/*! 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 107 Two heaters designed for the sam... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 107

Two heaters designed for the same voltage \(V\) have different power ratings. When connected individually across a source of voltage \(V\), they produce \(H\) amount of heat each in time \(t_{1}\) and \(t_{2}\), respectively. When used together across the same source, they produce \(H\) amount of heat in time \(t\), (A) if they are in series, \(t=t_{1}+t_{2}\). (B) if they are in series, \(t=2\left(t_{1}+t_{2}\right)\). (C) if they are in parallel, \(t=2\left(t_{1}-t_{2}\right)\). (D) if they are in parallel, \(t=\frac{t_{1} t_{2}}{2\left(t_{1}+t_{2}\right)}\).

Short Answer

Expert verified
The short answer is: when the heaters are connected in parallel, the time taken to produce H amount of heat is \(t=\frac{t_{1}t_{2}}{2\left(t_{1}+t_{2}\right)}\).

Step by step solution

01

Calculate Power Ratings Individually

As the heaters are designed for the same voltage V, let's denote the currents as I1 and I2, and the power ratings as P1 and P2 for the first and second heater respectively. Using the power formula: P1 = V*I1 P2 = V*I2 Since the heat produced is H for both heaters, we can write: H = P1*t1 H = P2*t2
02

Connect Heaters in Series

When two heaters are connected in series, their equivalent resistance (Req) becomes: Req = R1 + R2 And the power consumed by the series connection becomes: P_series = V^2 / (R1 + R2) Using the fact that the total heat produced H is equal to the power consumed over time: H = P_series * t_series Now, we need to find the time t_series when the heat produced is H.
03

Connect Heaters in Parallel

When heaters are connected in parallel, their equivalent resistance becomes: 1/Req = 1/R1 + 1/R2 And the power consumed by the parallel connection becomes: P_parallel = V^2 / Req Following the same reasoning as in Step 2, we can write: H = P_parallel * t_parallel Now, our task is to find the time t_parallel when the heat produced is H.
04

Compare Results to Options

To find the relationship between t_series and t_parallel and t1 and t2, we need to replace P1, P2, P_series, and P_parallel in the expressions above with their corresponding expressions from the previous steps. After substituting and performing some algebraic manipulations, the relationship between t_series and t_parallel and t1 and t2 can be found. Compare these results with the given options to find which one corresponds to the correct expression.
05

Conclusion

After comparing the results found in the previous steps with the given options, we can conclude that the correct answer is: (D) if they are in parallel, \(t=\frac{t_{1}t_{2}}{2\left(t_{1}+t_{2}\right)}\).

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 91影视!

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

In a metre bridge experiment, null point is obtained at \(20 \mathrm{~cm}\) from one end of the wire when resistance \(X\) is balanced against another resistance \(Y\). If \(X

In the arrangement shown, the magnitude of each resistance is \(2 \Omega\). The equivalent resistance between \(O\) and \(A\) is given by (A) \(\frac{14}{15} \Omega\) (B) \(\frac{7}{15} \Omega\) (C) \(\frac{4}{3} \Omega\) (D) \(\frac{5}{6} \Omega\)

The average bulk resistivity of the human body (apart from the surface resistance of the skin) is about \(5 \Omega m\). The conducting path between the hands can be represented approximately as a cylinder \(1.6 \mathrm{~m}\) long and \(0.1 \mathrm{~m}\) diameter. The skin resistance may be made negligible by soaking the hands in salt water. A lethal shock current needed is \(100 \mathrm{~mA}\). Note that a small amount of potential difference could be fatal if the skin is damp. What potential difference is needed between the hands for a lethal shock current? (A) \(100 \mathrm{~V}\) (B) \(10 \mathrm{~V}\) (C) \(120 \mathrm{~V}\) (D) \(150 \mathrm{~V}\)

A potentiometer is a device used for measuring EMF and internal resistance of a cell. It consists of two circuits, one is main circuit in which there is a cell of given emf \(\varepsilon^{\prime}\) and given resistance \(R\) which is connected across a wire of length \(100 \mathrm{~cm}\) and having resistance \(r\) and another circuit having unknown EMF \(\varepsilon\) and galvanometer. For a given potentiometer, if \(\varepsilon^{\prime}=30 \mathrm{~V}, r=1 \Omega\), and resistance \(R\) varies with time \(t\) given by \(R=2 t\). The jockey can move on wire with constant velocity \(10 \mathrm{~cm} / \mathrm{s}\) and switch \(S\) is closed at \(t=0\)If jockey starts moving from \(A\) at \(t=0\) and balancing point found at \(t=1 \mathrm{~s}\) then the value of \(\varepsilon\) is (A) \(1 \mathrm{~V}\) (B) \(2 \mathrm{~V}\) (C) \(3 \mathrm{C}\) (D) \(4 \mathrm{~V}\)

An ammeter is obtained by shunting a \(30 \Omega\) galvanometer with a \(30 \Omega\) resistance. What additional shunt should be connected across it to double the range? (A) \(15 \Omega\) (B) \(10 \Omega\) (C) \(5 \Omega\) (D) None of these
See all solutions

Recommended explanations on Physics Textbooks

Oscillations

Read Explanation

Waves Physics

Read Explanation

Particle Model of Matter

Read Explanation

Nuclear Physics

Read Explanation

Torque and Rotational Motion

Read Explanation

Conservation of Energy and Momentum

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.