Find I_{1} in the following circuit. V_{1} = V_{2} = 1 V:

Option 4 : 0.2 A

**Concept:**

Nodal Analysis:

Nodal analysis is a method of analyzing networks with the help of KCL equations.

For a network of N nodes, the number of simultaneous equations to be solved to get the unknowns

= Number of KCL equations

= N - 1

= one less than the number of nodes

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:**

**Given:**

V_{1 }= V_{2} = 1 V

Applying Nodal analysis at point A:

\(\frac{V_1}{5} \ + \ \frac{V_1 \ - \ V_2}{10} \ = \ I_1\)

As V_{1} = V_{2} = 1 V

\(I_1=\frac{V_1}{5} \ + 0\)

\(I_1=\frac{1}{5}=0.2 \ A\)

Hence **option (4) is the correct answer.**

Find I_{b}?

Option 1 : -3 mA

__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,

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

What will be the values of Vs and is respectively in the circuit shown below?

Option 1 : 14 V, -1 A

**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,

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

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

\(2 = {i_s} + 3\)

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

In nodal analysis, the preferred reference node is a node that is connected to

1. ground

2. many parts of the network

3. the highest voltage source

Which of the above is/are correct?Option 2 : 2 only

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:**

Nodal method of circuit analysis is based on

a) KVL and Ohm’s law

b) KVL, KCL and Ohm’s law

c) KCL and KVL

d) KCL and Ohm’s law

Which of the following options is correct?

Option 4 : d) is correct but b) is wrong

Nodal Analysis is based on:

1. KCL

2. Ohm's law

**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

**Ohm's law:**

At constant temperature, the current flowing through a resistance is directly proportional to the potential difference across its end.

V = R I

Where V is the potential difference, R is resistance and I is current flowing.

What will be the Potential at node P?

Option 1 : 21.44 V

__Concept:__

Nodal Analysis:

Nodal analysis is a method of analyzing networks with the help of KCL equations.

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:__

**Applying Nodal analysis at point P:**

**Let V _{1} be the voltage across P**

\(\frac{V_1}{68} \ + \ \frac{V_1 \ - \ 19}{64} \ + \ \frac{V_1 \ - \ 48}{64}=0\)

64V_{1} = -136V_{1} + 4556

\(V_1=\frac{4556}{200}=22.7 V\)

**V _{1} ≈ 21.44 V**

Hence **option (1)** is the correct answer.

Find the value of current i_{1} in the following circuit.

Option 2 : \(\frac{{ - 4}}{3}A\)

By applying nodal analysis at point A,

\( - 2 + \frac{v}{3} - 4{i_1} + \frac{v}{2} = 0\)

And v = 3i_{1}

⇒ -2 + i_{1} – 4i_{1} + 1.5i_{1} = 0

⇒ -1.5i_{1} = 2

\( \Rightarrow {i_1} = - \frac{4}{3}A\)

Option 1 : voltage source

__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, V_{1} and V_{2} 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.

In the circuit shown in the figure, the magnitude of the current (in amperes) through R2 is ________.

**Concept:**

Nodal Analysis is based on:

1. KCL

2. Ohm's law

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

**Calculation:**

Apply Nodal Analysis at Node A:

\(\frac{{{V_A} - 60}}{5} - .04\;{V_x} + \frac{{{V_A}}}{8} = 0\)

\(\frac{{{V_A}}}{5} + \frac{{{V_A}}}{8} - .04\;{V_x} = 12\) ---(1)

Now,

Apply voltage divider across Resistor R_{3},

\({V_x} = \frac{{5{V_A}}}{8}\) ---(2)

Put this value in equation (1)

\(\frac{{{V_A}}}{5} + \frac{{{V_A}}}{8} - \frac{{.2\;{V_A}}}{8} = 12\)

On solving we’ll get,

V_{A} = 40 V & V_{x} = 25 V

Current through R_{2} will be:

\({I_{{R_2}}} = \frac{{{V_A} - {V_x}}}{3}\)

\({I_{{R_2}}} = 5A\)

Option 4 : One less than the number of nodes

Nodal Analysis:

Nodal analysis is a method of analyzing network with the help of KCL equations.

For a network of N nodes, the number of simultaneous equations to be solved to get the unknowns

= Number of KCL equations

= N - 1

= one less than the number of nodes

Mesh Analysis:

Mesh analysis is a method of analyzing network with the help of KVL equations.

For a network having N nodes and B branches, the number of simultaneous equations to be solved to get the unknowns

= Number of KVL equations

= number of independent loop equations

= B - N + 1In the circuit shown below, what will be the voltage at the dependent source (current source)?

Option 2 : 2 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:**

Let Node Voltage across current dependent current source be V.

By applying KCL at node having V Volt,

\(10=i_x+2i_x+\dfrac{V}{1}+\dfrac{V\ +\ 8}{2}\)

\(10=3i_x+\dfrac{3V\ +\ 8}{2}\) .......(1)

V = 2i_{x} .......(2)

Substitute (2) in (1)

\(10=3i_x+\dfrac{3(2i_x)\ +\ 8}{2}\)

12i_{x} + 8 = 20

i_{x} = 1 A

V = 2i_{x }= **2 V**

What will be the value of V_{a} in the given circuit?

Option 2 : 142.8 V

__Concept:__

Nodal Analysis:

Nodal analysis is a method of analyzing networks with the help of KCL equations.

For a network of N nodes, the number of simultaneous equations to be solved to get the unknowns

= Number of KCL equations

= N - 1

= one less than the number of nodes

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:__

Applying Nodal analysis across 10 Ω resistor:

As Va is the voltage across 10 Ω resistor

\(\frac{V_a}{10} \ + \ \frac{V_a }{25} =20\)

35V_{a} = 20 × 250

\(V_a=\frac{1000}{7}\)

**Va = 142.8 V**

Hence option (2) is the correct answer.

In the circuit shown below, what is the dissipated power in the 4 Ω resistor?

Option 1 : 4 W

__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:**

Let the node voltages at X be V_{x} and** **reference voltage be V_{y} i.e V_{y} = 0 V

Applying KCL at node X,

\(2=\dfrac{V_x\ -\ 12}{12}+\dfrac{V_x\ +\ 6}{6}+\dfrac{V_x}{4}\)

\(2=\dfrac{(V_x-12)\ +\ 2\ (V_x\ +\ 6)\ +\ 3\ V_x}{12}\)

V_{x} - 12 + 2V_{x} + 12 + 3V_{x }= 24

6V_{x }= 24

V_{x }= 4 V

Power dessipation across 4 Ω = \(\dfrac{V_s^2}{R}= \dfrac{4^2}{4}\)

=** 4 W**

Obtain v_{0} for the given network in terms of v_{1}, i_{2} & v_{3}.

Option 3 : v_{0} = 0.2 v_{1} + 8 i_{2} - 0.2 v_{3}

__Concept:__

Nodal Analysis:

Nodal analysis is a method of analyzing networks with the help of KCL equations.

For a network of N nodes, the number of simultaneous equations to be solved to get the unknowns

= Number of KCL equations

= N - 1

= one less than the number of nodes

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:__

Let the voltage across i_{2} is v_{2}.

By applying the nodal analysis, we get

⇒ v_{2} – v_{1} - 40i_{2} + 4v_{2} – 4v_{3} = 0

⇒ 5v_{2} = v_{1} + 4v_{3} + 40i_{2}

⇒ v_{2} = 0.2v_{1} + 0.8v_{3} + 8i_{2}

By applying KVL at the output,

-v_{2} + v_{3} + v_{0} = 0

⇒ v_{0} = v_{2} – v_{3}

⇒ v_{0} = 0.2 v_{1} + 0.8v_{3} + 8i_{2} – v_{3}

What is the value of the current flowing in 3 Ω resistor?

Option 4 : 4 A

By re-drawing the dc circuit, we get

Let voltage across node 'A' is V_{A}

Now applying nodal analysis at node 'A', we get

\(\frac{{{V_A} - 20}}{2} + \frac{{{V_A}}}{5+3} - 10 = 0\;\)

4V_{A} - 80 + V_{A} - 80 = 0

5V_{A} = 160

V_{A} = 32 volts

Now current flowing through 3Ω resistor can be calculated as

\(I = \frac{{{V_A}}}{8} = \frac{{32}}{8}\)

**I = 4 A**

The value of current in 80 Ω resistor of above circuit is

Option 1 : 0.5 A

\(\frac{{50 - {V_A}}}{4} = \frac{{{V_A}}}{{80}} + \frac{{{V_A}}}{{20}}\)

\(\frac{{50 - {V_A}}}{4} = \frac{{{V_A} + 4{V_A}}}{{80}}\)

(50 – V_{A})20 = 5 V_{A}

1000 = 25 V_{A}

\({V_A} = \frac{{1000}}{{25}}\)

= 40 V

\({I_1} = \frac{{40}}{{80}}\)

= 0.5 A

Determine the voltmeter reading in the given circuit.

Option 2 : 7.5 V

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 i_{1} = 0

4V_{a} + 15 = 80 i_{1} -----(1)

From the circuit we can say,

V_{a }/ 20 = - i_{1}

V_{a} = - 20 i_{1}

∴ equation(1) reduces as

4 × (- 20 i_{1}) + 15 = 80 i_{1}

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

Let, the voltage reads by the voltmeter = V_{0}

It is the voltage value of the dependent source.

V0 = 80 i_{1}

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

V0** = 7.5 V**

In the circuit given below, all resistor values are in ohms. Find the voltage across 10 ohms resistor.

Option 1 : 1 V

**Concept:**

__Nodal Analysis:__

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

What will be the voltage of point A with respect to B?

Option 4 : -10 V

Kirchhoff’s Current Law (KCL):

It states that the amount of current flowing into a node or junction is equal to the sum of the currents flowing out of it

Nodal Analysis:

- 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:**

Let node voltage at point A be V_{A} and reference voltage be at node B i.e V_{B }= 0 V.

By applying KCL,

\(\dfrac{V_A\ -\ 15}{5}+\dfrac{V_A}{10}+6=0\)

\(\dfrac{2(V_A\ -\ 15)\ +\ V_A\ +\ 60}{10}=0\)

2V_{A} - 30 + V_{A} + 60 = 0

3V_{A} = - 30

V_{A} = - 10 V

The voltage of point A with respect to B i.e V_{A} - V_{B}

= (-10 - 0) V

= **- 10 V**

In the figure shown, the value of the current I (in Amperes) is __________.

__ Concept__:

Kirchhoff’s Current Law: The total current entering a junction or a node is equal to the charge leaving the node as no charge is lost.

- The algebraic sum of every current entering and leaving the node has to be null. This property of Kirchhoff law is commonly known as Conservation of charge.
- Node: Node is a point in a network where two or more circuit elements are connected.

I1 + i2 + i3 + i4 + i5 = 0

__ Calculation__:

The given circuit is redrawn as:

Applying KCL at node V, we can write:

\(\frac{{V - 5}}{5} - 1 + \frac{V}{{5 + 10}} = 0\)

\(\frac{{V - 5}}{5} + \frac{V}{{15}} = 1\)

3(V - 5) + V = 15 × 1

3V – 15 + V = 15

4V = 30

V = 7.5 V

Now, the required current I will be:

\(I = \frac{V}{{15}} = \frac{{7.5}}{{15}}\)

\(I = \frac{{75}}{{150}} = \frac{1}{2}\)

I = 0.5 A