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An inductor \((L=400 \mathrm{mH}),\) a capacitor \((C=4.43 \mu \mathrm{F}),\) and a resistor \((R=500 \Omega)\) are connected in series. A \(50.0-\mathrm{Hz}\) AC source produces a peak current of 250 \(\mathrm{mA}\) in the circuit. (a) Calculate the required peak voltage \(\Delta V_{\text { max }}\) (b) Determine the phase angle by which the current leads or lags the applied voltage.

Short Answer

Expert verified
The peak voltage \(\Delta V_{\text{max}}\) calculated using impedance is required to create the peak current. The phase angle \(\phi\) is calculated from the reactances and resistance, which determines if the current leads or lags the voltage.

Step by step solution

01

Determine Reactances

First calculate the inductive reactance (XL) using the formula: \(X_L = 2\pi f L\). Similarly, calculate the capacitive reactance (XC) using the formula: \(X_C = \frac{1}{2\pi f C}\), where \(f\) is the frequency, \(L\) is the inductance, and \(C\) is the capacitance.
02

Calculate Impedance

Calculate the total impedance (Z) of the circuit using the formula: \(Z = \sqrt{R^2 + (X_L - X_C)^2}\). The impedance gives us information about the total opposition that the circuit offers to the alternating current.
03

Compute Peak Voltage

Use Ohm's Law for AC circuits to compute the peak voltage: \(\Delta V_{\text{max}} = I_{\text{max}} \cdot Z\), where \(I_{\text{max}}\) is the peak current.
04

Calculate Phase Angle

Determine the phase angle \(\phi\) using the formula: \(\phi = \arctan\left(\frac{X_L - X_C}{R}\right)\). The result will be in radians. A positive angle here means the current leads the voltage (which occurs when the circuit is capacitive), while a negative angle implies the current lags the voltage (which occurs when the circuit is inductive).

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Key Concepts

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

Inductive Reactance
Inductive reactance is a measure of the opposition an inductor presents to the flow of alternating current (AC). It is directly proportional to both the frequency of the AC signal and the inductance of the coil. In essence, as the frequency or the inductance increases, so does the inductive reactance.

An easy way to understand inductive reactance is to visualize water flowing through a pipe with a spring-loaded piston inside. The piston's resistance to movement represents the inductive reactance. As the water (current) tries to change direction more quickly (higher frequency), the piston finds it harder to move back and forth, increasing the opposition to flow.

Mathematically, inductive reactance (\(X_L\)) is calculated using the formula: \(X_L = 2\pi f L\), where \(f\) is the AC frequency and \(L\) is the inductance. In our exercise, using this formula helps us find how much the inductor is resisting the AC flow, which is necessary for further analysis of the circuit.
Capacitive Reactance
In contrast to inductive reactance, capacitive reactance measures the opposition a capacitor presents to the change in voltage across it—effectively resisting changes in the electrical current's flow. Unlike inductive reactance, capacitive reactance decreases with an increase in frequency. This is because a capacitor charged at a higher frequency has less time to build up charge, and hence opposes the current flow less.

Using the piston analogy again, if the piston were instead a stretchy membrane that allowed water to accumulate on one side, the faster the water tries to change direction, the less opposition the membrane would present.

The formula for capacitive reactance (\(X_C\)) is given by: \(X_C = \frac{1}{2\pi f C}\), where \(C\) represents capacitance. By calculating this value, we can determine the extent of the opposition to the flow of AC caused by the capacitor in our example exercise.
Impedance
Impedance is a comprehensive measure that describes the total opposition that a circuit presents to alternating current. It not only considers the resistive elements, like the resistor in our exercise, but also the complex interaction of capacitive and inductive reactances. Impedance is important because it dictates how much current will flow for a given voltage.

Think of impedance as a combination of both resistance and reactance in an AC circuit. It's like a complex kind of friction that currents face when they flow through different components, affecting both the magnitude and the timing of the current.

The formula we used to calculate impedance (\(Z\)) is: \(Z = \sqrt{R^2 + (X_L - X_C)^2}\). It combines the resistance (\(R\)), inductive reactance (\(X_L\)), and capacitive reactance (\(X_C\)), giving us a complete picture of how the circuit opposes the electrical current.
Phase Angle
The phase angle in an AC circuit is a measure that describes how much the current waveform is ahead of or behind the voltage waveform. It is a crucial concept in AC analysis because it has significant effects on the efficiency and power transfer within a circuit.

If the phase angle is zero, the voltage and current are said to be 'in phase,' which means they reach their peak values simultaneously. Conversely, if the current leads or lags the voltage, then they are 'out of phase,' which results in a non-zero phase angle. The leading or lagging occurs due to the presence of capacitors and inductors, respectively.

We calculate the phase angle (\(\phi\)) using the formula: \(\phi = \arctan\left(\frac{X_L - X_C}{R}\right)\). If the angle is positive, the current leads the voltage (the circuit acts like a capacitor), and if it is negative, the current lags (the circuit behaves as an inductor). Our exercise guides us to compute this value to determine the current-voltage relationship in the given AC circuit.

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Most popular questions from this chapter

An RLC circuit is used in a radio to tune into an FM station broadcasting at 99.7 \(\mathrm{MHz}\) . The resistance in the circuit is \(12.0 \Omega,\) and the inductance is 1.40\(\mu \mathrm{H}\) . What capacitance should be used?

An audio amplifier, represented by the AC source and resistor in Figure P33.7, delivers to the speaker alternating voltage at audio frequencies. If the source voltage has an amplitude of \(15.0 \mathrm{V}, R=8.20 \Omega,\) and the speaker is equivalent to a resistance of \(10.4 \Omega,\) what is the time- averaged power transferred to it?

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A resistor \(R,\) inductor \(L,\) and capacitor \(C\) are connected in series to an AC source of rms voltage \(\Delta V\) and variable frequency. Find the energy that is delivered to the circuit during one period if the operating frequency is twice the resonance frequency.

A \(10.0-\Omega\) resistor, \(10.0-\mathrm{mH}\) inductor, and \(100-\mu \mathrm{F}\) capacitor are connected in series to a \(50.0-\mathrm{V}\) (rms) source having variable frequency. Find the energy that is delivered to the circuit during one period if the operating frequency is twice the resonance frequency.

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