Question
Download Solution PDFTwo circuits, the impedance of which are given by Z1 = (4 + j3) Ω and Z2 = (8 - j6) Ω are connected in parallel. If the total current supplied is 15 A, what is the value of the total admittance of the circuit?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFThe correct answer is option 4):(Y = 0.24 - j0.06 mho)
Concept:
Admittance is a measure of how easily a circuit or device will allow a current to flow. It is defined as the reciprocal of impedance,
Z = \(1 \over Y\)
When two admittance connected in parallel then admittance
Y = Y1 + Y2
Calculation:
Given
Z1 = (4 + j3) Ω
Y1 = \(1\over Z_1\)
=\(1 \over 4+j3\)
= \(0.16 -0.12 i\)
Z2 = (8 - j6) Ω
Y2 = \(1 \over Z_2\)
= \( 1\over (8 - j6)\)
= 0.08 + 0.06i
Y = Y1 + Y2
= 0.08 + 0.06i + \(0.16 -0.12 i\)
= 0.24 - j0.06 mho
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