A series circuit containing a 295 microfarad capacitor and a coil whose resistance and inductance are 3 ohms and 4.42 mH, respectively are supplied by the following series connected generators: 35 V at 60 Hz, 10 V at 180 Hz and 8 V at 240 Hz. Determine the power factor of the circuit.

Question

A series circuit containing a 295 microfarad capacitor and a coil whose resistance and inductance are 3 ohms and 4.42 mH, respectively are supplied by the following series connected generators: 35 V at 60 Hz, 10 V at 180 Hz and 8 V at 240 Hz. Determine the power factor of the circuit.

Answer 1

I will outline the solution.

Figure Z at each frequency.
Then, determine each current at those three frequencies.
Then, take the real part of each current, multiply it by the resistance, that is real power.

Then, take the imaginary part of each current, add, that is the imaginary part. Then, to get apperent power, you need to combine the imag with the real, get the complex current, figure its magnitude, apparent power= I^2*Z

pf=realpower/apparent power.

Answer 2

To determine the power factor of the circuit, we need to calculate the impedance and the phase angle of the circuit.

First, let's calculate the impedance (Z) of the circuit. The impedance of the circuit is given by:

Z = √((R^2) + (Xl - Xc)^2)

Where:
R = resistance = 3 ohms
Xl = inductive reactance = 2πfL (in rad/s)
Xc = capacitive reactance = 1 / (2πfC) (in rad/s)
f = frequency in Hz
L = inductance = 4.42 mH = 4.42 x 10^-3 H
C = capacitance = 295 microfarads = 295 x 10^-6 F

Let's calculate Xl and Xc for each frequency.

For 60 Hz:
Xl = 2π(60)(4.42 x 10^-3) = 0.5264 ohms
Xc = 1 / (2π(60)(295 x 10^-6)) = 9.18 ohms

For 180 Hz:
Xl = 2π(180)(4.42 x 10^-3) = 1.5794 ohms
Xc = 1 / (2π(180)(295 x 10^-6)) = 3.06 ohms

For 240 Hz:
Xl = 2π(240)(4.42 x 10^-3) = 2.1058 ohms
Xc = 1 / (2π(240)(295 x 10^-6)) = 2.27 ohms

Now we can calculate the impedance (Z) for each frequency.

For 60 Hz:
Z = √((3^2) + (0.5264 - 9.18)^2) = √(9 + 69.36) = √78.36 = 8.85 ohms

For 180 Hz:
Z = √((3^2) + (1.5794 - 3.06)^2) = √(9 + 1.0816) = √10.0816 = 3.17 ohms

For 240 Hz:
Z = √((3^2) + (2.1058 - 2.27)^2) = √(9 + 0.0289) = √9.0289 = 3.00 ohms

The power factor (PF) of the circuit is the cosine of the phase angle (θ), which can be calculated using the formula:

θ = arctan((Xl - Xc) / R)

For 60 Hz:
θ = arctan((0.5264 - 9.18) / 3) = arctan(-2.3179) ≈ -66.51°

For 180 Hz:
θ = arctan((1.5794 - 3.06) / 3) = arctan(-0.4939) ≈ -26.03°

For 240 Hz:
θ = arctan((2.1058 - 2.27) / 3) = arctan(-0.0541) ≈ -3.10°

The power factor (PF) can be calculated using the cosine of the phase angle (θ):

PF = cos(θ)

For 60 Hz:
PF = cos(-66.51°) ≈ 0.41

For 180 Hz:
PF = cos(-26.03°) ≈ 0.89

For 240 Hz:
PF = cos(-3.10°) ≈ 0.998

Therefore, the power factor of the circuit at 60 Hz is approximately 0.41, at 180 Hz is approximately 0.89, and at 240 Hz is approximately 0.998.

Answer 3

To determine the power factor of the circuit, we first need to calculate the total impedance of the circuit.

The impedance of a series RLC circuit is given by the equation:

Z = √(R^2 + (Xl - Xc)^2)

Where:
Z is the total impedance
R is the resistance of the coil
Xl is the inductive reactance (2πfL) where f is the frequency and L is the inductance of the coil
Xc is the capacitive reactance (1/2πfC) where f is the frequency and C is the capacitance of the capacitor

Let's calculate the impedance for each generator frequency:

At 60 Hz:
Xl = 2πfL = 2π * 60 * (4.42 * 10^(-3)) = 0.527 ohms
Xc = 1/(2πfC) = 1/(2π * 60 * (295 * 10^(-6))) = 9.58 ohms (approximately)

Z1 = √(R^2 + (Xl - Xc)^2) = √((3)^2 + (0.527 - 9.58)^2) = √(9 + (-9.053)^2) = √(9 + 81.948) = √(90.948) = 9.53 ohms (approximately)

At 180Hz:
Xl = 2πfL = 2π * 180 * (4.42 * 10^(-3)) = 2.999 ohms (approximately)
Xc = 1/(2πfC) = 1/(2π * 180 * (295 * 10^(-6))) = 1.106 ohms (approximately)

Z2 = √(R^2 + (Xl - Xc)^2) = √((3)^2 + (2.999 - 1.106)^2) = √(9 + 1.893) = √(10.893) = 3.3 ohms (approximately)

At 240 Hz:
Xl = 2πfL = 2π * 240 * (4.42 * 10^(-3)) = 3.999 ohms (approximately)
Xc = 1/(2πfC) = 1/(2π * 240 * (295 * 10^(-6))) = 0.832 ohms (approximately)

Z3 = √(R^2 + (Xl - Xc)^2) = √((3)^2 + (3.999 - 0.832)^2) = √(9 + 10.46) = √(19.46) = 4.41 ohms (approximately)

Now we can calculate the power factor using the formula:

Power factor (PF) = cos(θ) = R/Z

Where:
PF is the power factor
θ is the phase angle between the current and voltage
R is the resistance of the circuit
Z is the total impedance of the circuit

Let's calculate the power factor:

At 60 Hz:
PF1 = (3 ohms) / (9.53 ohms) = 0.314 (approximately)

At 180 Hz:
PF2 = (3 ohms) / (3.3 ohms) = 0.909 (approximately)

At 240 Hz:
PF3 = (3 ohms) / (4.41 ohms) = 0.68 (approximately)

Therefore, the power factors of the circuit at 60 Hz, 180 Hz, and 240 Hz are approximately 0.314, 0.909, and 0.68, respectively.