Network Theory Questions and Answers – Norton’s Theorem Involving Dependent and Independent Sources

This set of Network Theory Questions & Answers for Exams focuses on “Norton’s Theorem Involving Dependent and Independent Sources”.

1. The circuit shown in figure has a load equivalent to _________
The circuit shown in figure has a load equivalent to 83 Ω
a) \(\frac{4}{3}\) Ω
b) \(\frac{8}{3}\) Ω
c) 4 Ω
d) 2 Ω
View Answer

Answer: b
Explanation: Applying KCL in the given circuit, we get, \(\frac{V}{4} + \frac{V-2I}{2}\) = I
Or, \(\frac{3V-4I}{4}\) = I
Or, 3V = 8I
∴ \(\frac{V}{I} = \frac{8}{3}\) Ω.

2. In the following circuit, the value of Norton’s resistance between terminals a and b are ___________
The value of Norton’s resistance between terminals a & b are RN = 90 Ω
a) RN = 1800 Ω
b) RN = 270 Ω
c) RN = 90 Ω
d) RN = 90 Ω
View Answer

Answer: d
Explanation: 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.

3. For the circuit shown in figure below, the value of Norton’s resistance is _________
The value of Norton’s resistance is 100 Ω for the circuit
a) 100 Ω
b) 136.4 Ω
c) 200 Ω
d) 272.8 Ω
View Answer

Answer: a
Explanation: IX = 1 A, VX = Vtest
Vtest = 100(1-2IX) + 300(1-2IX – 0.01VS) + 800
Or, Vtest = 1200 – 800IX – 3Vtest
Or, 4Vtest = 1200 – 800 = 400
Or, Vtest = 100V
∴ RN = \(\frac{V_{test}}{1}\) = 100 Ω.
advertisement
advertisement

4. For the circuit shown in the figure below, the Norton Resistance looking into X-Y is __________
The Norton Resistance looking into X-Y is 2 Ω for the circuit
a) 2 Ω
b) \(\frac{2}{3}\)
c) \(\frac{5}{3}\)
d) 2 Ω
View Answer

Answer: d
Explanation: \(R_N = \frac{V_{OC}}{I_{SC}}\)
VN = VOC
Applying KCL at node A, \(\frac{2I-V_N}{1} + 2 = I + \frac{V_N}{2}\)
Or, I = \(\frac{V_N}{1}\)
Putting, 2VN – VN + 2 = VN + \(\frac{V_N}{2}\)
Or, VN = 4 V.
∴ RN = 4/2 = 2Ω.

5. In the figure given below, the value of Resistance R by Norton’s Theorem is ___________
The value of Resistance R by Norton’s Theorem is 20 for the circuit
a) 40
b) 20
c) 50
d) 80
View Answer

Answer: b
Explanation: \(\frac{V_P-100}{10} + \frac{V_P}{10}\) + 2 = 0
Or, 2VP – 100 + 20 = 0
∴ VP = 80/2 = 40V
∴ R = 20Ω (By Norton’s Theorem).
Note: Join free Sanfoundry classes at Telegram or Youtube

6. In the figure given below, the Norton Resistance, as seen at the terminals P-Q, is given by __________
Find the Norton Resistance, as seen at the terminals P-Q
a) 5 Ω
b) 7.5 Ω
c) 5 Ω
d) 7.5 Ω
View Answer

Answer: a
Explanation: For finding VN,
For finding VN VN = 4×1010+10 = 2V is given in figure
VN = \(\frac{4 × 10}{10+10}\) = 2V
For finding RN,
The Norton Resistance, as seen at the terminals P-Q, is given by 5 Ω
RN = 10 || 10
= \(\frac{10×10}{10+10}\) = 5 Ω.

7. 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 & Q in the following circuit is 1
a) 1
b) 1 + s + \(\frac{1}{s}\)
c) 2 + s + \(\frac{1}{s}\)
d) 3 + s + \(\frac{1}{s}\)
View Answer

Answer: a
Explanation: To calculate the Norton resistance, all the current sources get open-circuited and voltage sources get short-circuited.
∴ RN = (\(\frac{1}{s}\) + 1) || (1+s)
= \(\frac{\left(\frac{1}{s} + 1\right)×(1+s)}{\left(\frac{1}{s} + 1\right)+(1+s)}\)
= \(\frac{\frac{1}{s}+1+1+s}{\frac{1}{s}+1+1+s}\) = 1
So, RN = 1.
advertisement

8. 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 ____________
The value of Norton resistance for the network N is 30 kΩ if VAB = 4 V for RL = 10 kΩ
a) 16 kΩ
b) 30 kΩ
c) 3 kΩ
d) 50 kΩ
View Answer

Answer: b
Explanation: When RL = 10 kΩ and VAB = 4 V
Current in the circuit \(\frac{V_{AB}}{R_L} = \frac{4}{10}\) = 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 = \(\frac{1}{2}\) = 0.5 mA
VN = 0.5(RN + 2)
= 0.5 RN + 1
∴ 0.1RN = 3
Or, RN = 30 kΩ.

9. For the circuit given below, the Norton’s resistance across the terminals A and B is _____________
The Norton’s resistance across the terminals A & B is 7 kΩ in given circuit
a) 5 Ω
b) 7 kΩ
c) 1.5 kΩ
d) 1.1 kΩ
View Answer

Answer: b
Explanation: 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.
advertisement

10. A circuit is given in the figure below. The Norton equivalent as viewed from terminals x and x’ is ___________
Find the Norton equivalent as viewed from terminals x & x’
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
View Answer

Answer: b
Explanation: We, draw the Norton equivalent of the left side of xx’ and source transformed right side of yy’.
The Norton equivalent as viewed from terminals x & x’ is 6 Ω & 0.833 A
Vxx’ = VN = \(\displaystyle\frac{\frac{4}{8} + \frac{8}{24}}{\frac{1}{8} + \frac{1}{24}}\) = 5V
∴ RN = 8 || (16 + 8)
= \(\frac{8×24}{8+24}\) = 6 Ω
∴ \(I_N = \frac{V_N}{R_N} = \frac{5}{6}\) = 0.833 A.

11. For the circuit given in figure below, the Norton equivalent as viewed from terminals y and y’ is _________
The Norton equivalent as viewed from terminals x & x’ is 6 Ω & 0.833 A
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
View Answer

Answer: d
Explanation: We draw the Norton equivalent of the left side of xx’ and source transformed right side of yy’.
The Norton equivalent as viewed from terminals x & x’ is 6 Ω & 0.833 A
Norton equivalent as seen from terminal yy’ is
Vyy’ = VN =\(\displaystyle\frac{\frac{4}{24} + \frac{8}{8}}{\frac{1}{24} + \frac{1}{8}}\) = 5V
= \(\frac{0.167+1}{0.04167+0.125}\) = 7 V
∴ RN = (8 + 16) || 8
= \(\frac{24×8}{24+8}\) = 6 Ω
∴ IN = \(\frac{V_N}{R_N} = \frac{7}{6}\) = 1.167 A.

12. In the figure given below, the power loss in 1 Ω resistor using Norton’s Theorem is ________
The power loss in 1 Ω resistor using Norton’s Theorem is 9.26 W
a) 9.76 W
b) 9.26 W
c) 10.76 W
d) 11.70 W
View Answer

Answer: b
Explanation: Let us remove the 1 Ω resistor and short x-y.
At Node 1, assuming node potential to be V, \(\frac{V-10}{5}\) + ISC = 5
But ISC = \(\frac{V}{2}\)
∴ \(\frac{V-10}{5} + \frac{V}{2}\) = 5
Or, 0.7 V = 7
That is, V= 10 V
∴ ISC = \(\frac{V}{2}\) = 5 A
To find Rint, all constant sources are deactivated. Rint = \(\frac{(5+2)×2}{5+2+2} = \frac{14}{9}\) = 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.

13. The value of RN from the circuit given below is ________
The value of RN from the circuit given below is 12.12 Ω
a) 3 Ω
b) 1.2 Ω
c) 5 Ω
d) 12.12 Ω
View Answer

Answer: d
Explanation: VX = 3\(\frac{V_X}{6}\) + 4
Or, VX = 8 V = VOC
If terminal is short-circuited, VX = 0.
ISC = \(\frac{4}{3+3}\) = 0.66 A
∴ RN = \(\frac{V_{OC}}{I_{SC}}\) = \(\frac{8}{0.66}\) = 12.12 Ω.

14. The current I, as shown in the figure below, is ________
The current I, as shown in the figure below is 1 A in given circuit
a) 3 A
b) 2 A
c) 1 A
d) 0
View Answer

Answer: c
Explanation: The 3 Ω resistance is an extra element because voltage at node B is independent of the 3 Ω resistance.
I1 = \(\frac{3}{2+1}\) = 1 A (B -> A)
The net current in 2 Ω resistance is I = 1 – I1
= 2 – 1 = 1 A (A -> B).

15. While computing the Norton equivalent resistance and the Norton equivalent current, which of the following steps are undertaken?
a) Both the dependent and independent voltage sources are short-circuited and both the dependent and independent current sources are open-circuited
b) Both the dependent and independent voltage sources are open-circuited and both the dependent and independent current sources are short-circuited
c) 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
d) 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
View Answer

Answer: d
Explanation: 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.

Sanfoundry Global Education & Learning Series – Network Theory.

To practice all exam questions on Network Theory, here is complete set of 1000+ Multiple Choice Questions and Answers.

If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]

advertisement
advertisement
Subscribe to our Newsletters (Subject-wise). Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

Youtube | Telegram | LinkedIn | Instagram | Facebook | Twitter | Pinterest
Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

Subscribe to his free Masterclasses at Youtube & discussions at Telegram SanfoundryClasses.