a. zero
b. Xl=V/current, xl=2pi f L solve for L
c. pfactor= real power/apparent power=2.7V/(5.8V)=2.7/5.8
An inductive circuit, in parallel with a non-inductive
resistor of 20 Ω, is connected across a 50 Hz supply.
The currents through the inductive circuit and the
non-inductive resistor are 4.3 A and 2.7 A respectively, and the current taken from the supply is 5.8 A.
Find: (a) the active power absorbed by the inductive
branch; (b) its inductance; and (c) the power factor of
the combined network. Sketch the phasor diagram�
3 answers
b. V = I*R = 2.7 * 20 = 54 Volts.
Xl = V/I = 54/4.3 = 12.6 Ohms.
Xl = 2pi*F*L = 12.6.
6.28*50*L = 12.6,
L = -----henrys.
c. Tan A = Xl/R = 12.6/20.
A = 32.2 degrees.
Pf = Cos32.2 =
Xl = V/I = 54/4.3 = 12.6 Ohms.
Xl = 2pi*F*L = 12.6.
6.28*50*L = 12.6,
L = -----henrys.
c. Tan A = Xl/R = 12.6/20.
A = 32.2 degrees.
Pf = Cos32.2 =
a. P = I^2R = (5.8)^2 . 20 = 672.8 W...... Note: a pure inductor does not dissipate energy thus the average power is equal to zero, but it's not zero across the resitor
B.) Inductance (L)..... Remember, Ir = V/R thus V = (2.7)20 = 54 V.
IL = V/XL, because voltage is the same for both branches; XL = (54)/4.3= 12.6but XL = 2πfL = 2π{50}L =12.6 thus L = 0.04 H
C.) Phase Angle = tan^-1(R/Xl) = tan^-1( IL/IR) = TAN^-1 (4.3/2.7) =57.9° THUS POWER FACTOR= COS(57.9) = 0.53
B.) Inductance (L)..... Remember, Ir = V/R thus V = (2.7)20 = 54 V.
IL = V/XL, because voltage is the same for both branches; XL = (54)/4.3= 12.6but XL = 2πfL = 2π{50}L =12.6 thus L = 0.04 H
C.) Phase Angle = tan^-1(R/Xl) = tan^-1( IL/IR) = TAN^-1 (4.3/2.7) =57.9° THUS POWER FACTOR= COS(57.9) = 0.53
4 answers