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Chapter 21: Problem 30

Suppose that an ideal capacitor and an ideal inductor are connected in series in an ac circuit. (a) What is the phase difference between \(v_{\mathrm{C}}(t)\) and \(v_{\mathrm{L}}(t) ?\) [Hint: since they are in series, the same current \(i(t)\) flows through both. \(]\) (b) If the rms voltages across the capacitor and inductor are \(5.0 \mathrm{V}\) and \(1.0 \mathrm{V}\), respectively, what would an ac voltmeter (which reads rms voltages) connected across the series combination read?

Short Answer

Expert verified
Calculate the RMS voltage across the series combination of the capacitor and inductor, given the individual RMS voltages across each component. Answer: The phase difference between the voltage across the capacitor and the voltage across the inductor is \(\pi\) or 180 degrees. The RMS voltage across the series combination is approximately 5.10 V.

Step by step solution

01

Recall the voltage relations for capacitors and inductors in an AC circuit

In an AC circuit with sinusoidal voltage and current, the voltage across a capacitor is given by \(v_C(t) = V_C\sin(\omega t - \frac{\pi}{2})\) and the voltage across an inductor is given by \(v_L(t) = V_L\sin(\omega t + \frac{\pi}{2})\), where \(V_C\) and \(V_L\) are the peak voltages of the capacitor and inductor, respectively, \(\omega\) is the angular frequency, and \(t\) is time.
02

Determine the phase difference

Comparing the expressions for \(v_C(t)\) and \(v_L(t)\), we can see that the phase difference between the voltage across the capacitor and the inductor is \(\Delta \phi = \left(\frac{\pi}{2} + \frac{\pi}{2}\right) = \pi\) or 180 degrees. This means that the voltage across the capacitor and inductor are exactly out of phase with each other. #RMS Voltage Across the Series Combination#
03

Write expressions for RMS voltages

The given RMS voltages across the capacitor and inductor are 5.0 Volts and 1.0 Volts, respectively. Recall that the RMS voltage (\(V_\textrm{RMS}\)) is related to the peak voltage (\(V_\textrm{peak}\)) by the equation \(V_\textrm{RMS} = \frac{V_\textrm{peak}}{\sqrt{2}}\). This allows us to find the peak voltages across the capacitor and inductor: \(V_C = 5.0\sqrt{2}\) and \(V_L = 1.0\sqrt{2}\).
04

Determine the total peak voltage across series combination

Since the voltage across the capacitor and inductor are out of phase, we need to use the Pythagorean theorem to find the total peak voltage across the series combination, \(V_\textrm{total peak}\): $$V_\textrm{total peak} = \sqrt{V_C^2 + V_L^2} = \sqrt{(5.0\sqrt{2})^2 + (1.0\sqrt{2})^2}$$
05

Find the RMS voltage across the series combination

To find the RMS voltage across the series combination, we need to convert the total peak voltage to its RMS equivalent using the formula \(V_\textrm{RMS} = \frac{V_\textrm{peak}}{\sqrt{2}}\). $$V_\textrm{total RMS} = \frac{V_\textrm{total peak}}{\sqrt{2}} = \frac{\sqrt{(5.0\sqrt{2})^2 + (1.0\sqrt{2})^2}}{\sqrt{2}} \approx 5.10 \mathrm{V}$$ So the AC voltmeter connected across the series combination would read approximately 5.10 V.

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Most popular questions from this chapter

Two ideal inductors \((0.10 \mathrm{H}, 0.50 \mathrm{H})\) are connected in series to an ac voltage source with amplitude \(5.0 \mathrm{V}\) and frequency \(126 \mathrm{Hz}\). (a) What are the peak voltages across each inductor? (b) What is the peak current that flows in the circuit?
A variable capacitor with negligible resistance is connected to an ac voltage source. How does the current in the circuit change if the capacitance is increased by a factor of 3.0 and the driving frequency is increased by a factor of \(2.0 ?\)
An inductor has an impedance of \(30.0 \Omega\) and a resistance of $20.0 \Omega\( at a frequency of \)50.0 \mathrm{Hz}$. What is the inductance? (Model the inductor as an ideal inductor in series with a resistor.)
In an RLC circuit, these three clements are connected in series: a resistor of \(20.0 \Omega,\) a \(35.0-\mathrm{mH}\) inductor, and a 50.0 - \(\mu\) F capacitor. The ac source of the circuit has an rms voltage of \(100.0 \mathrm{V}\) and an angular frequency of $1.0 \times 10^{3} \mathrm{rad} / \mathrm{s} .$ Find (a) the reactances of the capacitor and inductor, (b) the impedance, (c) the rms current, (d) the current amplitude, (e) the phase angle, and (f) the rims voltages across each of the circuit elements. (g) Does the current lead or lag the voltage? (h) Draw a phasor diagram.
The circuit shown has a source voltage of \(440 \mathrm{V}\) ms, resistance \(R=250 \Omega\), inductance \(L=0.800 \mathrm{H},\) and capacitance $C=2.22 \mu \mathrm{F} .\( (a) Find the angular frequency \)\omega_{0}$ for resonance in this circuit. (b) Draw a phasor diagram for the circuit at resonance. (c) Find these rms voltages measured between various points in the circuit: \(V_{a b}, V_{b c}, V_{c d}, V_{b d},\) and \(V_{c d},\) (d) The resistor is replaced with one of \(R=125 \Omega\). Now what is the angular frequency for resonance? (e) What is the rms current in the circuit operated at resonance with the new resistor?
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