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

Z = √(R² + (ωL - 1/ωC)²)

where R is the resistance, L is the inductance, C is the capacitance, and ω is the angular frequency (2πf, where f is the frequency in hertz).

If the frequency is halved, then ω is divided by 2. If the resistance is doubled, then R is multiplied by 2. Substituting these values into the impedance equation, we get:

Z = √((2R)² + ((ω/2)L - 1/(ω/2)C)²)

Simplifying this expression, we can see that the impedance does not remain constant, halve, or double. Therefore, the answer is that the new impedance of the circuit cannot be determined without additional information about the values of L and C.

• When the frequency of the voltage applied to a series RL circuit is increased, the phase angle of the circuit will decrease.
• In a series RL circuit, the current lags the voltage because of the inductive reactance of the inductor.
• The amount of lag is determined by the phase angle, which is the angle between the voltage and current waveforms.
• As the frequency of the voltage applied to the circuit increases, the inductive reactance of the inductor decreases, which causes the phase angle to decrease as well.
• This means that the current waveform gets closer in phase to the voltage waveform, and the amount of lag between them decreases.
• Therefore, the answer is that the phase angle of the circuit will decrease when the frequency of the voltage applied to a series RL circuit is increased.

• When the frequency is decreased, the impedance of a parallel RL circuit decrease.
• In a parallel RL circuit, the total impedance (Z) is determined by the resistance (R) and the inductive reactance (XL) of the inductor.
• The inductive reactance is given by the formula XL = 2πfL, where f is the frequency of the applied voltage and L is the inductance of the inductor.
• When the frequency of the applied voltage is decreased, the inductive reactance of the inductor also decreases.
• As the frequency is decreased, the inductive reactance of the circuit decreases while the resistance remains constant.
• This results in a decrease in the total impedance of the circuit.

In an RL circuit, the impedance is determined by both the resistance and the inductive reactance combined.

The impedance of the circuit is given by the formula \(Z = \sqrt {{R^2} + {\omega ^2}{L^2}}\)

where

R is the resistance

XL is the inductive reactance.

Therefore, the correct answer is resistance and inductive reactance.

• When the resistor voltage in a series RL circuit becomes less than the inductor voltage, the phase angle Increase.
• When the resistor voltage in a series RL circuit becomes less than the inductor voltage, it means that the current in the circuit is lagging behind the voltage.
• This results in an increase in the phase angle between the voltage and current, as the phase angle represents the phase difference between the two quantities.

To calculate the power factor, we need to find the impedance of the circuit, which is given by:

Z = R + jX

where R is the resistance and X is the inductive reactance.

In this case, R = 1.5 kΩ and X = 2.2 kΩ. Therefore, the impedance Z is:

Z = 1.5 kΩ + j(2.2 kΩ) = (1.5 + j2.2) kΩ

To calculate the power factor, we need to find the phase angle of the impedance, which is given by:

θ = tan?¹(X/R)

In this case, θ = tan?¹(2.2/1.5) ≈ 56.3°

The power factor (PF) is defined as the cosine of the phase angle, i.e., PF = cos(θ).

PF = cos(56.3°) ≈ 0.564

So the power factor is approximately 0.564.

The current drawn by the coil can be calculated using Ohm's law, which states that:

I = V/Z

where I is the current, V is the voltage, and Z is the impedance of the coil.

In this case, the coil has zero resistance, so the impedance is purely inductive and is given by:

Z = jωL

where

j is the imaginary unit
ω is the angular frequency (2πf)
L is the inductance of the coil (20 H)
f is the frequency of the power line (60 Hz).

Substituting the values, we get:

Z = j(2π × 60 Hz × 20 H) = j7.54 kΩ

The impedance is purely reactive, meaning the current will be 90 degrees out of phase with the voltage.

Therefore, the current drawn by the coil can be calculated as follows:

I = V/Z = 110 V / j7.54 kΩ = -j14.6 mA

Note that a current is a complex number with a magnitude of 14.6 mA and a phase angle of -90 degrees (or 270 degrees).

Taking the magnitude of the current, we get:

|I| = 14.6 mA

The impedance of the parallel RL circuit is

\({\rm{Z = }}\frac{{R{X_L}}}{{\sqrt {{R^2} + {X^2}_L} }}\)

Substituting R = 140 Ω and XL = 60 Ω, we get:

\(\frac{{(140)(60)}}{{\sqrt {{{(140)}^2} + {{(60)}^2}} }}\) = = 55.15 Ω

The magnitude of the total impedance is approximately 55.15Ω

In an AC circuit, the total power (or apparent power) is given by the product of the voltage and current, which is expressed in units of volt-amperes (VA).

The apparent power can also be expressed in terms of the real power (in watts) and the reactive power (in volt-amperes reactive or VAR), as:

Apparent power = Real power + Reactive power

For a purely inductive load, the real power is zero, so the apparent power is equal to the reactive power.

Therefore, if the reactive power is 12 VAR, the apparent power is also 12 VA.

The impedance of a series RL circuit is given by:

Z = R + jX

where R is the resistance, X is the reactance of the inductor, and j is the imaginary unit.

In this case, R = 47 Ω and X = 120 Ω. Substituting these values, we get:

Z = 47 + j120

To express the impedance in polar form, we need to find the magnitude and phase angle of Z. The magnitude of Z is given by:

|Z| = sqrt(R² + X²) = sqrt(47² + 120²) = 129 Ω

The phase angle of Z is given by:

θ = arctan(X/R) = arctan(120/47) = 68.6 degrees

Therefore, the impedance expressed in polar form is:

Z = 129 ∠68.6° Ω