EE115
Engineering Student
ee115w5.docxSeries and Parallel Inductive Reactance
Consider the series inductive reactive circuit:
Calculate the following: (Express all answers in magnitude/phase angle form)
XLT = XL1 + XL2 + XL3 +…+etc., XL = 2πfL, IT = VT / ZT, tan θz =
L2 = 150mH, XL2 = (6.28 * 1 kHz * 150mH) = 6.28 * (1*103) * (150*10-3) = 942Ω∠90ᵒ
L1 = 80mH, XL1 = (6.28 * 1 kHz * 80mH) = 6.28 *(1*103) * (80*10-3) = 502.4Ω∠90ᵒ
Zeq =
Zeq = = =
= 450Ω + 1446.4Ω =1.514kΩ
tan θz = = = tan-1 (3.214) = 72.72ᵒ
IT = VT / ZT = 15V/1.514kΩ = 9.9mA∠72.72ᵒ
V = I * R VR1 = 9.9mA * 150Ω = 1.49V VR2 = 9.9mA * 300Ω = 2.97V
VL2 = 9.9mA * 942Ω = 9.32V VL1 = 9.9mA * 502.4Ω = 4.97V
a. Zeq = 1.514kΩ∠72.72ᵒ
b. IT = 9.9mA∠-72.72ᵒ
c. XL2 = 942Ω∠90ᵒ
d. XL1 = 502.4Ω∠90ᵒ
e. VR1 = 1.49V∠-72.72ᵒ
f. VR2 = 2.97V∠-72.72ᵒ
g. VL1 = 4.97V∠17.28ᵒ
h. VL2 = 9.32V∠17.28ᵒ
2. Consider the parallel inductive reactive circuit:
Calculate the following: (Express all answers in magnitude/phase angle form)
REQ = 1/ + + + …etc. XLEQ = 1/ + + + …etc. XL = 2πfL IT =
tan θ1 =
L1 = 80mH, XL1 = (6.28 * 1 kHz * 80mH) = 6.28 *(1*103) * (80*10-3) = 502.4Ω∠90ᵒ
L2 = 150mH, XL2 = (6.28 * 1 kHz * 150mH) = 6.28 * (1*103) * (150*10-3) = 942Ω∠90ᵒ
ZEQ = 1/ + + = 1/ + + = =
(91.479 +27.92j) = 95.64∠16.97ᵒ
a. Zeq = 95.64Ω∠16.97ᵒ
b. IT = = .156A∠-16.97ᵒ
c. XL2 = 942Ω∠90ᵒ
d. XL1 = 502.4Ω∠90ᵒ
e. IR1 = = .01A
f. IR2 = = .05A
g. IL1 = = 29.85mA∠-90ᵒ
h. IL2 = = 15.92mA∠-90ᵒ
