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

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.

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

I_N = 6A

Step 2: Norton's equivalent resistance:

18 V is replaced by the short circuit.

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

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

R_N = 2Ω 

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.

Method I:

Norton Equivalent current (I_SC) 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 I_SC = - I

I_SC = 6A

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

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

Where V_th is the open-circuit voltage at the load

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

V_OC = V_th = 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:

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

R_N = 0.5 Ω

I_N = 6A

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.

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.

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

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

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

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

I_s = 10 A

R_N = 15 Ω

R_L = 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.

The correct option is (3)

Calculation:

The given circuit is:

Equivalent Nortan Resistance (R_N) = R_ab

R_N = R_ab = [(8 + 4 + 8) || 5] Ω 

= [20 || 5] Ω 

= \(\frac{20\times 5}{20+5}\)

R_N = R_ab = 4 Ω 

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.

Calculation:

The given circuit is

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

∴ R_N = 3 + 3||6

= 3 + 2 = 5 Ω

Let find V_Th : Apply nodal analysis

V_Th

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

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

⇒ 3 V_Th = 30 V

∴ V_Th = 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)

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 (R_th).

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 I_s.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