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
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:
Answer (Detailed Solution Below)
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:
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:
6Ω resistance is connected in parallel with terminal AB.
By short-circuiting the load terminal AB, 6Ω is also short-circuited.
\(I_N = {18 \over 3}\)
IN = 6A
Step 2: Norton's equivalent resistance:
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?
Answer (Detailed Solution Below)
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
Answer (Detailed Solution Below)
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:
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:
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.
Answer (Detailed Solution Below)
Norton's Theorem Question 4 Detailed Solution
The correct answer is option 4.
Thevenin Theorem
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
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
Answer (Detailed Solution Below)
Norton's Theorem Question 5 Detailed Solution
Concept:
Calculation of Norton current:
For calculating Norton's current, the load terminal must be short-circuited:
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.
2Ω and 4Ω are in parallel.
\(R_1 = {2\times 4 \over 2+4}={8 \over 6}\Omega\)
\(R_{AB} = {8 \over 6}+2 \)
\(R_{AB} = {10\over 3}\Omega\)
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:
Answer (Detailed Solution Below)
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 “.
Below is Norton Equivalent of the above circuit
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.
Answer (Detailed Solution Below)
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.
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.
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:
By using source transformation, the above circuit becomes:
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:
Answer (Detailed Solution Below)
Norton's Theorem Question 8 Detailed Solution
The correct option is (3)
Calculation:
The given circuit is:
Equivalent Nortan Resistance (RN) = Rab
RN = Rab = [(8 + 4 + 8) || 5] Ω
= [20 || 5] Ω
= \(\frac{20\times 5}{20+5}\)
RN = Rab = 4 Ω
Norton's Theorem Question 9:
Norton equivalent of the circuit given below is
Answer (Detailed Solution Below)
Norton's Theorem Question 9 Detailed Solution
Calculation:
The given circuit is
For RN calculation, voltage source should be short-circuited and current should be open-circuited
∴ RN = 3 + 3||6
= 3 + 2 = 5 Ω
Let find VTh : Apply nodal analysis
VTh
\(\rm \frac{V_{Th}+10}{6}+\frac{V_{Th}-20}{3}=0\)
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
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
Answer (Detailed Solution Below)
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
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