/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 135 If two identical heaters each ra... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

If two identical heaters each rated as \((1000 \mathrm{~W}-220 \mathrm{~V})\) are connected in parallel to \(220 \mathrm{~V}\), then the total power consumed is : (a) \(200 \mathrm{~W}\) (b) \(2500 \mathrm{~W}\) (c) \(250 \mathrm{~W}\) (d) \(2000 \mathrm{~W}\)

Short Answer

Expert verified
(d) 2000 W

Step by step solution

01

Understand Parallel Connection

In a parallel connection, each heater is directly connected to the voltage source. Therefore, each heater operates independently at the rated voltage of \(220 \mathrm{~V}\). Given each heater is rated for \(1000 \mathrm{~W}\) at \(220 \mathrm{~V}\), they will each consume their full power rating.
02

Calculate Power for One Heater

One heater rated at \(1000 \mathrm{~W}\) is connected to \(220 \mathrm{~V}\). The power consumed in this case by one heater remains \(1000 \mathrm{~W}\) because the voltage supply matches the heater's rated voltage.
03

Find Total Power Consumption for Two Heaters

Since both heaters are identical and the voltage source is \(220 \mathrm{~V}\), the total power consumed is simply the sum of the power consumed by each heater. Thus, the total power is \(1000 \mathrm{~W} + 1000 \mathrm{~W} = 2000 \mathrm{~W}\).

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 91Ó°ÊÓ!

Key Concepts

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

Power Calculation
Power calculation is an essential aspect of working with electrical circuits. When talking about electrical devices like heaters, power is the rate at which electrical energy is converted into other forms (such as heat). Power is calculated using the formula \( P = VI \), where \( P \) is power in watts, \( V \) is voltage in volts, and \( I \) is current in amperes. Another form of the power formula is \( P = I^2R \) or \( P = \frac{V^2}{R} \), depending on known values.
For a single device like a heater connected to its rated voltage source, the power consumed is exactly the amount it is designed to handle, which means all its rated power is utilized. In the given problem, each heater is consuming \(1000\) watts when connected to a \(220\) volt supply, calculated simply since the rated power matches the supply voltage.
Parallel Circuit Analysis
Understanding parallel circuit analysis is crucial, particularly when dealing with multiple electrical components such as heaters. In a parallel circuit, all components share the same voltage across them but they branch independently. Each branch is connected directly to the source, receiving full voltage.
In our example, two identical heaters are connected in parallel, meaning both heaters receive the full \(220\) volts. A parallel connection allows each heater to operate at its full power since none depends on the power consumed by the other. This feature of parallel circuits prevents components from sharing loads, unlike series circuits where the current is the same but the voltage divides.
  • The total power consumption in a parallel circuit is the sum of individual power consumptions. Each electrical device runs separately even though it shares the same source voltage.
  • This understanding simplifies calculations, since each component operates independently at the source voltage.
Electrical Rating
An electrical rating is a specification that tells us the voltage and power at which a device is designed to operate safely and efficiently. It gives the maximum voltage and power level that a device should handle for optimal performance. We often see this rating on household appliances.
For a heater rated \(1000\) watts at \(220\) volts, it means the heater will consume \(1000\) watts only when supplied with exactly \(220\) volts. Supplying lower or higher voltage may cause inefficient performance or damage. Hence, while setting up electrical circuits, the electrical rating helps us ensure devices receive the correct power supply.
  • The rating is essential for predicting power consumption and ensuring safe operation within electrical systems.
  • It provides valuable guidance for the correct setup, especially in circuits with multiple appliances, like parallel circuits.
ratings are key for both safety and efficiency, ensuring devices work optimally within specified parameters.

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 steady current of \(10 \mathrm{~A}\) is passed through a water voltameter for \(300 \mathrm{sec}\). Assume relative molecular mass of \(\mathrm{H}_{2}\) is \(2.016\) and molar volume \(=22.4\) litre (volume of 1 mole of an ideal gas at STP.). Estimate the volume of \(\mathrm{H}_{2}\) evolved at standard temperature and pressure (Given: Faraday's constant \(=96500 \mathrm{C} / \mathrm{mole}\) ) (a) \(0.178\) litre (b) \(0.278\) litre (c) \(1.78\) litre (d) \(2.78\) litre

A silver and copper voltameters are connected in parallel to a \(12 \mathrm{~V}\) battery of negligible resistance. At what rate is energy being delivered by the battery, if in 30 minutes, \(1 \mathrm{~g}\) of silver and \(1.8 \mathrm{~g}\) of copper are deposited ? (Assume electrochemical equivalent of silver \(=11.2 \times 10^{-7} \mathrm{~kg} / \mathrm{C}\), electrochemical equivalent of copper \(\left.=6.6 \times 10^{-7} \mathrm{~kg} / \mathrm{C}\right)\) (a) \(42.2 \mathrm{~J} / \mathrm{s}\) (b) \(40.4 \mathrm{~J} / \mathrm{s}\) (c) \(24.1 \mathrm{~J} / \mathrm{s}\) (d) \(20.4 \mathrm{~J} / \mathrm{s}\)

Which of the following arrangements is correct on the basis of conductivity of materials ? (a) silver \(>\) copper \(>\) aluminium \(>\) tungsten \(>\) water (b) silver \(>\) aluminium \(>\) copper \(>\) water (c) copper \(>\) silver \(>\) tungsten \(>\) water (d) water \(>\) silver \(>\) tungsten \(>\) water

An electric bulb is marked \(100 \mathrm{~W}, 230 \mathrm{~V}\). The resistance of the bulb is : (a) \(229 \Omega\) (b) \(329 \Omega\) (c) \(429 \Omega\) (d) \(529 \Omega\)

A wire with a resistance of \(20 \Omega\) is drawn out so that its length becomes thrice its original length. The new resistance is: (a) \(60 \Omega\) (b) \(\frac{2 \theta}{3} \Omega\) (c) \(180 \Omega\) (d) \(18 \Omega\)

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.