Inductive Reactance Quiz By Gibilisco

A vector is a mathematical quantity that represents both magnitude and direction. Magnitude refers to the size or length of the vector, while direction indicates the orientation or angle at which the vector is pointing. Therefore, a vector must possess both magnitude and direction to be properly defined. The other options mentioned, such as resistance, inductance, and reactance, are not relevant to the definition of a vector.

When the number of turns in a coil increases, the reactance at a constant frequency increases. This is because reactance is directly proportional to the inductance of the coil, and the inductance of a coil is determined by factors such as the number of turns, the radius of the coil, and the permeability of the core material. When the number of turns increases, the inductance also increases, resulting in a higher reactance.

XL is the symbol for inductive reactance, which is the opposition to the flow of alternating current in an inductor. It is given by the formula XL = 2πfL, where f is the frequency and L is the inductance. In this case, the frequency is 990 kHz and the inductance is 689 µH. Plugging these values into the formula, we get XL = 2π(990 kHz)(689 µH) = 4.28 kΩ.

The frequency can be determined using the formula: reactance (in ohms) = 2πfL (where f is the frequency in hertz and L is the inductance in henries). Rearranging the formula, we get f = reactance / (2πL). Plugging in the values, we get f = 33 Ω / (2π * 400 µH). Simplifying this equation gives us f = 0.013 kHz, which is equivalent to 13 kHz.

The inductive reactance (XL) of an inductor is given by the formula XL = 2πfL, where f is the frequency and L is the inductance. In this question, the given XL is 555 Ω and the frequency f is 132 kHz. By rearranging the formula, we can solve for L: L = XL / (2πf). Substituting the given values, we get L = 555 Ω / (2π * 132 kHz) ≈ 670 µH. Therefore, the correct answer is 670 µH.

As the frequency of an AC wave gets lower, the value of XL for a particular coil of wire decreases. This is because XL represents the inductive reactance of the coil, which is directly proportional to the frequency of the AC wave. When the frequency decreases, the inductive reactance decreases as well. This can be explained by the fact that at lower frequencies, the coil has more time to build up and collapse its magnetic field, resulting in a lower opposition to the flow of current. Therefore, the value of XL decreases as the frequency of the AC wave decreases.

The inductive reactance of a coil is given by the equation XL = 2πfL, where XL is the inductive reactance, f is the frequency, and L is the inductance. Rearranging the equation, we can solve for L by dividing both sides by 2πf. Therefore, L = XL / (2πf). Plugging in the given values, we get L = 200 Ω / (2π * 500 Hz) = 0.0637 H. Since the answer options are given in millihenries, we convert the inductance to millihenries by multiplying by 1000, resulting in 63.7 mH.

Each point in the RL plane corresponds to a unique combination of resistance and inductive reactance. This is because the RL plane represents the impedance of a circuit, which is the combination of resistance and reactance. The resistance represents the opposition to the flow of current, while the inductive reactance represents the opposition to the change in current caused by the inductor. Therefore, each point in the RL plane represents a specific value of resistance and inductive reactance that characterizes the impedance of the circuit.

As the ratio of inductive reactance to resistance (XL/R) decreases in an RL circuit, it means that the resistance is becoming relatively larger compared to the inductive reactance. This indicates that the effect of the inductive component is decreasing, leading to a decrease in the phase angle. Therefore, the correct answer is "decreases".

The phase angle in an RL circuit can be calculated using the formula tan(θ) = XL / R, where XL is the inductive reactance and R is the resistance. In this case, the inductive reactance can be calculated using the formula XL = 2πfL, where f is the frequency and L is the inductance. Plugging in the values given (f = 200 kHz and L = 100 µH), we can calculate XL. Then, using the formula tan(θ) = XL / R and plugging in the values for XL and R (XL = calculated value and R = 100 Ω), we can solve for θ. The resulting value is approximately 51.5°.

In an RL circuit, the phase angle is the angle between the voltage and current waveforms. The phase angle can be calculated using the formula arctan(Lω/R), where L is the inductance, ω is the angular frequency (2πf), and R is the resistance. Given that the inductance is 88 mH, the resistance is 95 Ω, and the frequency is 800 Hz, we can calculate the angular frequency as 2π(800) = 1600π rad/s. Substituting these values into the formula, we get arctan((88×10^-3)(1600π)/(95)) ≈ 78°. Therefore, the correct answer is 78°.

The given inductor has an inductance value of 88 mH and an inductive reactance of 100 Ω. The formula to calculate the inductive reactance is XL = 2πfL, where XL is the inductive reactance, f is the frequency, and L is the inductance. Rearranging the formula to solve for f, we get f = XL / (2πL). Plugging in the values, we find f = 100 Ω / (2π * 88 mH). Simplifying this equation gives us f ≈ 181 Hz. Therefore, the correct answer is 181 Hz.

The correct answer is "All of the above are true." Each specific complex impedance value defined in the form R + jXL corresponds to a specific point in the RL plane, corresponds to a specific inductive reactance, and corresponds to a specific resistance.

In a circuit containing inductive reactance but no resistance, the phase angle is equal to 90°. This is because inductive reactance is caused by the presence of an inductor in the circuit, which causes a phase shift between the voltage and current. The voltage lags behind the current by 90° in an inductive circuit. Therefore, the phase angle is constantly 90° in this scenario.

If the inductive reactance and the resistance in an RL circuit are equal, it means that the impedance of the circuit is purely resistive. In a purely resistive circuit, the phase angle between the voltage and current is always 0°. However, in this case, the options provided do not include 0° as a choice. Therefore, the phase angle cannot be determined solely based on the given information and it depends on the actual values of the resistance and the inductive reactance.

The reactance of an inductor can be calculated using the formula Xl = 2πfL, where Xl is the reactance, f is the frequency, and L is the inductance. In this case, the frequency is given as 1000Hz and the inductance is given as 100 mH. Plugging these values into the formula, we get Xl = 2π(1000)(100 x 10^-3) = 628 Ω. Therefore, the correct answer is 628 Ω.

The points in the RL plane for all resulting impedances will lie along a ray of indefinite length, pointing outward from the origin. This is because the resistance and inductive reactance are always in the ratio 3:1, which means they can vary but will always maintain this ratio. As a result, the impedance will vary along a line that extends indefinitely from the origin in the RL plane.

As the number of turns in a coil that carries AC increases without limit, the current in the coil will decrease, approaching zero. This is because as the number of turns increases, the inductance of the coil also increases. Inductance is the property of a coil that opposes changes in current. As the inductance increases, it resists the flow of current, causing the current to decrease. Eventually, as the number of turns approaches infinity, the inductance becomes very large and the current approaches zero. Therefore, the current in the coil will decrease, approaching zero.