Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A modern house is wired for \(115 \mathrm{~V}\), and the current is limited by circuit breakers to a maximum of \(200 .\) A. (For the purpose of this problem, treat these as DC quantities.) a) Calculate the minimum total resistance the circuitry in the house can have at any time. b) Calculate the maximum electrical power the house can consume.

Short Answer

Expert verified
Solution: a) Calculate the minimum total resistance: R = V / I R = (115 V) / (200 A) R_min = 0.575 ohms b) Calculate the maximum electrical power: P = V * I P_max = (115 V) * (200 A) P_max = 23,000 W or 23 kW Answer: The minimum total resistance is 0.575 ohms, and the maximum electrical power the house can consume is 23,000 W or 23 kW.

Step by step solution

01

a) Calculate the minimum total resistance

Using Ohm's law, we have V = I * R, where V is voltage, I is current, and R is resistance. We need to find R when V = 115 V and I = 200 A. To get R, we rewrite Ohm's law as: R = V / I Now, substituting the values V = 115 V and I = 200 A, we get: R = (115 \mathrm{~V})/(200\mathrm{~A}) Calculating this, we get the minimum total resistance the circuitry in the house can have at any time: R_min.
02

b) Calculate the maximum electrical power

To calculate the maximum electrical power, we use the power equation: P = V * I Using the same values of V and I as in part a (V = 115 V and I = 200 A), we have: P_max = (115 \mathrm{~V})(200\mathrm{~A}) Calculating this, we get the maximum electrical power the house can consume: P_max.

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electrical Resistance
Electrical resistance is a measure of how much a material opposes the flow of electric current. It's often compared to water flowing through a pipe; just as narrower or longer pipes create more resistance to water flow, certain materials can hinder electrical current more than others.

Ohm's law, which states that voltage (\textbf{V}) equals current (\textbf{I}) times resistance (\textbf{R}), gives us the formula to calculate resistance: \[ R = \frac{V}{I} \].
In the case of the household wiring example, a low resistance allows more current to flow. However, too low resistance can be dangerous because it could result in overheating and potentially start a fire. That's why houses have circuit breakers, to prevent currents becoming too high. More on that in it's own section!
Electrical Power
Electrical power, measured in watts (\textbf{W}), is the rate at which electrical energy is consumed or converted into another form of energy, like heat or light. The formula for electric power is given by \[ P = V \times I \], where \textbf{P} is the power, \textbf{V} is the voltage, and \textbf{I} is the current.

In our household example, the circuit breakers limit the current, ensuring that the power we use stays within safe limits to prevent electrical fires or damage to the electrical system and appliances. Understanding the concept of power is fundamental when designing or modifying electrical systems in homes to ensure safety and energy efficiency.
Circuit Breakers
Circuit breakers are crucial components in household electrical systems designed to prevent overloading, which could lead to electrical fires. These devices automatically stop the electrical flow when the current exceeds a safe level.

The current rating of a circuit breaker, like the given 200 A in our household example, is the maximum current it can handle before it trips off. Calculating the resistance and power helps in determining the appropriate circuit breaker rating for a system. Properly rated and functioning circuit breakers are essential to any safe electrical setup, working as a guard against potential electric hazards.

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

The resistivity of a conductor is \(\rho=1.00 \cdot 10^{-5} \Omega \mathrm{m}\). If a cylindrical wire is made of this conductor, with a crosssectional area of \(1.00 \cdot 10^{-6} \mathrm{~m}^{2},\) what should the length of the wire be for its resistance to be \(10.0 \Omega ?\)

Which of the arrangements of three identical light bulbs shown in the figure draws most current from the battery? a) \(A\) d) All three draw equal current. b) \(B\) e) \(\mathrm{A}\) and \(\mathrm{C}\) are tied for drawing the most current. c) \(C\)

Before bendable tungsten filaments were developed, Thomas Edison used carbon filaments in his light bulbs. Though carbon has a very high melting temperature \(\left(3599^{\circ} \mathrm{C}\right)\) its sublimation rate is high at high temperatures. So carbonfilament bulbs were kept at lower temperatures, thereby rendering them dimmer than later tungsten-based bulbs. A typical carbon-filament bulb requires an average power of \(40 \mathrm{~W}\), when 110 volts is applied across it, and has a filament temperature of \(1800^{\circ} \mathrm{C}\). Carbon, unlike copper, has a negative temperature coefficient of resistivity: \(\alpha=-0.0005^{\circ} \mathrm{C}^{-1}\) Calculate the resistance at room temperature \(\left(20^{\circ} \mathrm{C}\right)\) of this carbon filament.

A constant electric field is maintained inside a semiconductor. As the temperature is lowered, the magnitude of the current density inside the semiconductor a) increases. c) decreases. b) stays the same. d) may increase or decrease.

Two conducting wires have identical lengths \(L_{1}=L_{2}=\) \(L=10.0 \mathrm{~km}\) and identical circular cross sections of radius \(r_{1}=r_{2}=r=1.00 \mathrm{~mm} .\) One wire is made of steel (with resistivity \(\rho_{\text {steel }}=40.0 \cdot 10^{-8} \Omega \mathrm{m}\) ); the other is made of copper (with resistivity \(\left.\rho_{\text {copper }}=1.68 \cdot 10^{-8} \Omega \mathrm{m}\right)\) a) Calculate the ratio of the power dissipated by the two wires, \(P_{\text {copper }} / P_{\text {steel }},\) when they are connected in parallel; a potential difference of \(V=100 . \mathrm{V}\) is applied to them. b) Based on this result, how do you explain the fact that conductors for power transmission are made of copper and not steel?

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