DC-Lab-Exp-5

Title: Study of Branch current analysis, Mesh analysis and Nodal Analysis

experime nt is to be b e able to demonstrate the Abstract: The purpose of this experiment validity of the branch current curr ent analysis, mesh analysis and nodal n odal analysis technique through experimental measurements and observe the difference between theoretical and simulation results.

Introduction:

The main objective of this experiment was to verify the ohms law. In doing so, followings were performed: a) To design an electrical circuit with relevant parameters and sources. b) To set up the circuit with appropriate connections, sources, and instruments. c) To compare the measured value with with the theoretical estimated value. d) To find the reason for error in result, and to draw conclusion on how to overcome.

Theory and Methodology: The branch, mesh and nodal analysis techniques are used to solve complex c omplex networks, which are a re not in series or in parallel. The branch and mesh analysis techniques will determine the current of the network, while the nodal analysis approach will provide the potential levels of the nodes of the networks with wit h respect to some reference. The application appli cation of each technique follows a sequence of steps, each of which will result in a set of equations for the various various variables, whether they are current or voltage. i)Branch current analysis: The current through each branch of the network is called the branch current. Once the branch current is known, all other quantities, quantit ies, such as voltage or power can be determined. There are four steps to the branch current which are given below.

a) Assign a distinct current of arbitrary direction to each branch of the network. b) Indicate the polarities for each resistor as determined by the assumed current direction. c) Apply Kirchhoff’s voltage law around each closed, independent loop of the network. d) Apply Kirchhoff’s current law at the minimum number of nodes that will include all the branch currents of the network. e) Solve the resulting simultaneous linear equations for assumed branch currents.

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Example for branch current analysis (For Figure-1)

ii) Mesh analysis: The term mesh is derived from the similarities in appearance between the closed loop of a network and a wire mesh fence. A loop current is a branch current only when it is the only loop current assigned to that branch. There are five steps to the mesh current which are given below:

a) Assign a loop current to each independent, closed loop in a clockwise direction. b) The number of required equations is equal to the number of chosen independent, closed loops. Column 1 of each equation is i s formed by summing the resistance values of those resistors through which the loop current of interest passes and multiplying the result by that loop current. c) The mutual terms must be considered which are always subtracted from the first column. A mutual term is simply any resistive element having an additional loop current passing through it. It is possible to have more than one mutual term if the loop current of interest has an element in common with more than one other loop current. Each term is the product of the mutual resistor and the other loop current passing through the same element. d) The column to the right of the equality sign is the algebraic sum of the voltage sources through which the loop current of interest passes. Positive signs are assigned to those sources of voltage having a polarity such that the loop current passes from the negative to the positive terminal. A negative sign is assigned to those potentials for which the reverse is true.







Example for mesh analysis (For Figure-2)

iii) Nodal analysis: A node is a junction of two or more branches, where a branch is any combination of series elements. For a network of N nodes, therefore there will will existed (N-1) nodes with a fixed potential relative to the assigned reference node. There are five steps to the node voltages which are given below (format approach).

a) Choose a reference node node and assign a subscripted voltage label to the (N–1) remaining nodes of the network. b) The number of equations required for a complete solution is equal to the number of subscripted voltages (N–1). Column 1 of each equation is formed by summing the conductance tied to the node of interest and multiplying the result by that subscripted nodal voltage. c) The mutual terms must be considered which are always subtracted from the first column. It is possible to have more than one mutual term if the nodal voltage of current interest has an element in common with more than one other nodal voltage. Each mutual term is the product of the mutual conductance and the other nodal voltage tied to that conductance. d) The column to the right of the equality sign is the algebraic sum of of the current sources tied to the node of interest. A current source is assigned a positive sign if it supplies current to a node and a negative sign if it draws current from the node. e) Solve the resulting simultaneous equations for the desired voltages.







Example for nodal analysis (For Figure-3)

Apparatus:

1. 2. 3. 4. 5. 6. 7.

Trainer Board Voltmeter Ammeter AVO meter or Multimeter DC source Resistors Connecting Wires

Precautions:

1. All the apparatus were were checked. 2. Before connecting DC source in the trainer board that was checked. 3. While measuring current Digital multimeter was placed in series with the branch of the circuit where the current was to be measured, multimeter was in ammeter mode. 4. The DC source was not switched on while implementing the circuit in the trainer board 5. Voltmeter was connected in the parallel through the resistor. Ammeter was connected in the series through the resistor.







Circuit Diagram: XMM1 XMM2

R1

R2

1.527k

2.24k

V1 10 V

V2 5V

R3 4.78k XMM3

Figure:4

R1 1.527k

R3 4.78k

R2 2.24k

XMM1 XMM2

V2 10 V

V1 20 V

Figure:5

XMM2

R1

R6

R2

1.527k

2.24k

4.78k XMM1

R5 2.24k

R4 4.78k

V1 10 V

R3 V2

1.527k







XMM2

XMM1

R3 2.24k I1 6.55mA

R1 1.527k

R2 2.24k

R4 4.78k

I2 1.064mA

Figure:6 (converted) Experimental Procedure:

A – Branch-current Branch-current analysis: analysis: 1. The circuit was connected as shown in the figure 4. 3 branch-current equations was written to describe that circuit. Then the current was calculated through each branch of the network from those equations. Then the direction of the currents was indicate and current I 1,I2,I3 was measured. B – Mesh analysis 2. The circuit was connected as shown in the figure 5. 2 mesh equations was written to describe that circuit. Then the current was calculate through each loop of the network from these equations. Then the direction of the currents was indicate and current I 1,I2 was measured. C – Nodal analysis 3. The circuit was connected as shown in the figure 6. 3 nodal equations was written to describe that circuit. Then the nodal voltages was calculated from these equations and nodal voltages V1,V2 was measured.

R5 6.307k









Simulation:

Figure:4







Figure:6







Measurement: Data Table:

Table-1 (For Figure-4) Value of Resistors: R1=1.527kΩ, R2=2.245kΩ, R3=4.78kΩ. Value of Voltage Sources: E1=10V, E2=5V.

Table-2 (For Figure-5) Value of Resistors: R1=1.527kΩ, R2=2.24kΩ, R3=4.78kΩ. Value of Voltage Sources: E1=20V, E2=10V.







Table-3 (For Figure-6) Value of Resistors: R1=1.527kΩ, R2=4.78kΩ, R3=1.527kΩ, R4=4.78kΩ, R5=2.24kΩ, R6=2.24kΩ. Value of Voltage Sources: E1=10V, E2=5V.

Calculations:

R1=1.527k







-10 + I1R1 + I3R3 =0 =>I1R1 + I3R3 = 10 =>1.527I1 + 4.78I3 = 10 =>1.527I1 + 4.78( I1 + I2 ) =10

∴

=>6.31I1 + 4.78I2 = 10 [ putting value ofI3]………………(ii)

I3R3 + I2R2 -5 = 0 =>4.78I3 + 2.24I2 =5 =>4.78( I1 + I2 ) + 2.24I2 = 5 =>4.78I1 +7.02I2 = 5

∴

[ putting value of I3]………………(iii)

Using the Cramer's rule, we get,

64..3718 47..7082

D=

= {( 6.31 × 7.02 ) – ( 4.78 × 478 )}









 6.31 10  4.78 5 ( . )  ( × . )] [ .   × = . 

I2 =

= 0.724 mA

∴I3 = I1 +I2 = 2.159 +(-0.724) = 1.435 mA

R1=1.527k R2=2.24k R3=4.78k E1=20V E2=10V Now,







 10 −4.78  10 7.02 [10 ×7.02 −−4.78 ×10 ] = 21.43

I1=

= 5.50 mA

  6.307 10  −4.78 10

I2=

(.  × ){ ×(. )}] [ ×   = . =5.173 mA

R1=1.527K







V2/R23 + V2/R4 + (V2-V1)/R6 = 1.064 =>V2/6.307 + V2/4.78 + (V2-V1)/2.24 = 1.064 =>-30.15V1 + 54.99V2 = 71.85……………(ii)

Using the Cramer's rule, we get,

D=

−53.209.145 −514..59297 = (5.294 × 54.99) – {(-30.15) × (-1.527)} = 291.12 – 46.04 = 245.08







Results: For figure 4:

I1 = 2.159mA, I2 = 0.724mA I3 = 1.435mA

For figure 5:

I1 = 5.50mA I2 = 5.173mA

For figure 6:

V1 = 5.47V V2 = 4.31V

Discussion: 1. The trainer board and the multimeter was checked before the start of the experiment.

DC-Lab-Exp-5

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