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A coil of resistance 35.0\(\Omega\) and inductance 20.5 \(\mathrm{H}\) is in series with a capacitor and a \(200-\mathrm{V}\) (rms), \(100-\mathrm{Hz}\) source. The rms current in the circuit is 4.00 \(\mathrm{A}\) . (a) Calculate the capacitance in the circuit. (b) What is \(\Delta V_{\mathrm{rms}}\) across the coil?

Short Answer

Expert verified
The capacitance in the circuit is calculated to be approximately 78.2 \(\mu F\) and the \(\Delta V_{\mathrm{rms}}\) across the coil is approximately 2590 V (rms).

Step by step solution

01

Calculate the Impedance

First, determine the total impedance (Z) of the circuit using the rms voltage (Vrms) and the rms current (Irms). The formula for impedance in an AC circuit is \( Z = \frac{V_{rms}}{I_{rms}} \). Plug in the given values to find the impedance: \( Z = \frac{200 V}{4.00 A} = 50.0 \Omega \).
02

Calculate the Inductive Reactance

Next, calculate the inductive reactance (XL) of the coil using the formula \( X_L = 2\pi f L \), where f is the frequency and L is the inductance. Plug in the given values: \( X_L = 2\pi \cdot 100 Hz \cdot 20.5 H = 12910 \pi \Omega \).
03

Calculate the Capacitive Reactance

Using the impedance formula for an RLC series circuit, \( Z = \sqrt{R^2 + (X_L - X_C)^2} \), where XC is the capacitive reactance, solve for XC by rearranging the formula: \( X_C = X_L - \sqrt{Z^2 - R^2} \). Use the calculated values of Z and XL and the given resistance R to solve for XC.
04

Calculate the Capacitance

After calculating XC, determine the capacitance (C) using the formula \( X_C = \frac{1}{2\pi f C} \). Rearrange it to solve for C: \( C = \frac{1}{2\pi f X_C} \).
05

Calculate RMS Voltage Across the Coil

To find the rms voltage across the coil (\( V_{rms,coil} \)), use the formula that relates the rms current (Irms), the coil resistance (R), and the inductive reactance (XL): \( V_{rms,coil} = I_{rms} \cdot \sqrt{R^2 + X_L^2} \). Plug in the values to find the rms voltage across the coil.

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Key Concepts

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

Inductive Reactance
Inductive reactance (\(X_L\)) is a measure of the opposition that an inductor provides to alternating current (AC) due to the inductor's tendency to resist changes in current flow. It arises from the fact that an inductor generates a self-induced electromotive force (emf) as the current changes over time. This effect is governed by Faraday's law of electromagnetic induction.

The mathematical formula for inductive reactance is expressed as: \(X_L = 2\pi f L\), where \( f \) is the frequency of the AC source and \(L\) is the inductance of the coil. As frequency increases, so does the inductive reactance, making it harder for the current to change. This property is crucial in an RLC series circuit as it affects the overall impedance and the behavior of the circuit under different frequencies.

In our exercise, the inductive reactance was calculated at a frequency of 100 Hz with an inductance of 20.5 H, resulting in a value of \(12910\pi \)\(\Omega\). This significant reactance impacts how the circuit responds to the applied AC voltage and plays a part in determining the overall impedance.
Capacitive Reactance
Capacitive reactance (\(X_C\)) is the measure of a capacitor's opposition to the flow of alternating current (AC). It differs from inductive reactance because a capacitor allows current to flow more readily as the frequency of the AC increases. This is due to the capacitor charging and discharging at a faster rate with higher frequencies.

The formula for capacitive reactance is: \(X_C = \frac{1}{2\pi f C}\), where \( f \) is the frequency and \(C\) is the capacitance. As the frequency goes up, the capacitive reactance decreases. Conversely, a lower frequency results in a higher capacitive reactance, limiting the AC flow.

In our exercise scenario, to find the circuit's capacitance, the given impedance and previously calculated inductive reactance were used to calculate the capacitive reactance. This value then led to determining the required capacitance for the circuit when rearranged to \(C = \frac{1}{2\pi f X_C}\). Capacitive reactance plays a pivotal role alongside inductive reactance in shaping the impedance of the RLC circuit.
RLC Series Circuit
An RLC series circuit consists of a resistor (\(R\)), an inductor (\(L\)), and a capacitor (\(C\)) connected in series with an AC voltage source. The total impedance (\(Z\)) of such a circuit is the complex sum of the individual impedances of the resistor, inductor, and capacitor.

The formula for the impedance of an RLC series circuit is: \(Z = \sqrt{R^2 + (X_L - X_C)^2}\). The impedance is a function of resistance (\(R\)), inductive reactance (\(X_L\)), and capacitive reactance (\(X_C\)). The interplay between \(X_L\)) and \(X_C\)) is crucial, as they tend to cancel each other out. When \(X_L = X_C\), the circuit reaches resonance, where impedance is purely resistive, and the power factor is at its maximum.

In the step-by-step solution provided for the textbook exercise, by knowing the rms voltage and current, we derived the total impedance. Subsequently, calculating \(X_L\)) and \(X_C\)) allowed for the determination of capacitance in the circuit and the rms voltage across the coil. This approach demonstrates how the components of an RLC circuit are interdependent and must be considered together for accurate analysis.
RMS Voltage
RMS voltage, or root mean square voltage, is a way of expressing the average voltage of an alternating current (AC) source. It is particularly useful because it gives us a way to compare AC and direct current (DC) voltages in terms of the power they can deliver; specifically, it's the value of the DC voltage that would provide the same power to a resistor as the AC voltage in question.

The calculation for rms voltage typically involves the peak voltage of the AC signal, but in our exercise, the rms voltage was given as 200 V. Once we've established the rms current (\(I_{rms}\)) flowing through the circuit, we can determine the impedance (\(Z\)) using the formula \(Z = \frac{V_{rms}}{I_{rms}}\). Further, to find the voltage across individual components such as a coil, one can use the rms current and the respective component's impedance. For example, the rms voltage across the coil (\(V_{rms, coil}\)) in an RLC circuit can be found using the formula: \( V_{rms, coil} = I_{rms} \cdot \sqrt{R^2 + X_L^2}\).

Understanding RMS voltage is essential in AC circuit analysis as it provides a basis for calculating power, understanding impedance and ultimately determining the behavior of the circuit under varying electrical conditions.

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Most popular questions from this chapter

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?

In an RLC series circuit that includes a source of alternating current operating at fixed frequency and voltage, the resistance R is equal to the inductive reactance. If the plate separation of the capacitor is reduced to half of its original value, the current in the circuit doubles. Find the initial capacitive reactance in terms of R.

An \(80.0-\Omega\) resistor, a \(200-\mathrm{mH}\) inductor, and a \(0.150-\mu \mathrm{F}\)capacitor are connected in parallel across a 120 \(\mathrm{V}\) ( rms) source operating at 374 \(\mathrm{rad} / \mathrm{s}\) . (a) What is the resonant frequency of the circuit? (b) Calculate the rms current in the resistor, inductor, and capacitor. (c) What rms current is delivered by the source? (d) Is the current leading or lagging behind the voltage? By what angle?

What is the maximum current in a \(2.20-\mu \mathrm{F}\) capacitor when it is connected across (a) a North American electrical outlet having \(\Delta V_{\mathrm{rms}}=120 \mathrm{V}, f=60.0 \mathrm{Hz},\) and \((\mathrm{b})\) What If? a European electrical outlet having \(\Delta V_{\mathrm{rms}}=240 \mathrm{V},\) \(f=50.0 \mathrm{Hz}\) ?

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