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�
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
To find the answers to the given questions, we can use basic electrical formulas and concepts. Here's how we can approach each part of the question:
(a) To find the active power absorbed by the inductive branch, we can use the formula for power in an AC circuit:
Active Power (P) = Current (I)^2 × Resistance (R) × Power Factor (PF)
In this case, the non-inductive resistor absorbs the active power, so we can calculate it using the given values:
P (resistor) = (2.7 A)^2 × 20 Ω = 145.8 W
Since the inductive circuit doesn't absorb any active power (it only stores and releases reactive power), the active power absorbed by the inductive branch is zero.
(b) To find the inductance of the inductive circuit, we can use the formula for inductive reactance (XL):
Inductive Reactance (XL) = 2π × Frequency (f) × Inductance (L)
Given that the current in the inductive circuit is 4.3 A and the frequency is 50 Hz, we can rearrange the formula to find the inductance:
L = XL / (2πf)
L = (4.3 A) / (2π × 50 Hz) = 0.0137 H (rounded to four decimal places)
Therefore, the inductance of the inductive circuit is approximately 0.0137 H.
(c) The power factor is a measure of how effectively electrical power is being used. It is the cosine of the phase angle between the current and voltage waveforms in the circuit. To find the power factor of the combined network, we need to find the phase angle between the current and voltage.
From the given values, we know the current taken from the supply is 5.8 A, and we can assume it to be the total current flowing in the combined circuit. We are also given the current through the non-inductive resistor, which means it is in phase with the total current.
The current through the inductive circuit lags behind the voltage due to its inductive nature. Let's assume the phase angle between the voltage and current in the inductive circuit is θ.
Since the total current is the phasor sum of the currents in the resistor and inductive circuit, we can write:
5.8 A = √((2.7 A)^2 + (4.3 A)^2)
Solving for this equation gives us:
θ = arccos(2.7 A / 5.8 A) ≈ 61.49°
Therefore, the phase angle between the voltage and current in the inductive circuit is approximately 61.49°.
The power factor can be found by taking the cosine of this phase angle:
Power Factor = cos(61.49°) ≈ 0.496
So, the power factor of the combined network is approximately 0.496.
A sketch of the phasor diagram can be done by representing the voltage, current in the resistor, current in the inductive circuit, and the total current as vectors. The voltage vector would be horizontal, the current in the resistor vector would be vertical (in-phase with the voltage), and the current in the inductive circuit vector would be at an angle of 61.49° with the voltage vector. The total current vector would be the phasor sum of the two current vectors.
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 =
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