Berikut ini adalah pertanyaan dari aniszaf pada mata pelajaran TI untuk jenjang Sekolah Menengah Atas
found to be 2.3<0° amperes. Calculate the value of capacitor in microfarad, voltage across
the inductor and total power consumed (CLO1, C3)
(15 marks)
Jawaban dan Penjelasan
Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.
Penjelasan:
Resistance, R = 100 Ω
Inductance, L = 0.318 H
Applied voltage, V = 230<0° V
Frequency, f = 50 Hz
Current, I = 2.3<0° A
Reactance of inductor, XL = ωL = 2πfL = 2π × 50 × 0.318 = 100 Ω
Reactance of capacitor, XC = 1/ωC = 1/(2πfC)
Real part:
2.3 = 230 / (100 + 100 - 318.31C)
2.3 = 230 / (200 - 318.31C)
C = [230 / (2.3 × 118.31)] × 10^6
C = 8.2567 μF
Imaginary part:
0 = -230j / (100 + 100 - 318.31C)
0 = -230j / (200 - 318.31C)
318.31C = 200
C = 0.6289 μF
Reactance of inductor, XL = 2πfL = 2π × 50 × 3.18 = 1005.3 Ω
Real part:
2.3 = 230 / (100 + 1005.3 - 1/(2π × 50 × C))
2.3 = 230 / (1105.3 - 318.31C)
C = [230 / (2.3 × 318.31)] × 10^6
C = 361.46 μF
Imaginary part:
0 = -230j / (100 + 1005.3 - 1/(2π × 50 × C))
0 = -230j / (1105.3 - 318.31C)
318.31C = 1105.3
C = 3.47 μF
Hence, the value of capacitance of the circuit is 361.46 μF.
Voltage across the inductor, VL = XLI = 1005.3 × 2.3 = 2318.19 V
Total power consumed by the circuit, P = VI cos(θ), where θ is the phase angle between voltage and current.
cos(θ) = R/Z = 100 / √(100^2 + (1005.3 - 1/(2π × 50 × 361.46))^2) = 0.9848
P = 230 × 2.3 × 0
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Last Update: Mon, 10 Jul 23