Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 21: Problem 72

A capacitor is rated at \(0.025 \mu \mathrm{F}\). How much rms current flows when the capacitor is connected to a \(110-\mathrm{V}\) rms, \(60.0-\mathrm{Hz}\) line?

Short Answer

Expert verified
Based on the given information, the problem required finding the rms current flowing through a capacitor connected to an AC voltage source. To solve this, we first calculated the capacitive reactance using the formula \(X_C = \frac{1}{2\pi fC}\) and found it to be approximately \(106.22 \Omega\). Then, we used Ohm's law for AC circuits (\(I = \frac{V}{X_C}\)) to find the rms current, which was approximately \(1.036 A\).

Step by step solution

01

Calculate the capacitive reactance

We need to determine the capacitive reactance, denoted as \(X_C\), which can be calculated using the formula: \(X_C = \frac{1}{2\pi fC}\). Here, \(f\) represents the frequency of the AC source and \(C\) represents the capacitance. Plugging in the given values, we have: \(X_C = \frac{1}{2\pi (60.0Hz)(0.025\mu F)}\) Convert the capacitance in farads: \(0.025\mu F = 0.025 \times 10^{-6} F\) Now, calculate \(X_C\): \(X_C = \frac{1}{2\pi (60Hz)(0.025 \times 10^{-6} F)} \approx 106.22 \Omega\)
02

Calculate the rms current using Ohm's law for AC circuits

Ohm's law states that \(I = \frac{V}{X_C}\), where \(I\) is the rms current, \(V\) is the rms voltage, and \(X_C\) is the capacitive reactance calculated in step 1. Plugging in the values, we get: \(I = \frac{110V}{106.22 \Omega} \approx 1.036 A\) Therefore, the rms current flowing through the capacitor is approximately \(1.036 A\).

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.

Start your free trial

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

An RLC series circuit has \(L=0.300 \mathrm{H}\) and \(C=6.00 \mu \mathrm{F}\) The source has a peak voltage of \(440 \mathrm{V}\). (a) What is the angular resonant frequency? (b) When the source is set at the resonant frequency, the peak current in the circuit is \(0.560 \mathrm{A}\). What is the resistance in the circuit? (c) What are the peak voltages across the resistor, the inductor, and the capacitor at the resonant frequency?
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 \(2.0 ?\)
At what frequency does the maximum current flow through a series RLC circuit containing a resistance of \(4.50 \Omega,\) an inductance of \(440 \mathrm{mH},\) and a capacitance of \(520 \mathrm{pF} ?\)
A 6.20 -m \(H\) inductor is one of the elements in a simple RLC series circuit. When this circuit is connected to a \(1.60-\mathrm{kHz}\) sinusoidal source with an \(\mathrm{rms}\) voltage of \(960.0 \mathrm{V},\) an rms current of $2.50 \mathrm{A}\( lags behind the voltage by \)52.0^{\circ} .$ (a) What is the impedance of this circuit? (b) What is the resistance of this circuit? (c) What is the average power dissipated in this circuit?
At what frequency is the reactance of a 6.0 - \(\mu\) F capacitor equal to $1.0 \mathrm{k} \Omega ?$
See all solutions

Recommended explanations on Physics Textbooks

Mechanics and Materials

Read Explanation

Energy Physics

Read Explanation

Electrostatics

Read Explanation

Famous Physicists

Read Explanation

Medical Physics

Read Explanation

Quantum Physics

Read Explanation
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