Question
Download Solution PDFIn an AC circuit, there are three branches connected to a node. Branch 1 carries a current of 10∠0°A. and Branch 2 carries a current of 10∠90°A. Calculate the current in the third branch.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFExplanation:
In an AC circuit, when multiple branches are connected at a node, the currents in these branches must satisfy Kirchhoff's Current Law (KCL). According to KCL, the sum of currents entering a node is equal to the sum of currents leaving the node. This principle is crucial in analyzing and solving electrical circuits, especially when dealing with AC currents, which have both magnitude and phase components.
Given in the problem are the currents in two branches of an AC circuit:
Branch 1 carries a current of 10∠0∘A.
Branch 2 carries a current of 10∠90∘A.
Using KCL, the total current entering the node must equal the total current leaving the node. Therefore, the sum of the currents in all three branches should be zero:
I1+I2+I3=0
I3=−(10+j10)=−10−j10A.
Magnitude = 14.14A.
Phase angle = 45∘.
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