Electric Circuits Test 4

Result

Electric Circuits Test 4

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Question 1
1 / -0

An AC network has two sources. The application of superposition theorem gives the currents due to the two sources through a branch as 5∠45°A and 5∠90°A. The total current in that branch is:

A
10 A

B
2.32∠30°A

C
9.24∠67.5°A

D
0 A

Solution

The total current in that branch is I = I₁ + I₂
I₁ = 5∠45°A = 5 cos 45° + j5 sin 45° = \(\frac{5}{{\sqrt 2 }} + j\frac{5}{{\sqrt 2 }}\)
I₂ = 5∠90°A = 5 cos 90° + j5 sin 90° = j5
Total current = \(\frac{5}{{\sqrt 2 }} + j\frac{5}{{\sqrt 2 }} + j5 = 9.24\angle 67.5^\circ {\rm{A}}\)