A series combination of a resistor and a capacitor are connected to a \(110-\mathrm{V}\) rms, \(60.0-\mathrm{Hz}\) ac source. If the capacitance is \(0.80 \mu \mathrm{F}\) and the rms current in the circuit is $28.4 \mathrm{mA},$ what is the resistance?

Short Answer

Expert verified
Answer: The resistance in the circuit is approximately 2146.72Ω.

Step by step solution

01

Determine the impedance

Since we know the RMS voltage and RMS current, we can use the equation \(V = IZ\) to find the impedance. Rearranging for \(Z\), we get \(Z = \frac{V}{I}\). Plugging in the given values, \(Z = \frac{110\,\text{V}}{28.4\times10^{-3}\,\text{A}} = 3873.24\,\Omega\).
02

Determine the reactance of the capacitor

The reactance of the capacitor (\(X_C\)) can be found by using the formula \(X_C = \frac{1}{2\pi fC}\). Plugging in the given values for frequency and capacitance, we get \(X_C = \frac{1}{2\pi(60.0\,\text{Hz})(0.80\times10^{-6}\,\text{F})} = 3316.52\,\Omega\).
03

Determine the resistance

Finally, we can use the impedance formula, \(Z = \sqrt{R^2 + X_C^2}\), to find the resistance. Rearranging for \(R\), we get \(R = \sqrt{Z^2 - X_C^2}\). Plugging in the values for \(Z\) and \(X_C\), we get \(R = \sqrt{(3873.24\,\Omega)^2 - (3316.52\,\Omega)^2} = 2146.72\,\Omega\). The resistance in the circuit is approximately \(2146.72\,\Omega\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A series \(R L C\) circuit has \(R=500.0 \Omega, L=35.0 \mathrm{mH},\) and $C=87.0 \mathrm{pF} .$ What is the impedance of the circuit at resonance? Explain.
The FM radio band is broadcast between \(88 \mathrm{MHz}\) and $108 \mathrm{MHz}$. What range of capacitors must be used to tune in these signals if an inductor of \(3.00 \mu \mathrm{H}\) is used?
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 40.0 -mH inductor, with internal resistance of \(30.0 \Omega\) is connected to an ac source $$\varepsilon(t)=(286 \mathrm{V}) \sin [(390 \mathrm{rad} / \mathrm{s}) t]$$ (a) What is the impedance of the inductor in the circuit? (b) What are the peak and rms voltages across the inductor (including the internal resistance)? (c) What is the peak current in the circuit? (d) What is the average power dissipated in the circuit? (e) Write an expression for the current through the inductor as a function of time.
A parallel plate capacitor has two plates, each of area $3.0 \times 10^{-4} \mathrm{m}^{2},\( separated by \)3.5 \times 10^{-4} \mathrm{m} .$ The space between the plates is filled with a dielectric. When the capacitor is connected to a source of \(120 \mathrm{V}\) rms at \(8.0 \mathrm{kHz}\) an mms current of \(1.5 \times 10^{-4} \mathrm{A}\) is measured. (a) What is the capacitive reactance? (b) What is the dielectric constant of the material between the plates of the capacitor?
See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free