Nodal Analysis MCQ Quiz - Objective Question with Answer for Nodal Analysis - Download Free PDF

The correct answer is (n - 1).

Key Points

• In a frequency domain network with 'n - principal' nodes, one node is designated as the reference node or ground.
• The reference node is crucial for measuring the voltage at other nodes with respect to it.
• To solve for the desired results, we require (n - 1) node voltage equations.
• This is because we have 'n' nodes in total, and one of them is used as the reference, leaving us with (n - 1) nodes for which we need to find the voltages.
• In network analysis, the node voltage method is a fundamental technique where the number of node voltage equations is always (n - 1) for a network with 'n' nodes.
• The node voltage method simplifies the analysis of complex circuits by reducing the number of equations needed to solve for unknown voltages.
• It is important to correctly designate the reference node to ensure accurate and consistent results in the analysis.

 Additional Information

• Reference Node
  • The reference node, also known as the ground node, is a point in the circuit that is assumed to have a voltage of zero.
  • It is used as a common return path for current and a reference point for measuring other voltages in the circuit.
  • Choosing the correct reference node simplifies the analysis and calculation of node voltages.
• Node Voltage Method
  • The node voltage method involves writing Kirchhoff's Current Law (KCL) for each node except the reference node.
  • This method reduces the complexity of solving circuit equations by focusing on node voltages rather than branch currents.
  • It is particularly useful for analyzing circuits with multiple nodes and components.
• Kirchhoff's Current Law (KCL)
  • KCL states that the sum of currents entering a node is equal to the sum of currents leaving the node.
  • This law is fundamental in network analysis and is used to derive the node voltage equations.
  • By applying KCL at each node (except the reference node), we can form a system of linear equations to solve for the node voltages.

The correct answer is option"3".

Solution;-

Applying nodal analysis at node \(V_A\) -

\(\frac{V_A-12}{49}+\frac{V_A}{24}+\frac{V_A-6}{80}=0\) ......(i)

Solving eqn (i) We get,

\(V_A\) = 4.29 volts

But in the given options only the value near to 4.29 is 4.25 volts.

Calculation:

Let the voltage across the 12 ohm resistance be "V" Volts.

Let the current flowing through resistance "R" be "I"

Applying the KCL across the node voltage "V" we get,

2 = 1 + I

I = 1 A.

The voltage "V" can be calculated as 

\(\frac{V}{12} = 1 \)

V = 12 volts.

The current flowing through "R" can be calculated as

\(\frac{V -6}{R} = 1\)

putting the value of "V" we get value of "R" as

R = 6 Ω

The correct answer is option 2):(voltage source)

Concept:

• A supernode is basically a theoretical model in which we consider two non-reference nodes having an ideal voltage source present between them (N1 and N2 in our case), as a single node.
• In simple terms, we combine those two non-reference nodes to make a single node called a “supernode”

• In the above circuit diagram, V1 and V2 forms a supernode
•  

Additional Information

• A dependent source is a current or voltage source whose value is not fixed (i.e., independent) but rather which depends on some other circuit current or voltage.
• An independent current source is an energy source that pushes a constant flow of electrons through an electrical circuit regardless of the load presented to it
• A resistor is a passive two-terminal electrical component that implements electrical resistance as a circuit element.​

The correct answer is option 3) -12 V

Concept:

The nodal analysis:

• Nodal Voltage Analysis uses the “Nodal” equations of Kirchhoff’s first law to find the voltage potentials around the circuit.
• So by adding together all these nodal voltages the net result will be equal to zero.
• If there are “n” nodes in the circuit there will be “n-1” independent nodal equations and these alone are sufficient to describe and hence solve the circuit.
• At each node point write down Kirchhoff’s first law equation(KCL), that is: “the currents entering a node are exactly equal in value to the currents leaving the node” then express each current in terms of the voltage across the branch using Ohm's law.
• So the nodal analysis is primarily based on the application of KCL and Ohm's law.
• For “n” nodes, one node will be used as the reference node and all the other voltages will be referenced or measured with respect to this common node

Calculation:

The given circuit is 

The node voltage at b 12-V voltage source. 

v_{b ​} =−12 V 

In the nodal analysis, the preferred reference node is

• A node with the largest number of elements connected to it.
• A node which is connected to the maximum number of voltage sources, or
• A node of symmetry

Note:

Concept:

There are two types of Kirchoff’s Laws:

Kirchoff’s first law (KCL):

• This law is also known as junction rule or current law (KCL). According to it the algebraic sum of currents meeting at a junction is zero i.e. Σ i = 0.

​

• In a circuit, at any junction, the sum of the currents entering the junction must be equal to the sum of the currents leaving the junction i.e., i1 + i3 = i2 + i4
• This law is simply a statement of “conservation of charge” as if current reaching a junction is not equal to the current leaving the junction, the charge will not be conserved.

Kirchoff’s second law (KVL):

• This law is also known as loop rule or voltage law (KVL) and according to it “the algebraic sum of the changes in potential in the complete traversal of a mesh (closed-loop) is zero”, i.e. Σ V = 0.
• This law represents “conservation of energy” as if the sum of potential changes around a closed loop is not zero, unlimited energy could be gained by repeatedly carrying a charge around a loop.
• If there are n meshes in a circuit, the number of independent equations in accordance with the loop rule will be (n - 1).

Calculation:

Apply KVL to the outer loop,

- V_S + 2× 2 + 10 = 0 

V_S = 14 V

Apply KCL at node,

\(2 = {i_s} + \frac{{10 - 5}}{5} + \frac{{10}}{5}\)

\(2 = {i_s} + \frac{{15}}{5}\)

\(2 = {i_s} + 3\)

\({i_s} = - 1\;A\)

Super node:

If a voltage source is connected between two non-reference nodes, then we combine the two nodes as to yield super node.

Example:

In the above circuit diagram, V₁ and V₂ forms supernode.

Note: Always apply mesh analysis of the node

Super mesh:

If a current source is present at the common boundary of two meshes, then we create a super mesh by avoiding the current source and any element connected to it in series.

Example:

In the above circuit diagram, there is a current source in between two mesh and hence it forms a super mesh.

Note: Always apply nodal analysis at super mesh.

Consider node ‘a’ as mentioned above,

Assume voltage at node ‘a’ as V_a,

Apply KCL at node ‘a’, (assuming all currents are leaving the node)

\(\frac{{{V_a}}}{{20}} + 0.25 + \frac{{\left( {{V_a} - 80{i_1}} \right)}}{{60}} = 0\)

\(\frac{{3{\rm{Va\;}} + {\rm{\;}}15{\rm{\;}} + {\rm{\;}}{\rm{Va\;}}-{\rm{\;}}80{{\rm{i}}_1}}}{{60}}{\rm{ = 0}}\)

3V_a + 15 + V_a – 80 i₁ = 0

4V_a + 15 = 80 i₁   -----(1)

From the circuit we can say,

V_a / 20 = - i₁

V_a = - 20 i₁

∴ equation(1) reduces as

4 × (- 20 i₁) + 15 = 80 i₁

\({I_{1\;}} = \frac{{15}}{{160{\rm{ }}}} A\)

Let, the voltage reads by the voltmeter = V₀

It is the voltage value of the dependent source.

V0  = 80 i₁

\({{\rm{V}}_0}{\rm{\;}} = {\rm{\;}}80{\rm{\;}} \times \frac{{15}}{{160}}\)

V0  = 7.5 V

Concept:

Nodal Analysis:

• Nodal Voltage Analysis uses the “Nodal” equations of Kirchhoff’s first law to find the voltage potentials around the circuit. So by adding together all these nodal voltages the net result will be equal to zero.
• If there are “n” nodes in the circuit there will be “n-1” independent nodal equations and these alone are sufficient to describe and hence solve the circuit.
• At each node point write down Kirchhoff’s first law equation(KCL), that is: “the currents entering a node are exactly equal in value to the currents leaving the node” then express each current in terms of the voltage across the branch using Ohm's law.
• So the nodal analysis is primarily based on the application of KCL and Ohm's law.
• For “n” nodes, one node will be used as the reference node and all the other voltages will be referenced or measured with respect to this common node

Calculation:

From the above circuit,

V = 20 - 2 I        ------- (i) 

Apply KCL at node,

I + 2 = (20 - 2 I - 5I) / 3

3I + 6 = 20 - 7 I

⇒ 10 I = 14

⇒ I = 1.4 A

Concept:

Tellegen’s theorem:

• According to Tellegen’s theorem, the summation of instantaneous powers for the n number of branches in an electrical network is zero.
• Let n number of branches in an electrical network have I1, I2, I3, ….. In respective instantaneous currents through them.
• These branches have instantaneous voltages across them are V1, V2, V3, ….. Vn respectively.
• According to Tellegen’s theorem, \(\mathop \sum \limits_{k = 1}^n {V_k}.{I_k} = 0\)
• ∑ power delivered = ∑ power absorbed
• It is based on the conservation of energy.
• It is applicable to both linear and non-linear circuits.

Calculation:

Apply KCL at node 1

\(I_1=\frac{V_1}{1}+\frac{V_1-V_2}{0.5}\)

\(\frac{15-V_1}{1}=\frac{V_1}{1}+\frac{V_1-V_2}{0.5}\)

7.5 - 0.5 V₁ = 0.5 V₁ + V₁ - V₂

7.5 = 2 V₁ - V₂   -------  (1)

Apply KCL at node 2

\(I_2=\frac{V_2}{2}+\frac{V_2-V_1}{0.5}\)

\(\frac{20-V_2}{1}=\frac{V_2}{2}+\frac{V_2-V_1}{0.5}\)

20 = 3.5 V₂ - 2 V₁ --------  (2)

By solving 1 and 2 we get

V₂ = 11 V and V₁ = 9.25 V

I₁ = 5.75 A and I₂ = 9 A

Power delivered by 15 V source = 15 × 5.75 = 86.25 W

Power delivered by 20 V source = 20 × 9 = 180 W

By Tellegon's theorem 

Total power consumed = total power delivered = 180 + 86.25 = 266.25 W

The given circuit can easily be solved by using Nodal Analysis:

Procedure:

1. Identified the number of nodes in the circuit along with the reference node.

2. Identified the direction of current (incoming or outgoing) to each node except the reference node.

3. Apply KCL to each node except the reference node.

4. Solved the KCL equation.

Application:

The given circuit can be drawn as,

Apply KCL at node A,

I₁ = I₂ + 2 .... (1)

Let the voltage at node A = V

Hence,

\(I_1=\frac{100-V}{10}\)

\(I_2=\frac{V}{10}\)

\(2=\frac{V}{R}\)

\(\rightarrow R=\frac{V}{2}\) .... (2)

From equation (1),

\(\frac{100-V}{10}=\frac{V}{10}+2\)

or, \(10-2=\frac{2V}{10}\)

or, 2V = 80

or, V = 40 volts

From equation (2),

R = 40/2 = 20 Ω

Nodal Analysis:

• Nodal Voltage Analysis uses the “Nodal” equations of Kirchhoff’s first law to find the voltage potentials around the circuit. So by adding together all these nodal voltages the net result will be equal to zero.
• If there are “n” nodes in the circuit there will be “n-1” independent nodal equations and these alone are sufficient to describe and hence solve the circuit.
• At each node point write down Kirchhoff’s first law equation(KCL), that is: “the currents entering a node are exactly equal in value to the currents leaving the node” then express each current in terms of the voltage across the branch using Ohm's law.
• So the nodal analysis is primarily based on the application of KCL and Ohm's law.
• For “n” nodes, one node will be used as the reference node and all the other voltages will be referenced or measured with respect to this common node.

Let a network and find voltage V using nodal analysis

Applying nodal analysis at node V

\(\frac{{V - {V_1}}}{{{R_1}}} + \frac{{V - {V_2}}}{{{R_2}}} + \frac{{V - {V_3}}}{{{R_3}}} = 0\)

Explanation:

Apply Nodal At V_a;

\(\frac{{{V_a} - 3}}{1} + \frac{{{V_a}}}{1} + \frac{{{V_a} - 2\;{V_a}}}{1} = 0\)

V_a + V_a + V_a – 2V_a = 3

V_a = 3V

Now,

\({i_b} = \frac{{{V_a} - 2{V_a}}}{{1k}} = \frac{{3 - 6}}{1} = - 3\;mA\)

• Loop: A closed path in a circuit where more than two meshes can occur is known as Loop i.e. there may be many meshes in a loop, but a mesh does not contain on one loop.
   
• Mesh:
  • The number of independent loops for a network with n nodes and b branches is = b - n + 1
  • A mesh is a closed path in a circuit with no other paths inside it. In other words, a loop with no other loops inside it.
     
• Node: A point or junction where two or more circuit elements (resistor, capacitor, inductor, etc.) meet is called Node. In other words, a point of connection between two or more branches is known as a Node.
   
• Branch: That part or section of a circuit located between two junctions is called the branch. In a branch, one or more elements can be connected, and they have two terminals.

Application:

The given circuit diagram represent the node and branch in circuit,

Total Branch = 1, 2, 3, 4 = 4 nos

Total Node = A, B, C = 3 nos

Hence,

Sum of branches and nodes = 7

Concept:

Nodal Analysis:

• Nodal Voltage Analysis uses the “Nodal” equations of Kirchhoff’s first law to find the voltage potentials around the circuit. So by adding together all these nodal voltages the net result will be equal to zero.

Let a network and find voltage V using nodal analysis

Applying nodal analysis at node V

\(\frac{{V - {V_1}}}{{{R_1}}} + \frac{{V - {V_2}}}{{{R_2}}} + \frac{{V - {V_3}}}{{{R_3}}} = 0\)

By solving it we can find voltage V

Calculation:

Apply Nodal at node A, we get:

\(\frac{{{V_A}}}{{40}} - 0.25 + \frac{{{V_A} - 5}}{{10}} = 0\)

\(\frac{{{V_A}}}{{40}} + \frac{{{V_A}}}{{10}} = 0.75\)

5V_A = 0.75 × 40

V_A = 6V

The voltage across 10 Ω is,

V_{10 Ω} = V_A – V_B = 6 – 5 = 1V