Norton's Theorem MCQ Quiz in मल्याळम - Objective Question with Answer for Norton's Theorem - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Last updated on Mar 12, 2025

നേടുക Norton's Theorem ഉത്തരങ്ങളും വിശദമായ പരിഹാരങ്ങളുമുള്ള മൾട്ടിപ്പിൾ ചോയ്സ് ചോദ്യങ്ങൾ (MCQ ക്വിസ്). ഇവ സൗജന്യമായി ഡൗൺലോഡ് ചെയ്യുക Norton's Theorem MCQ ക്വിസ് പിഡിഎഫ്, ബാങ്കിംഗ്, എസ്എസ്‌സി, റെയിൽവേ, യുപിഎസ്‌സി, സ്റ്റേറ്റ് പിഎസ്‌സി തുടങ്ങിയ നിങ്ങളുടെ വരാനിരിക്കുന്ന പരീക്ഷകൾക്കായി തയ്യാറെടുക്കുക

Latest Norton's Theorem MCQ Objective Questions

Top Norton's Theorem MCQ Objective Questions

Norton's Theorem Question 1:

Nortan equivalent circuit for the network between A and B is current source of ____ Amps with parallel resistance of ______ Ohms:

F2 Savita Engineering 20-4-22 D31

  1. 2A, 6Ω
  2. 2A, 3Ω
  3. 6A, 2Ω
  4. 6A, 9Ω

Answer (Detailed Solution Below)

Option 3 : 6A, 2Ω

Norton's Theorem Question 1 Detailed Solution

Concept:

Norton Theorem:

Any linear bilateral network can be resolved into a simplified circuit known as a Norton equivalent circuit which consists of a Norton current source, Norton equivalent resistance, and load resistance connected in parallel.

The Norton equivalent circuit is given below:

F2 Savita Engineering 20-4-22 D32

Step1: Finding out Norton's current source:

  • Short circuit the load and find out the short circuit current by applying KCL or KVL.
  • This short circuit current value is the value of the Norton current source.

 

Step2: Finding out Norton's equivalent resistance:

  • Open circuit the load terminal.
  • Replace all independent voltage and current sources with short and open circuits respectively.

 

Calculation:

Step 1: Norton's current source:

F2 Savita Engineering 20-4-22 D31

6Ω resistance is connected in parallel with terminal AB.

By short-circuiting the load terminal AB, 6Ω is also short-circuited.

F2 Savita Engineering 20-4-22 D30

\(I_N = {18 \over 3}\)

IN = 6A

Step 2: Norton's equivalent resistance:

F2 Savita Engineering 20-4-22 D27

18 V is replaced by the short circuit.

Now, 3Ω and 6Ω are connected in parallel.

\(R_N= {{3\times 6} \over {3+6}}\)

RN = 2Ω 

Norton's Theorem Question 2:

Which one of the following theorems becomes important if the circuit has sources operating at different frequencies?

  1. Norton theorem
  2. Thevenin theorem
  3. Superposition theorem
  4. Maximum power transfer theorem

Answer (Detailed Solution Below)

Option 3 : Superposition theorem

Norton's Theorem Question 2 Detailed Solution

Solution:

  • The Super Position Theorem becomes important for the circuits having sources operating at different frequencies.
  • In this case, since the impedances depend on frequency, we must have a different frequency-domain circuit for each frequency.
  • Both thecircuit should be analyse separately and all the calculations for current, voltage and power should be done accordingly.

 

Important Points The power consumed by a element of the circuit can be added directly to know the total power consumed if the circuit is operating under different frequencies.

Mistake Points Do not ry to add current and voltage calculated with two different frequencies in phasor domain. Add them in time domain only to avoid the mistake.

Norton's Theorem Question 3:

For the linear network shown below, V – I characteristic is given in the figure. The value of Norton equivalent current and resistance respectively are

F2 S.B Madhu 28.04.20 D2

  1. 3 A, 2 Ω
  2. 6 Ω, 2 Ω
  3. 6 A, 0.5 Ω
  4. 3 A, 0.5 Ω

Answer (Detailed Solution Below)

Option 3 : 6 A, 0.5 Ω

Norton's Theorem Question 3 Detailed Solution

Method I:

Norton Equivalent current (ISC) is the short circuit current across the load.

The circuit to evaluate the Norton current is drawn as:

F2 S.B Madhu 28.04.20 D3

From the given curve, when V = 0, I = - 6 A

Since ISC = - I

ISC = 6A

Since the Norton and Thevenin equivalents are dual of each other, we can write:

\({{R}_{th}}=\frac{{{V}_{th}}}{{{I}_{sc}}}\)

Where Vth is the open-circuit voltage at the load

From the given characteristic curve, when I = 0 (Open-circuit), Vth is:

VOC = Vth = 3 V

∴ The Norton equivalent resistance will be:

\({{R}_{N}}={{R}_{th}}=\frac{3}{6}=0.5~\text{ }\!\!\Omega\!\!\text{ }\)

Method II:

From the given graph, the equation of the line is:

I = 2V – 6     ...1)

For the given linear network across the load, the Norton equivalent circuit can be drawn as:

F2 S.B Madhu 28.04.20 D1

From the above figure, we can write:

\({{I}_{N}}+I=\frac{V}{{{R}_{N}}}\)

V = Voltage across the load

\(I=\frac{V}{{{R}_{N}}}-{{I}_{N}}\)

Comparing this with equation (1), we can write:

\(\frac{1}{{{R}_{N}}}=2\)

RN = 0.5 Ω

IN = 6A

Norton's Theorem Question 4:

Select the correct option from the difference between the Network theorms in electrical engineering.

1. Thevenin’s theorem provides an equivalent voltage source and an equivalent series resistance.

2. Norton’s theorem provides an equivalent Current source and an equivalent parallel resistance.

  1. 1 is true and 2 is false 
  2. 2 is true and 1 is false 
  3. Both are false 
  4. Both are true 

Answer (Detailed Solution Below)

Option 4 : Both are true 

Norton's Theorem Question 4 Detailed Solution

The correct answer is option 4.

Thevenin Theorem

SSC JE EE 1

Thevenin's theorem states that any linear circuit can be simplified to an equivalent circuit consisting of a single voltage source and resistance connected in series with the load.

Norton Theorem

SSC JE EE 2

Norton's theorem states that any linear circuit can be simplified to an equivalent circuit consisting of a single current source and a resistance that is connected in parallel with the load.

Norton's Theorem Question 5:

In the following circuit, the values of Norton's current IN and Norton's resistance RN across AB are

F3 Savita Engineering 20.05.2022 D6

  1. 3 A, 10/3 Ω
  2. 10 A, 4 Ω
  3. 1⋅5 A, 6 Ω
  4. 1⋅5 A, 4 Ω

Answer (Detailed Solution Below)

Option 1 : 3 A, 10/3 Ω

Norton's Theorem Question 5 Detailed Solution

Concept:

Calculation of Norton current:  

For calculating Norton's current, the load terminal must be short-circuited:

F3 Savita Engineering 20.05.2022 D7

Let us assume the voltage at node 'C' be V.

Applying KCL at node C:

\( {V-15\over 2}+ {V\over 4}+{V\over 2}=0\)

\( {2V-30+V+2V\over 4}=0\)

\(5V=30\)

V = 6

\(I_N = {6 \over 2}= 3A\)

Calculation of Norton resistance:  

For calculating Norton's resistance, the load terminal must be open-circuited and the independent voltage and current source must be replaced by short and open circuits respectively.

F3 Savita Engineering 20.05.2022 D8

2Ω and 4Ω are in parallel.

\(R_1 = {2\times 4 \over 2+4}={8 \over 6}\Omega\)

F3 Savita Engineering 20.05.2022 D9

\(R_{AB} = {8 \over 6}+2 \)

\(R_{AB} = {10\over 3}\Omega\)

Norton's Theorem Question 6:

A Norton circuit with 10 A current source and 15 Ω resistance is connected across a resistance of 5 Ω. The current in 5 Ω resistance will be:

  1. 5 A
  2. 2.5 A
  3. 7.5 A
  4. 10 A

Answer (Detailed Solution Below)

Option 3 : 7.5 A

Norton's Theorem Question 6 Detailed Solution

Concept:

Norton’s Theorem states that “Any linear circuit containing several energy sources and resistances can be replaced by a single constant current source in parallel with a single resistor “.

quesImage7403

Below is Norton Equivalent of the above circuit

F1 Harish Battula 21-5-2021 Swati D7

IN = Norton Current

RN = Norton Resistance

RL = Load Resistance

Steps to follow for Norton’s Theorem:

  • Calculate Norton’s current source by removing the load resistor from the original circuit and replacing it with a short circuit.
  • Calculating the current through a shorted wire.
  • Calculating the Norton resistance by removing all power sources in the original circuit (voltage sources shorted and current sources open).
  • Calculating total resistance between the open connection points.
  • Draw the Norton equivalent circuit, with the Norton current source in parallel with the Norton resistance.
  • The load resistor re-attaches between the two open points of the equivalent circuit.
 

Calculation:

Given

Is = 10 A

RN = 15 Ω

RL = 5 Ω 

From current division rule

current across load = \({I_s}\times {R_N\over R_N+R_L} = {10}\times {15\over 15+5}\)

= 7.5 A

Additional Information

 Steps to follow for Norton’s Theorem:

  • Calculate Norton’s current source by removing the load resistor from the original circuit and replacing it with a short circuit.
  • Calculating the current through a shorted wire.
  • Calculating the Norton resistance by removing all power sources in the original circuit (voltage sources shorted and current sources open).
  • Calculating total resistance between the open connection points.
  • Draw the Norton equivalent circuit, with the Norton current source in parallel with the Norton resistance.
  • The load resistor re-attaches between the two open points of the equivalent circuit.

Norton's Theorem Question 7:

A Norton equivalent of a circuit consists of a 4 A current source in parallel with a 2 Ω resistor. The Thevenin's equivalent of this circuit is a ________ V source in series with a 2 Ω resistor.

  1. 2
  2. 0.5
  3. 6
  4. 8

Answer (Detailed Solution Below)

Option 4 : 8

Norton's Theorem Question 7 Detailed Solution

Concept:

Thevenin’s theorem: Any two-terminal bilateral linear DC circuits can be replaced by an equivalent circuit consisting of a voltage source and a series resistor. 

SSC JE EE 1

Norton’s Theorem: Any two-terminal bilateral linear DC circuits can be replaced by an equivalent circuit consisting of a current source and a parallel resistor.

SSC JE EE 2

Now we can say that Norton’s theorem is the converse of Thevenin’s theorem.

Application:

A Norton equivalent of a circuit consists of a 4 A current source in parallel with a 2 resistor. It can be represented as follows:

F1&F2 U.B Shashi 19 07 2019 D 1

By using source transformation, the above circuit becomes:

F2 S.B Madhu 11.03.20 D3

Now, it is the Thevenin's equivalent of the given circuit with 8 V source in series with a 2 Ω resistor.

Norton's Theorem Question 8:

Find RN for the following circuit:

F6 Madhuri Engineering 18.08.2022 D20

  1. 25 Ω
  2. 20 Ω
  3. 4 Ω
  4. 5 Ω

Answer (Detailed Solution Below)

Option 3 : 4 Ω

Norton's Theorem Question 8 Detailed Solution

The correct option is (3)

Calculation:

The given circuit is:

F6 Madhuri Engineering 18.08.2022 D20

Equivalent Nortan Resistance (RN) = Rab

R= Rab = [(8 + 4 + 8) || 5] Ω 

= [20 || 5] Ω 

\(\frac{20\times 5}{20+5}\)

R= Rab = 4 Ω 

Norton's Theorem Question 9:

Norton equivalent of the circuit given below is

F1 Raviranjan 13-1-22 Savita D7

  1. F1 Raviranjan 13-1-22 Savita D8
  2. F1 Raviranjan 13-1-22 Savita D9
  3. F1 Raviranjan 13-1-22 Savita D10
  4. F1 Raviranjan 13-1-22 Savita D11

Answer (Detailed Solution Below)

Option 4 : F1 Raviranjan 13-1-22 Savita D11

Norton's Theorem Question 9 Detailed Solution

Calculation:

The given circuit is

F1 Raviranjan 13-1-22 Savita D7

For RN calculation, voltage source should be short-circuited and current should be open-circuited

F1 Raviranjan 13-1-22 Savita D12

∴ RN = 3 + 3||6

= 3 + 2 = 5 Ω

Let find VTh : Apply nodal analysis

F1 Raviranjan 13-1-22 Savita D7VTh

\(\rm \frac{V_{Th}+10}{6}+\frac{V_{Th}-20}{3}=0\)

or, \(\rm \frac{V_{Tn}+10+2V_{Tn}-40}{6}=0\)

⇒ 3 VTh = 30 V

∴ VTh = 10 V

∴ Norton's current \(\rm I_N=\frac{V_{Tn}}{R_{Tn}}=\frac{10}{5}=2\ A\)

∴ Norton's equivalent will be

F1 Raviranjan 13-1-22 Savita D11

Therefore, correct option is (4)

Norton's Theorem Question 10:

The Norton’s equivalent of the below circuit at the left terminal x - y is

mad19

  1. mad20

  2. mad21

  3. mad22

  4. mad23

  5. quesOptionImage2122

Answer (Detailed Solution Below)

Option 2 :

mad21

Norton's Theorem Question 10 Detailed Solution

Concept:

To solve Problems with Dependent Sources:

To find Voc : Calculate the open circuit voltage across load terminals. This open circuit voltage is called Thevenin’s voltage (Vth). Norton's current is the ratio of Thevenin’s voltage (Vth) & Thevenin’s Resistance (Rth).

To find Rth

To calculate Thevenin’s Resistance we should  replace all independent current sources by Open circuit and Independent voltage sources by Short circuit (keep dependent sources as it is).

Circuits which have only dependent sources can’t function on their own so Vth and Isc doesn’t exists but still they exhibit resistance, that resistance can be indirectly determined by V/I method by placing an active source across the terminals.

1. Place a voltage source of 1V across the terminal and find the current (IT) flowing   through it. Then,  

\({{\rm{R}}_{{\rm{th}}}} = {{\rm{R}}_{\rm{N}}} = \frac{{1{\rm{V}}}}{{{{\rm{I}}_{\rm{T}}}}}\)

(or) 

2. Place a current source of 1A across the terminals and find the voltage (Vt) across the current source. Then,

 \({{\rm{R}}_{{\rm{th}}}} = {{\rm{R}}_{\rm{N}}} = \frac{{{{\rm{V}}_{\rm{T}}}}}{{1{\rm{A}}}}\)

Calculation:

The loop current in left hand loop is 0.15 V0, where V0 is open circuit voltage

Taking KVL we get

10 + (0.15 V0) 5 – V0 = 0

\( \Rightarrow {V_0} = \frac{{10}}{{0.25}} = 40\;V\)

In order to apply Norton’s theorem, xy terminals are short to find Is.c

mad25

Since V0 = 0 the dependent current source is also zero

\( \Rightarrow {I_{s.c}} = \frac{{10}}{{5 + 5}} = 1\;A\)

Hence internal resistance or Norton's equivalent resistance will be:

\({R_{th}} = \frac{{{V_o}}}{{{I_{s.c}}}} = 40\;{\rm{\Omega }}\)

∴ The Norton equivalent circuit is

mad26

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