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An inductor is connected to an AC power supply having a maximum output voltage of \(4.00 \mathrm{~V}\) at a frequency of \(300.0 \mathrm{~Hz}\). What inductance is needed to keep the rms current less than \(2.00 \mathrm{~mA}\) ?

Short Answer

Expert verified
The inductance needed to keep the rms current less than \(2.00 \mathrm{~mA}\) is approximately \(1.06 \mathrm{~H}\).

Step by step solution

01

Identify Given Values

The exercise provides the following information: Maximum output voltage \(V = 4.00 \mathrm{~V}\), Frequency \(f = 300.0 \mathrm{~Hz}\), maximum rms current \(I = 2.00 \mathrm{~mA}\).
02

Convert Units

Convert the rms current from milliamps to amps. \(1 \mathrm{~mA} = 0.001 \mathrm{~A}\), hence the current \(I = 2 \mathrm{~mA} = 0.002 \mathrm{~A}\).
03

Calculate Inductive Reactance

According to Ohm's law, \(V = I \cdot R\), we rearrange for \(R = V / I\). This represents the inductive reactance in this case, since we only have an inductor in the circuit. Substituting provided values leads to \(X_L = 4.00 \mathrm{~V} / 0.002 \mathrm{~A} = 2000 \mathrm{~\Omega}\).
04

Solve for Inductance

Knowing that \(X_L = 2\pi fL\), we solve for \(L\) which leads to \(L = X_L / (2\pi f)\). Substituting \(X_L = 2000 \mathrm{~\Omega}\) and \(f = 300.0 \mathrm{~Hz}\) leads to \(L = 2000 \mathrm{~\Omega} / (2\pi x 300.0 \mathrm{~Hz}) \approx 1.06 \mathrm{~H}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

AC Power Supply
An AC (Alternating Current) power supply provides voltage in a waveform that alternates in polarity over time. This means that the direction of current flow changes periodically. The voltage provided by an AC power source is constantly changing, typically in a sinusoidal manner. This type of power supply is distinct from a DC (Direct Current) power supply, where the voltage remains constant.

AC power supplies are common in household and industrial settings. The voltage can be described using several terms:
  • Peak Voltage (Vmax or Vpeak): The maximum voltage value reached during one cycle of the waveform.
  • RMS (Root Mean Square) Voltage (Vrms): A measure of the effective value of an AC voltage, often used in calculations involving power. The RMS voltage is related to the peak voltage by the formula: \( V_{rms} = \frac{V_{max}}{\sqrt{2}} \).
AC voltage frequency, such as the 300 Hz given in the exercise, indicates how fast the waveform repeats per second. In this scenario, the inductor connects to an AC source with a sinusoidal voltage having a maximum value of 4.00 V.
Inductive Reactance
Inductive reactance is a property of inductors in an AC circuit that opposes the change in current. It is represented by \(X_L\) and is measured in ohms (\(\Omega\)). Unlike resistance, which affects both AC and DC, inductive reactance only affects AC circuits.

Inductive reactance can be calculated with the formula:
\[ X_L = 2\pi f L \]
Here:
  • \(X_L\) is the inductive reactance,
  • \(f\) is the frequency of the AC supply, and
  • \(L\) is the inductance of the coil.
The higher the frequency or the larger the inductance, the greater the inductive reactance. This means that the inductor will "resist" the alternating current more strongly, reducing the amount of current that flows for a given voltage. In our problem, the inductive reactance is calculated to be 2000 \(\Omega\).
RMS Current
RMS (Root Mean Square) current is a measure used in AC circuits that describes the amount of current effectively delivered throughout a cycle. RMS values enable us to compare AC and DC circuits on equivalent terms.

To understand RMS current, consider it as the equivalent value of DC current that would produce the same power effect in the circuit. It's an important metric because alternating current varies constantly, and RMS offers a standardized measure.

RMS current is calculated differently than regular current in DC. For a sinusoidal waveform, which is very common in power supplies, the formula for RMS current is:
\[ I_{rms} = \frac{I_{peak}}{\sqrt{2}} \]
In our exercise, we aimed for an RMS current maximum of 2.00 mA, or 0.002 A, in the given AC power circuit.
Inductance Calculation
The inductance (L) of an inductor is a measure of how effectively it can oppose changes in current, reflecting its ability to store energy in a magnetic field. Inductance is measured in henries (H).

To find inductance when given the inductive reactance and the frequency of the AC power source, use the formula:
\[ L = \frac{X_L}{2\pi f} \]
In this equation:
  • \(L\) is the inductance,
  • \(X_L\) is the inductive reactance, measured in ohms, and
  • \(f\) is the frequency, measured in hertz.
In the given exercise, calculating using \(X_L\) of 2000 \(\Omega\) and a frequency of 300 Hz, the inductance required was approximately 1.06 H. This tells us the inductor should be capable of limiting the RMS current to 2.00 mA under the specified conditions.

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Most popular questions from this chapter

A speeder tries to explain to the police that the yellow warning lights on the side of the road looked green to her because of the Doppler shift. How fast would she have been traveling if yellow light of wavelength \(580 \mathrm{~nm}\) had been shifted to green with a wavelength of \(560 \mathrm{~nm}\) ? Note: For speeds less than \(0.03 c\), Equation \(21.32\) will lead to a value for the change of frequency accurate to approximately two significant digits.

An AC voltage of the form \(\Delta v=(90.0 \mathrm{~V})\) sin \((350 t)\) is applied to a series \(R L C\) circuit. If \(R=50.0 \Omega, C=25.0 \mu \mathrm{F}\), and \(L=0.200 \mathrm{H}\), find the (a) impedance of the circuit, (b) rms current in the circuit, and (c) average power delivered to the circuit.

A \(40.0-\mu F\) capacitor is connected to a \(50.0-\Omega\) resistor and a generator whose rms output is \(30.0 \mathrm{~V}\) at \(60.0 \mathrm{~Hz}\). Find (a) the rms current in the circuit, (b) the rms voltage drop across the resistor, (c) the rms voltage drop across the capacitor, and (d) the phase angle for the circuit.

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A resistor \(\left(R=9.00 \times 10^{2} \Omega\right)\), a capacitor \((C=0.250 \mu \mathrm{F})\), and an inductor \((L=2.50 \mathrm{H})\) are connected in series across a \(2.40 \times 10^{2}-\mathrm{Hz}\) AC source for which \(\Delta V_{\max }=\) \(1.40 \times 10^{2} \mathrm{~V}\), Calculate (a) the impedance of the circuit, (b) the maximum current delivered by the source, and (c) the phase angle between the current and voltage. (d) Is the current leading or lagging the voltage?

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