Identical resistors are connected to separate 12 V AC sources. One source operates at 60 Hz, the other at 120 Hz. In which circuit, if either, does the resistor dissipate the greater average power?
Answer to Problem 1CQ
Explanation of Solution
The expression to find the average power dissipated by a resistor in an AC circuit is given as follows.
Here,
R is the resistance circular path of radius.
The rms voltage value of 60 Hz AC source and 120 Hz AC source is 12 V.
From (1), it can be defined that the average power dissipation is directly proportional to square of the rms voltage and does not depend on the operating frequency. So, the average power dissipation remains constant irrespective of frequency changes with same rms voltage.
Therefore, the average power dissipated by a resistor in both circuits (a circuit with 60 Hz AC source and a circuit with 120 Hz Ac source) is same.
Conclusion:
Hence, the power dissipated in both circuits is same.
Want to see more full solutions like this?
Chapter 26 Solutions
College Physics: A Strategic Approach (4th Edition)
Additional Science Textbook Solutions
Campbell Essential Biology (7th Edition)
Applications and Investigations in Earth Science (9th Edition)
Genetic Analysis: An Integrated Approach (3rd Edition)
Organic Chemistry (8th Edition)
Anatomy & Physiology (6th Edition)
Microbiology with Diseases by Body System (5th Edition)
- Find the total capacitance in micro farads of the combination of capacitors shown in the figure below. HF 5.0 µF 3.5 µF №8.0 μLE 1.5 µF Ι 0.75 μF 15 μFarrow_forwardthe answer is not 0.39 or 0.386arrow_forwardFind the total capacitance in micro farads of the combination of capacitors shown in the figure below. 2.01 0.30 µF 2.5 µF 10 μF × HFarrow_forward
- I do not understand the process to answer the second part of question b. Please help me understand how to get there!arrow_forwardRank the six combinations of electric charges on the basis of the electric force acting on 91. Define forces pointing to the right as positive and forces pointing to the left as negative. Rank in increasing order by placing the most negative on the left and the most positive on the right. To rank items as equivalent, overlap them. ▸ View Available Hint(s) [most negative 91 = +1nC 92 = +1nC 91 = -1nC 93 = +1nC 92- +1nC 93 = +1nC -1nC 92- -1nC 93- -1nC 91= +1nC 92 = +1nC 93=-1nC 91 +1nC 92=-1nC 93=-1nC 91 = +1nC 2 = −1nC 93 = +1nC The correct ranking cannot be determined. Reset Help most positivearrow_forwardPart A Find the x-component of the electric field at the origin, point O. Express your answer in newtons per coulomb to three significant figures, keeping in mind that an x component that points to the right is positive. ▸ View Available Hint(s) Eoz = Η ΑΣΦ ? N/C Submit Part B Now, assume that charge q2 is negative; q2 = -6 nC, as shown in (Figure 2). What is the x-component of the net electric field at the origin, point O? Express your answer in newtons per coulomb to three significant figures, keeping in mind that an x component that points to the right is positive. ▸ View Available Hint(s) Eoz= Η ΑΣΦ ? N/Carrow_forward
- 1. A charge of -25 μC is distributed uniformly throughout a spherical volume of radius 11.5 cm. Determine the electric field due to this charge at a distance of (a) 2 cm, (b) 4.6 cm, and (c) 25 cm from the center of the sphere. (a) = = (b) E = (c)Ẻ = = NC NC NCarrow_forward1. A long silver rod of radius 3.5 cm has a charge of -3.9 ис on its surface. Here ŕ is a unit vector ст directed perpendicularly away from the axis of the rod as shown in the figure. (a) Find the electric field at a point 5 cm from the center of the rod (an outside point). E = N C (b) Find the electric field at a point 1.8 cm from the center of the rod (an inside point) E=0 Think & Prepare N C 1. Is there a symmetry in the charge distribution? What kind of symmetry? 2. The problem gives the charge per unit length 1. How do you figure out the surface charge density σ from a?arrow_forward1. Determine the electric flux through each surface whose cross-section is shown below. 55 S₂ -29 S5 SA S3 + 9 Enter your answer in terms of q and ε Φ (a) s₁ (b) s₂ = -29 (C) Φ զ Ερ (d) SA = (e) $5 (f) Sa $6 = II ✓ -29 S6 +39arrow_forward
- No chatgpt pls will upvotearrow_forwardthe cable may break and cause severe injury. cable is more likely to break as compared to the [1] ds, inclined at angles of 30° and 50° to the vertical rings by way of a scaled diagram. [4] I 30° T₁ 3cm 3.8T2 cm 200 N 50° at it is headed due North and its airspeed indicat 240 km/h. If there is a wind of 100 km/h from We e relative to the Earth? [3]arrow_forwardCan you explain this using nodal analysis With the nodes I have present And then show me how many KCL equations I need to write, I’m thinking 2 since we have 2 dependent sourcesarrow_forward
- Glencoe Physics: Principles and Problems, Student...PhysicsISBN:9780078807213Author:Paul W. ZitzewitzPublisher:Glencoe/McGraw-HillPhysics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage Learning
- College PhysicsPhysicsISBN:9781938168000Author:Paul Peter Urone, Roger HinrichsPublisher:OpenStax CollegeCollege PhysicsPhysicsISBN:9781285737027Author:Raymond A. Serway, Chris VuillePublisher:Cengage LearningPhysics for Scientists and Engineers, Technology ...PhysicsISBN:9781305116399Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning