Nodal Analysis MCQ Quiz in తెలుగు - Objective Question with Answer for Nodal Analysis - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

We observe that the polarity of the output voltage V₀ is negative at the top and positive at the bottom.

Let the voltage across each parallel branch is V, i.e.

By applying nodal analysis, we can write:

\(\frac{{V - {V_1}}}{R} + \frac{{V - {V_2}}}{{2R}} + \frac{{V - {V_3}}}{{3R}} = 0\)

6(V – V₁) + 3(V – V₂) + 2(V – V₃) = 0

11V = 6V₁ + 3V₂ + 2V₃

\(V = \frac{1}{{11}}\left[ {6{V_1} + 3{V_2} + 2V\_3} \right]\)

\(V = \frac{1}{{11}}\left[ {6\left( {100} \right) + 3\left( {80} \right) + 2\left( { - 60} \right)} \right]\)

\(V = 65.45\;V\)

By applying the nodal analysis, we get

\(\frac{V}{{10}} + \frac{{V - 200}}{{10}} + \frac{{V - 200}}{5} = 0\)

⇒ 4 V – 200 – 400 = 0

\(V = \frac{{600}}{4} = 150\;V\)

\(I = \frac{{150 - 200}}{5} = - \frac{{50}}{5} = - 10A\)

But the direction is opposite so hence it will be 10 A.

Alternative method:

We can solve this problem by applying the superposition theorem.

Case I:

By source transformation, we get

∴ I’ = -10 A

Case II: 

10 I’’ – 200 = 0

I’’ = 20

Hence, I = I’ + I’

= -10 + 20 = 10 A

Concept:

1) Node:

In an electrical network, junction of branches at a common point is called Node.

2) Junction: 

A junction is a point where at least three circuit paths meet.

Calculation:

In the given network, nodes are: A, B, C, and D

total number of nodes = 4

Junctions are: B and D

Hence total number of junctions = 2

When two current sources are in parallel with different magnitudes can be represented as a single current source by an algebraic sum of both the current sources

The equivalent circuit diagram will be:

To make the current from the 10 V voltage source zero:

Apply nodal at node N

i + 5 = i₁ 

If the current from the voltage source is zero then-current i₁ = 5 A 

The voltage across the parallel branch will be constant.

Thus R = V / I

R = 10 / 5 = 2 Ω 

Therefore, the resistance will be 2 Ω to make the current from the voltage source equals to zero.

Apply KVL in the Loop;

10 - 3I + 2V_b - 2I - 4I = 0  ----(1)

From the circuit,

V_b = 2I

\(I=\frac{V_b}{2}\)  ---(2)

Put value of I in equation (1),

\(10-3\frac{V_b}{2}+2V_b-V_b-2V_b=0\)

\(10-5\frac{V_b}{2}=0\)

\(10=5\frac{V_b}{2}\)

V_b = 4V

Concept of KCL:

According to Kirchhoff’s current law (KCL), the algebraic sum of the electric currents meeting at a common point is zero. I.e. the sum of currents entering a node is

equal to the sum of currents leaving the node. It is based on the conservation of charge.

By applying KCL,

i₁ + i₅ = i₂ + i₃ + i₄

Application:

Let node Voltage across 12 Ω  be V

By applying KCL

I₁ = I₂ + I₃ + I₄

\(\frac{60-V}{7}=\frac{V}{12}+\frac{V}{6}+\frac{V}{12}\)

\(\frac{60-V}{7}=\frac{V+2V+V}{12}\)

\(\frac{60-V}{7}=\frac{4V}{12}=\frac{V}{3}\)

3 × (60 - V) = 7V

180 - 3V = 7V

10V = 180

V = 18 Volts

I₁ = \(\frac{60-V}{7}=\frac{60-18}{7}\) = 6 A

I₂ = \(\frac{V}{12}=\frac{18}{12}\) = 1.5 A

I₃ = \(\frac{V}{6}=\frac{18}{6}\) = 3 A

I₄ = \(\frac{V}{12}=\frac{18}{12}\) = 1.5 A

Calculation:

Given, V_a = 10V

Applying KCL at node 'A' 

\( {10-11 \over 5}+ {10 \over R}+ {10 \over 100}=0\)

\( {-1 \over 5}+ {10 \over R}+ {1 \over 10}=0\)

\( {10 \over R}=0.2-0.1\)

R = 100 Ω 

Concept:

Nodal analysis:

Nodal analysis or branch current method is used to determine the voltage (potential difference) between the nodes.

Nodal analysis is similar to the KCL method.

Calculation:

Given circuit is

By applying nodal analysis at node V_x

\( - 1 + \frac{{{V_x}}}{2} + \frac{{{V_x} - 2{V_x}}}{1} = 0\)

⇒ - 2 + V_x - 2 V_x = 0

⇒ V_x = - 2 V

Therefore the value of V_x in the given circuit diagram is - 2 volt

V_ab = V_a – V_b

For R_x = ∞ (Open-circuit), the circuit is redrawn as:

Using voltage division rule, we can write:

\({V_a} = 15 \times \frac{{360}}{{360 + 90}}\;V\)

V_a = 12 V

Applying KVL from node ‘b’ to 15 V ground, we get:

V_b – 15 = 0 (∵ I = 0)

V_b = 15 V

Now, V_ab = V_a - V_b

V_ab = 12 – 15 V

V_ab = -3 V

Writing node equations at node A and B

\(\frac{{{V_a} - 5}}{1} + \frac{{{V_a} - 0}}{1} = 0\)

2V_a – 5 = 0 ⇒ V_a = 2.5 V

\(\frac{{{V_b} - 4{V_{ab}}}}{3} + \frac{{{V_b} - 0}}{1} = 0\)

\({V_{ab}} = {V_a} - {V_b} = 2.5 - {V_b}\)

\( \Rightarrow \frac{{{V_b} - 4\left( {2.5 - {V_b}} \right)}}{3} + {V_b} = 0\)

\( \Rightarrow {V_b} - 10 + 4{V_b} + 3{V_b} = 0\)

\( \Rightarrow 8{V_b} = 10 \Rightarrow {V_b} = 1.25\;V\)

\(i = \frac{{{V_b}}}{1} = 1.25\ A\)