Norton’s Theorem MCQ [Free PDF] – Objective Question Answer for Norton’s Theorem Quiz

21. Using Norton’s Theorem calculate the equivalent load as shown in the figure

Using Norton's Theorem calculate the equivalent load as shown in the figure 

A. 4/3 Ω
B. 4/3 Ω
C. 4 Ω
D. 2 Ω

Answer: B

Applying KCL in the given circuit, we get

\(\frac{V}{4} + \frac{V − 2I}{2}\) = I

Or

3V − 4I/4 = I

Or, 3V = 8I

∴ V/I = 8/3 Ω.

 

22. In the following circuit, the value of Norton’s resistance between terminals a and b are ___________

In the following circuit, the value of Norton’s resistance between terminals a and b are ___________

A. RN  = 1800 Ω
B. RN  = 270 Ω
C. RN  = 90 Ω
D. RN  = 90 Ω

Answer: D

By writing loop equations for the circuit, we get,

VS  = VX, IS  = IX

VS  = 600(I1 – I2) + 300(I1 – I2) + 900 I1

= (600 + 300 + 900) I1 – 600I2 – 300I3

= 1800I1 – 600I2 – 300I3

I1  = IS, I2  = 0.3 VS

I3  = 3IS  + 0.2VS

VS  = 1800IS – 600(0.01VS) – 300(3IS  + 0.01VS)

= 1800IS – 6VS – 900IS – 3VS

10VS  = 900IS

For Voltage, VS  = RN IS  + VOC

Here VOC  = 0

So, Resistance RN  = 90Ω.

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23. For the circuit shown in the figure below, the value of Norton’s resistance is _________

For the circuit shown in figure below, the value of Norton’s resistance is _________

A. 100 Ω
B. 136.4 Ω
C. 200 Ω
D. 272.8 Ω

Answer: A

IX  = 1 A, VX  = Vtest

Vtest  = 100(1 − 2IX) + 300(1 − 2IX – 0.01VS) + 800

Vtest  = 1200 – 800IX – 3Vtest

4Vtest  = 1200 – 800 = 400

Vtest  = 100V

∴ RN = V/1 = 100 Ω.

 

24. For the circuit shown in the figure below, the Norton Resistance looking into X − Y is __________

For the circuit shown in the figure below, the Norton Resistance looking into X-Y is __________

A. 2 Ω
B.2/3 Ω
C. 5/3 Ω
D. 2 Ω

Answer: D

RN = VOC/ISC

VN  = VOC

Applying KCL at node A

\(\frac{2I − V_N}{1} + 2 = I + VN/2

Or, I = VN/1

Putting, 2VN – VN  + 2 = VN + VN/2

Or, VN  = 4 V.

∴ RN  = 4/2 = 2Ω.

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25. In the figure given below, the value of Resistance R by Norton’s Theorem is ___________

In the figure given below, the value of Resistance R by Norton’s Theorem is ___________

A. 40
B. 20
C. 50
D. 80

Answer: B

\(\frac{V_P − 100}{10} + \frac{V_P}{10}\) + 2 = 0

2VP – 100 + 20 = 0

∴ VP  = 80/2 = 40V

∴ R = 20Ω (By Norton’s Theorem).

 

26. In the figure given below, the Norton Resistance, as seen at the terminals P − Q, is given by __________

In the figure given below, the Norton Resistance, as seen at the terminals P-Q, is given by __________

A. 5 Ω
B. 7.5 Ω
C. 5 Ω
D. 7.5 Ω

Answer: A

For finding VN,

In the figure given below, the Norton Resistance, as seen at the terminals P-Q, is given by __________

VN = (4 × 10)/(10 + 10) = 2V

For finding RN,

In the figure given below, the Norton Resistance, as seen at the terminals P-Q, is given by __________

RN  = 10 || 10

= (10×10)/(10 + 10) = 5 Ω.

 

27. The Norton equivalent impedance Z between the nodes P and Q in the following circuit is __________

The Norton equivalent impedance Z between the nodes P and Q in the following circuit is __________

A. 1
B. 1 + s + 1/s
C. 2 + s + 1/s
D. 3 + s + 1/s

Answer: A

To calculate the Norton resistance, all the current sources get open − circuited and voltage sources get short − circuited.

∴ RN = (1/s + 1) || (1 + s)

$\frac{{\left( {\frac{1}{s} + 1} \right) \times (1 + s)}}{{\left( {\frac{1}{s} + 1} \right) + (1 + s)}}$

 

$\frac{{\frac{1}{s} + 1 + 1 + s}}{{\frac{1}{s} + 1 + 1 + s}}$

So, RN  = 1.

 

28. In the circuit given below, it is given that VAB  = 4 V for RL  = 10 kΩ and VAB  = 1 V for RL = 2kΩ. The value of Norton resistance for the network N is ________

In the circuit given below, it is given that VAB = 4 V for RL = 10 kΩ and VAB = 1 V for RL = 2kΩ. The value of Norton resistance for the network N is

A. 16 kΩ
B. 30 kΩ
C. 3 kΩ
D. 50 kΩ

Answer: B

When RL  = 10 kΩ and VAB  = 4 V

Current in the circuit

VAB/RL = 4/10 = 0.4 mA

Norton voltage is given by

VN  = I (RN  + RL)

= 0.4(RN  + 10)

= 0.4RN  + 4

Similarly, for RL  = 2 kΩ and VAB  = 1 V

So, I = 1/2 = 0.5 mA

VN  = 0.5(RN  + 2)

= 0.5 RN  + 1

∴ 0.1RN  = 3

Or, RN  = 30 kΩ.

 

29. For the circuit given below, Norton’s resistance across the terminals A and B is _________

For the circuit given below, Norton’s resistance across the terminals A and B is

A. 5 Ω
B. 7 kΩ
C. 1.5 kΩ
D. 1.1 kΩ

Answer: B

Let VAB  = 1 V

5 VAB  = 5

Or, 1 = 1 × I1 or, I1  = 1

Also, 1 = − 5 + 1(I – I1)

∴ I = 7

Hence, R = 7 kΩ.

 

30. A circuit is given in the figure below. The Norton equivalent as viewed from terminals x and x’s is ___________

A circuit is given in the figure below. The Norton equivalent as viewed from terminals x and x’ is

A. 6 Ω and 1.333 A
B. 6 Ω and 0.833 A
C. 32 Ω and 0.156 A
D. 32 Ω and 0.25 A

Answer: B

We, draw the Norton equivalent of the left side of xx’ and source transformed right side of yy’.

A circuit is given in the figure below. The Norton equivalent as viewed from terminals x and x’ is

Vxx’  = VN = \(\displaystyle\frac{\frac{4}{8} + \frac{8}{24}}{\frac{1}{8} + \frac{1}{24}}\) = 5V

 

∴ RN  = 8 || (16 + 8)

= (8×24)/(8 + 24) = 6 Ω

∴ (IN = VN/RN = 5/6 = 0.833 A.

 

31. For the circuit given in the figure below, the Norton equivalent as viewed from terminals y and y’ is _________

A circuit is given in the figure below. The Norton equivalent as viewed from terminals x and x’ is

A. 32 Ω and 0.25 A
B. 32 Ω and 0.125 A
C. 6 Ω and 0.833 A
D. 6 Ω and 1.167 A

We draw the Norton equivalent of the left side of xx’ and source transformed right side of yy’.Norton equivalent as seen from terminal yy’ is

A circuit is given in the figure below. The Norton equivalent as viewed from terminals x and x’ is

Vxx’  = VN = \(\displaystyle\frac{\frac{4}{8} + \frac{8}{24}}{\frac{1}{8} + \frac{1}{24}}\) = 5V

= (0.167 + 1)/(0.04167 + 0.125) = 7 V
∴ RN  = (8 + 16) || 8
= (24×8)/(24 + 8) = 6 Ω

∴IN = VN/RN = 7/6 = 1.167 A.

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32. In the figure given below, the power loss in 1 Ω resistor using Norton’s Theorem is ________

In the figure given below, the power loss in 1 Ω resistor using Norton’s Theorem is

A. 9.76 W
B. 9.26 W
C. 10.76 W
D. 11.70 W

Answer: B
Let us remove the 1 Ω resistor and short x − y.
At Node 1, assuming node potential to be V,

(V − 10)/5 + ISC  = 5

But ISC = V/2

∴ (V − 10)/5 + V/2 = 5

Or, 0.7 V = 7

That is, V = 10 V

∴ ISC = V/2 = 5 A

To find Rint, all constant sources are deactivated.

Rint = (5 + 2)×2/(5 + 2 + 2) = 14/9 = 1.56 Ω

Rint  = 1.56 Ω; ISC  = IN  = 5A

Here, I = IN

\(\frac{R_{int}}{R_{int} + 1} = 5 × \frac{1.56}{1.56 + 1}\) = 3.04 A

 

∴ Power loss = (3.04)2 × 1 = 9.26 W.

 

33. The value of RN from the circuit given below is ________

The value of RN from the circuit given below is

A. 3 Ω
B. 1.2 Ω
C. 5 Ω
D. 12.12 Ω

Answer: D

VX = 3VX/6 + 4

VX  = 8 V = VOC

If terminal is short − circuited, VX  = 0.

ISC = 4/3 + 3) = 0.66 A

∴ RN = VOC/ISC = 8/0.66 = 12.12 Ω.

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34. By using the Norton equivalent theorem find the current I, as shown in the figure below, is ________

By using Norton equivalent theorem find the current I, as shown in the figure below, is ________

A. 3 A
B. 2 A
C. 1 A
D. 0

Answer: C

The 3 Ω resistance is an extra element because the voltage at node B is independent of the 3 Ω resistance.

I1 = 3/2 + 1 = 1 A (B − > A.

The net current in 2 Ω resistance is

I = 1 – I1

= 2 – 1 = 1 A (A − > B..

 

35. While computing the Norton equivalent resistance and the Norton equivalent current, which of the following steps are undertaken?

  1. Both the dependent and independent voltage sources are short-circuited and both the dependent and independent current sources are open-circuited
  2. Both the dependent and independent voltage sources are open-circuited and both the dependent and independent current sources are short-circuited
  3. The dependent voltage source is open-circuited keeping the independent voltage source untouched and the dependent current source is short-circuited keeping the independent current source untouched
  4. The dependent voltage source is short-circuited keeping the independent voltage source untouched and the dependent current source is open-circuited keeping the independent current source untouched

Answer: D

While computing the Norton equivalent voltage consisting of both dependent and independent sources, we first find the equivalent resistance called the Norton resistance by opening the two terminals.

Then while computing the Norton current, we short-circuit the dependent voltage sources keeping the independent voltage sources untouched, and open-circuiting the dependent current sources keeping the independent current sources untouched.

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