Answer:
- branch 1: i = 25.440∠-32.005°; pf = 0.848 lagging
- branch 2: i = 21.466∠63.435°; pf = 0.447 leading
- total: i = 31.693∠10.392° leading; pf = 0.984 leading
Explanation:
To calculate the currents in the parallel branches, we need to know the impedance of each branch. That will be the sum of the resistance and reactance.
The inductive reactance is ...
The capacitive reactance is ...
<u>Branch 1</u>
The impedance of branch 1 is ...
Z1 = 8 +j4.99984 Ω
so the current is ...
I1 = V/Z = 240/(8 +j4.99984) ≈ 25.440∠-32.005°
The power factor is cos(-32.005°) ≈ 0.848 (lagging)
<u>Branch 2</u>
The impedance of branch 2 is ...
Z2 = 5 -j10 Ω
so the current is ...
I2 = 240/(5 +j10) ≈ 21.466∠63.435°
The power factor is cos(63.436°) ≈ 0.447 (leading)
<u>Total current</u>
The total current is the sum of the branch currents. A suitable calculator can add these vectors without first converting them to rectangular form.
It = I1 +I2 = (21.573 -j13.483) +(9.6 +j19.2)
It ≈ 31.173 +j5.717 ≈ 31.693∠10.392°
The power factor for the circuit is cos(10.392°) ≈ 0.984 (leading)
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The phasor diagram of the currents is attached.
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<em>Additional comment</em>
Given two vectors, their sum can be computed several ways. One way to compute the sum is to use the Law of Cosines. In this application, the angle between the vectors is the supplement of the difference of the vector angles: 84.560°.
