A \(35-\Omega\) electric stove burner consumes \(1.5 \mathrm{kW}\) of power. At what voltage does it operate?

Short Answer

Expert verified
The required voltage for the operation of the electric stove is approximately \(226.27 \) volts.

Step by step solution

01

Convert power to watts

The power is given in kilowatts. Convert this to watts by multiplying by 1000. Therefore, \( P = 1.5 \times 1000 = 1500W \). It is essential to convert power to watts because the units in the formula for power must match.
02

Insert known values into the formula

Use the formula \(P=V^2/R\) and substitute the given values for power (P) and resistance (R). This gives: \[1500 = V^2/35\].
03

Solve for Voltage (V)

Rearrange the equation to solve for the Voltage V: \[V= \sqrt{1500 \times 35}\]

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.

Electric Power
Understanding electric power is fundamental when learning about how various appliances consume energy. Electric power describes the rate at which electrical energy is transferred by an electric circuit. The standard unit for power is the watt (W). When you use an electrical device, the power it uses is converted into work or heat. For instance, a stove burner transforms electrical power into heat used for cooking.

The power of a device can be calculated by multiplying the voltage (V) across the device by the current (I) flowing through it, which is expressed in the formula: \(P = VI\). However, when the resistance (R) is known instead of the current, the formula can be modified to \(P = V^2 / R\), which demonstrates how power relates to voltage and resistance. This formula is particularly handy when only the power consumption and resistance are available, as seen in the exercise where an electric stove burner’s power consumption helps calculate its operating voltage.
Electric Resistance
The concept of electric resistance is critical for understanding how the flow of electric current is impeded within a conductor. Resistance, measured in ohms (Ω), is a material's tendency to resist the flow of charge. This property is an inherent characteristic of the material and geometrical attributes of the conductor.

An analogy often used for resistance is water flowing through a pipe; just as a narrower pipe restricts water flow, a material with high resistance restricts the flow of electric charge. In our stove burner example, the resistance is provided as \(35-Ω\), which tells us how much the burner impedes the electric current. This value is essential when coupled with the known power consumption to calculate the voltage required for the burner to operate. Knowing resistance can also help in determining the safety and efficiency of electrical devices since materials with high resistance can produce more heat, which needs to be managed for safe operation.
Voltage Calculation
Voltage calculation is an essential skill in electrical engineering that helps you determine the potential difference required to power an appliance. Voltage, which is measured in volts (V), refers to the potential difference between two points in a circuit and is a driving force that pushes electric charges to move through a conductor.

In the exercise, we're required to find the voltage, given the electric power consumption and the resistance of the stove burner. By rearranging the formula \(P = V^2 / R\) to solve for 'V', which stands for voltage, we get \(V = \sqrt{ P \times R }\). Following the steps provided, we insert the known values and simplify to compute the voltage. This approach demonstrates how to deduce the voltage needed for an appliance when you know how much power it consumes, and its resistance.

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 brownout occurs when an electric utility can't supply enough power to meet demand. Rather than cut off some customers completely, the utility reduces the voltage across its system. Brownouts are most likely on hot summer days, when heavy air-conditioning loads drive up demand for electricity. In a particular brownout, the utility reduces the voltage by \(10 \%.\) Which of the following occurs in the conductors of the preceding problem during the brownout? a. Both the electric field and electron drift speed decrease. b. The electric field decreases but the electron drift speed doesn't. c. The current is carried by fewer electrons. d. The electrons undergo more frequent collisions.

An implanted pacemaker supplies the heart with 72 pulses per minute, each pulse providing \(6.0 \mathrm{V}\) for \(0.65 \mathrm{ms}\). The resistance of the heart muscle between the pacemaker's electrodes is \(550 \Omega\). Find (a) the current that flows during a pulse, (b) the energy delivered in one pulse, and (c) the average power supplied by the pacemaker.

A gold film in an integrated circuit measures \(2.5 \mu \mathrm{m}\) thick by \(0.18 \mathrm{mm}\) wide. It carries a current density of \(0.68 \mathrm{MA} / \mathrm{m}^{2}\) What's the total current?

You put a \(1.5-\mathrm{V}\) battery across a piece of material, and a \(100-\mathrm{mA}\) current flows. With a 9 - \(V\) battery, the current increases to 400 mA. Is the material ohmic or not?

A brownout occurs when an electric utility can't supply enough power to meet demand. Rather than cut off some customers completely, the utility reduces the voltage across its system. Brownouts are most likely on hot summer days, when heavy air-conditioning loads drive up demand for electricity. In a particular brownout, the utility reduces the voltage by \(10 \%.\) Metallic conductors like lightbulb filaments and electric stove burners have resistance that increases with increasing temperature. During the brownout, the current in such devices a. decreases by \(10 \%.\) b. decreases by more than \(10 \%.\) c. decreases by less than \(10 \%.\) d. You can't tell without knowing more about how the resistance varies.

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