth + RL) (Rth + RL – 2 RL) = 0

Rth - RL = 0

Rth = RL or RL = Rth

Therefore, the condition for maximum power dissipation across the load is RL=RTh. That means, if the value of load resistance is equal to the value of source resistance that is Thevenin’s resistance, then the power dissipated across the load will be of maximum value.

Substitute RL = Rth and PL = PL,max

PL,max = V th 2 { Rth / (Rth + Rth ) 2 }

= V th 2 ( Rth / 4 Rth 2 )

= V th 2 / 4 Rth

PL,max = V th 2 / 4 RL [RL = Rth]

Efficiency of Maximum power transfer

= PL,max/ Ps

PL,max = maximum amount of power transferred to the load

Ps = amount of power generated by the source.

The amount of power generated by the source is

Ps = I2 Rth + I2 RL

Ps = 2 I2 Rth ; since RL = Rth

Substitute I = Vth / 2 Rth

Ps = 2(Vth / 2Rth) 2 Rth

Ps = 2(Vth / 4Rth) 2 Rth

Ps = Vth 2 / 2 Rth

Substituting with PL max and Ps

= (Vth 2 / 4 Rth) / (Vth 2 / 2 Rth)

max = ½

Problem:

Find the maximum power that can be delivered to the load resistor RL of the circuit shown in the following figure.

Step 1 – Using Thevenin’s circuit we get

Thevenin’s voltage Vth = 200/3 and Thevenin’s resistance Rth = 40/3

Step 2 – Replace the left side of terminals A & B of the given circuit with the above Thevenin’s equivalent circuit. The resultant circuit diagram will be

Step 3 − We can find the maximum power that will be delivered to the load resistor, RL by using the following formula.

P L,max = Vth 2 / 4 Rth

Substitute VTh=200/3 and RTh=40/3Ω

P L,max = (200/3) 2 / 4 (40/3)

Therefore, the maximum power that will be delivered to the load resistor RL of the given circuit is 250/3 .

Problem:

For the circuit shown find the value of RL that absorbs maximum power from the circuit and the corresponding power under this condition.

Load resistance RL is disconnected from the terminals ‘a’ and ‘b’ and the corresponding circuit diagram.

The above circuit is equivalently represented by a Thevenin circuit and the corresponding Thevenin voltage VTh and Thevenin resistance RTh are calculated by following the steps given below:

Considering 20V only

From the above circuit the current through ‘b-c’ branch =20/ 20 = 1𝐴 (𝑓𝑟𝑜𝑚 𝑎 𝑡𝑜 𝑏). whereas the voltage across the ‘b-a’ branch vba =1 ×10 =10 volt . Since ’b’ is higher potential than ‘a’ ∴vab = − 10volt.

Considering only 10v source only

No current is flowing through ‘cb’-branch. Vab = 5v (‘a’ is higher potential than ‘b’). Consider only 2 A current source only

To compute Rth

Replace all voltage and current sources by their internal resistance of the circuit.

Rth = Rab = ((5+5)||10) + (10||10) = 5+5 =10Ω

RL = Rth = 10Ω

The maximum power dissipated to RL is

Pmax = Vth 2 / 4 Rth = ¼ 25 /10 = 0.625 watts

References :