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If current 1(1) = 3A sin omegat and 1(2)...

If current `1_(1) = 3A` sin `omegat and 1_(2)= 4A cos omegat, then 1_(3)` is

A

5A `sin (omegat + 53^(@))`

B

`5Asin (omegat + 37^(@))`

C

`5Asin (omegat + 45^(@))`

D

`5Asin (omegat + 30^(@))`

Text Solution

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The correct Answer is:
A
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