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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

A \(150-\Omega\) resistor is in series with a \(0.75-\mathrm{H}\) inductor in an ac circuit. The rms voltages across the two are the same. (a) What is the frequency? (b) Would each of the rms voltages be half of the rms voltage of the source? If not, what fraction of the source voltage are they? (In other words, \(V_{R} / \ell_{m}=V_{L} / \mathcal{E}_{m}=?\) ) (c) What is the phase angle between the source voltage and the current? Which leads? (d) What is the impedance of the circuit?
A variable capacitor is connected in series to an inductor with negligible internal resistance and of inductance \(2.4 \times 10^{-4} \mathrm{H} .\) The combination is used as a tuner for a radio. If the lowest frequency to be tuned in is \(0.52 \mathrm{MHz}\). what is the maximum capacitance required?
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