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Test: Method of Analysis - Electronics and Communication Engineering (ECE) MCQ


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15 Questions MCQ Test - Test: Method of Analysis

Test: Method of Analysis for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Test: Method of Analysis questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The Test: Method of Analysis MCQs are made for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Method of Analysis below.
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Test: Method of Analysis - Question 1

v1 = ?

Detailed Solution for Test: Method of Analysis - Question 1

Applying the nodal analysis

Test: Method of Analysis - Question 2

va = ?

Detailed Solution for Test: Method of Analysis - Question 2

v= 2(3+1) + 3(1) =11V

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Test: Method of Analysis - Question 3

v1 = ?

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Test: Method of Analysis - Question 4

va = ?

Detailed Solution for Test: Method of Analysis - Question 4

Test: Method of Analysis - Question 5

v2 = ?

Detailed Solution for Test: Method of Analysis - Question 5

Test: Method of Analysis - Question 6

ib = ?

Detailed Solution for Test: Method of Analysis - Question 6

Test: Method of Analysis - Question 7

i1 = ?

Detailed Solution for Test: Method of Analysis - Question 7

sing Thevenin equivalent and source transform

Test: Method of Analysis - Question 8

i1 = ?

Detailed Solution for Test: Method of Analysis - Question 8

75 = 90ki1+ 10k(i1 - 7.5m)

150 = 100ki1 = i1 = 1.5 mA

Test: Method of Analysis - Question 9

i1 = ?

Detailed Solution for Test: Method of Analysis - Question 9

3 = 2i1 + 3(i1 - 4) 

i1 = 3 A

Test: Method of Analysis - Question 10

i1 = ?

Detailed Solution for Test: Method of Analysis - Question 10

45 = 2ki1 + 500(i1 + 15m)

= i1 = 15 mA

Test: Method of Analysis - Question 11

i1 = ?

Detailed Solution for Test: Method of Analysis - Question 11

6.6 = 50i1 + 100(i1 + 0.1) + 40(i1 - 0.06) + 60(i1 - 0.1) i1 = 0.02 A

Test: Method of Analysis - Question 12

For the circuit of Fig. the value of vs , that will result in v1 = 0, is

Detailed Solution for Test: Method of Analysis - Question 12

If v1 = 0, the dependent source is a short circuit
 

Test: Method of Analysis - Question 13

i1 , i2 = ?

Detailed Solution for Test: Method of Analysis - Question 13

ix = i1 - i2
15 = 4i1 -2(i1 - i2) + 6(i1 - i2)
⇒ 8i1 - 4i2 = 15   ..........(i)
-18 = 2i2 + 6(i2 -i1)
⇒ 3i1 -4i2 = 9    ..........(ii)
i1 =1.2 A, i2 = -1.35 A 

Test: Method of Analysis - Question 14

v1 = ?

Detailed Solution for Test: Method of Analysis - Question 14

14 = 3i1 + vy + 6 (i1 - 2 - 7) + 2vy + 2(i1 -7)
vy = 3(i1 - 2)
14 = 3i1 +9(i1 - 2)+ 6(i1 - 9)+2(i1 -7)
14 = 20i1 -18 - 54 - 14 ⇒ i1 = 5 A
v1 = 6(5 - 2 - 7) + 2 x 3(5 - 2)+2(5 - 7) = -10 V 

Test: Method of Analysis - Question 15

vx = ?

Detailed Solution for Test: Method of Analysis - Question 15

Let i1 and i2 be two loop current

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