Exercise problems: Solving the Electrical Network

EXERCISE PROBLEMS

LOOP CURRENT METHOD

1. In the circuit of figure, find the current I by the mesh method.

[Ans: 2.5 A]

2. Obtain the currents in all the branches of the network in figure by mesh method.

[Ans: I₁ = 0.44 A, I₂ = 0.37 A, I₃ = 1.32 A]

3. In the network shown determine the direction and magnitude of current flow in the milli ammeter A having a resistance of 10Ω.

[Ans: 26.7 mA]

4. A voltage source V₃ causes the mesh current I₁ to be zero. Find the value of V₃.

[Ans: 16.8 <133.2° volts]

5. In a circuit shown below, determine the power in the impedance 2 + j5 Ω cor.nected between A and B. Use loop current method to find the current and verify the answer by simplification of series parallel combination. 

[Ans: 2.69 watts]

6. In the network of the figure, find V2 such that the V2 source current will be zero.

[Ans: V2=4 / 180°V]

Node Voltage Method of Solving the Electrical Network

The network shown above, contains three principle nodes namely 1, 2 and 3. In node voltage method, one of the principle nodes is selected as reference or datum node (or zero potential node). At each of the other principal nodes, a voltage is assigned. This voltage is with reference to the reference node. At each principal node, KCL is applied and the equations are obtained. In this method, the node voltages are independent variables. By finding the values of these node voltages, we can calculate the various branch currents. The above network is re-drawn as shown below:

Node 3 is taken as reference.

V₁ and V₂ are the node voltages at nodes 1 and 2 respectively.

At node 1, applying KCL, we get

and at node 2 application of KCL yields

Rearranging the equations (1) & (2), we get

Putting equations (3) & (4) in the matrix form, we get

1-1 element contains sum of reciprocals of all resistances connected to 1. 2-2 element contains sum of reciprocals of all resistances connected to node 2. 1-2 and 2 - 1 elements are each equal to the negative of the sum of the reciprocals of the resistances of all branches joining nodes 1 and 2. (In the present case there is only one resistance between 1 and 2 i.e., R_C). Voltage matrix contains the independent variables V₁ and V₂.

Current matrix contains V_a/R_Aand V_b/R_E. These are called driving currents. The driving currents are positive if they flow towards the principal node. If the driving current flows away from principal node, it is taken as negative.

The above matrix can be written as below in the general form,

I₁ = algebraic sum of all driving currents meeting at node 1.

1₂ =algebraic sum of all driving currents meeting at node 2.

If the conductance's are given, instead of resistances, the above matrix is written in the following form,

[Note:

1. In applying node voltage method convert the practical voltage sources into equivalent current sources before putting in the matrix form.

2. This method can be applied for both AC and DC circuits.

3. If there are "n" nodes, then the number of equations required is "n - 1" i.e., the number of unknown node voltages is "n - 1"]

To solve DC network by Nodal Method (Nodal Analysis)

[Note: If the circuit consists of practical voltage sources and current sources, apply this method after converting the voltage source into current source i.e., in this method solution is easier or convenient if all the sources are current sources. Then, by inspection, we can obtain the nodal equations in the matrix form and proceed for solving the network. Also the student is to note whether the given passive element is resistance [ohms] or conductance [mhos or siemen]. ]

Example 9 Frame the nodal equations of the network of figure and hence find the difference of potential between nodes 2 and 4.

Solution: Let the node 4 be the reference i.e., zero potential node. Let V₁, V₂ and V₃ be the node voltages at the nodes 1, 2 and 3 respectively. By inspection,

It is required to find the difference of potential between nodes 2 and 4. i.e., V₂-V₄ = V₂-0= V₂. So, it is enough if we know the value of V₂.

Example 10 Compute the voltage at nodes A and B in the circuit of figure.1.92

Solution: First converting the voltage source into its equivalent current source and re-drawing the circuit the following figure is obtained.

There are two nodes A and B other than the reference node. Let V_A and V_B be the voltages of the nodes A and B respectively with respect to reference node. By inspection,

Example 11 Use nodal voltage method and hence find the power dissipated in the 10 ohms resistor on the circuit shown in figure.

Solution: Taking the node 4 as reference, and converting the voltage source into current source, the above network is re-drawn as below:

Therefore, the potential difference across 10 ohms = 'V10

= 24.16 volts

Power dissipated in 10Ω =(24.16)² /10

=58.4 Watts

Example 12 Find the currents I₁, I₂ and I₃ and the voltages V_a and V_b in the network of figure by using nodal analysis.

Solution: Converting the voltage source into current source, the following circuit is obtained.