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Test: Network Theorems (A.C.) - 2 - Electrical Engineering (EE) MCQ


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10 Questions MCQ Test GATE Electrical Engineering (EE) 2024 Mock Test Series - Test: Network Theorems (A.C.) - 2

Test: Network Theorems (A.C.) - 2 for Electrical Engineering (EE) 2024 is part of GATE Electrical Engineering (EE) 2024 Mock Test Series preparation. The Test: Network Theorems (A.C.) - 2 questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Network Theorems (A.C.) - 2 MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Network Theorems (A.C.) - 2 below.
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Test: Network Theorems (A.C.) - 2 - Question 1

For the circuit given, determine the Thevenin voltage as seen by RL.

Test: Network Theorems (A.C.) - 2 - Question 2

The Norton equivalent current is

Test: Network Theorems (A.C.) - 2 - Question 3

The Thevenin equivalent voltage is

Test: Network Theorems (A.C.) - 2 - Question 4

In order to get maximum power transfer from a capacitive source, the load must

Test: Network Theorems (A.C.) - 2 - Question 5

Determine VTH if R1 is changed to 3.3 kΩ.

Test: Network Theorems (A.C.) - 2 - Question 6

Referring to the given circuit, find ZTH if R is 15 kΩ and RL is 38 kΩ.

Test: Network Theorems (A.C.) - 2 - Question 7

Determine VTH when R1 is 180 Ω and XL is 90 Ω.

Test: Network Theorems (A.C.) - 2 - Question 8

Find the voltage across 2Ω resistor due to 20V source in the circuit shown below:

Detailed Solution for Test: Network Theorems (A.C.) - 2 - Question 8

The voltage at node A in the figure is
(V - 20)/20 + (V - 10)/10 + V/2 = 0
⇒ V = 3.07V. Now short circuiting 10V source,

(V - 20)/20 + V/2 + V/10 = 0
⇒ V = 1.5V.

Test: Network Theorems (A.C.) - 2 - Question 9

Find the equivalent thevenin’s resistance between terminals A and B in the circuit shown below.

Detailed Solution for Test: Network Theorems (A.C.) - 2 - Question 9

To find Rth, two voltage sources are removed and replaced with short circuit. The resistance at terminals AB then is the parallel combination of the 10Ω resistor and 5Ω resistor

⇒ Rth=(10 × 5)/15 = 3.33Ω.

Test: Network Theorems (A.C.) - 2 - Question 10

Find the current flowing between terminals A and B of the circuit shown below.

Detailed Solution for Test: Network Theorems (A.C.) - 2 - Question 10

The magnitude of the current in Norton’s equivalent circuit is equal to the current passing through the short circuited terminals that are I = 20/5 = 4A.

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