Test: Coupled Circuits Part - 1


10 Questions MCQ Test Networking Theory | Test: Coupled Circuits Part - 1


Description
This mock test of Test: Coupled Circuits Part - 1 for Electrical Engineering (EE) helps you for every Electrical Engineering (EE) entrance exam. This contains 10 Multiple Choice Questions for Electrical Engineering (EE) Test: Coupled Circuits Part - 1 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Coupled Circuits Part - 1 quiz give you a good mix of easy questions and tough questions. Electrical Engineering (EE) students definitely take this Test: Coupled Circuits Part - 1 exercise for a better result in the exam. You can find other Test: Coupled Circuits Part - 1 extra questions, long questions & short questions for Electrical Engineering (EE) on EduRev as well by searching above.
QUESTION: 1

In the circuit of fig. P1.9.1-2 i1 = 4 sin2 A, and i2 = 0.

Q. v1 =?

Solution:

QUESTION: 2

In the circuit of fig. P1.9.1-2 i1 = 4 sin2 A, and i2 = 0.

Q. v2 =?

Solution:

QUESTION: 3

Consider the circuit shown in Fig. P1.9.3-4

Q. If i1 = 0 and i2 = 2 sin4 A, the voltage v1 is

Solution:

QUESTION: 4

Consider the circuit shown in Fig. P1.9.3-4

Q. If i1 = e-2tV and i2 = 0, the voltage v2 is

Solution:

QUESTION: 5

Consider the circuit shown in fig. P19.5-6

Q. If current i1 = 3 cos4 A and i2 = 0, then voltage v1 and v2 are

Solution:


QUESTION: 6

Consider the circuit shown in fig. P19.5-6

Q. If current i1 = 0 and i2 = 4sin3tA, then voltage v1 and v2 are

Solution:



= 3di2/dt
= 36cos 3t V

QUESTION: 7

In the circuit shown in fig. P1.9.7-8, i1= 3 cos3t A and i= 4 sin3 A.

Q. v1 =?

Solution:


=-18 sin t + 12 cost = 6 (2 cos t - 3 sin t) V

QUESTION: 8

In the circuit shown in fig. P1.9.7-8, i1= 3 cos3t A and i= 4 sin3 A.

Q. v2 =?

Solution:


= 24 cos 3t - 9 sin 3t = 3 ( 8 cos 3t - 3 sin 3t) V

QUESTION: 9

In the circuit shown in fig. P1.9.9-10, i= 5 sin3 A and i= 3 cos3t A

Q. v1 =?

Solution:


= 45 cos 3t + 27 sin 3t = 9 (5cos 3t + 3sin 3t) V

QUESTION: 10

In the circuit shown in fig. P1.9.9-10, i= 5 sin3 A and i= 3 cos3t A

Q. v2 =?

Solution:


= 36 sin 3t + 45 cos 3t = 9 (4sin 3t + 5 cos 3t) V

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