A wire is in the form of a tetrahedron shown in figure. The resistance of each wire is \(\mathrm{R}\). What is the resistance of the frame between the corners \(\mathrm{A}\) and \(\mathrm{B}\). (A) \((2 \mathrm{R} / 3)\) (B) \(2 \mathrm{R}\) (C) \(\mathrm{R}\) (D) \((\mathrm{R} / 2)\)

Short Answer

Expert verified
The correct answer for the equivalent resistance between vertices A and B of the tetrahedron frame is (B) \(2R\).

Step by step solution

01

Label the vertices

Label the vertices of the tetrahedron frame as A, B, C, and D, where A and B are the corners we are interested in finding the resistance between.
02

Determine resistance combinations

The vertices form a total of 4 wire edges emanating from A (AB, AC, AD) and 3 from B (BC, BD, and BA, which is equal to AB). The resistances are in series with one another, and the resistances terminating at C and D are in parallel.
03

Express resistance of AC and AD in parallel

Since the resistance along AC and AD are R each and parallel, their equivalent resistance can be determined using the equation: \[R_{AC\_AD} =\frac{R \cdot R}{R + R} = \frac{R^2}{2R} = \frac{R}{2}\]
04

Express resistance of BC and BD in parallel

Similarly, the resistances between vertices B and C, and B and D, are in parallel, so we have: \[R_{BC\_BD} =\frac{R \cdot R}{R + R} = \frac{R^2}{2R} = \frac{R}{2}\]
05

Combine equivalent resistances in series

Now, we have the equivalent resistances (R, \(R_{AC\_AD}\), and \(R_{BC\_BD}\)) connected in series along the path from A to B. Therefore, their equivalent resistance is simply the sum of them: \[R_{AB} = R + \frac{R}{2} + \frac{R}{2} = R + R = 2R\]
06

Select the correct answer

The correct answer for the equivalent resistance between vertices A and B is (B) 2R.

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!

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 simple meter-bridge circuit, the both gaps are bridge by coils \(\mathrm{P}\) and \(\mathrm{Q}\) having the smaller resistance. A balance is obtained when the jockey key makes contact at a point of the bridge wire $40 \mathrm{~cm}\( from the \)\mathrm{P}\( end. On shunting the coil \)\mathrm{Q}$ with a resistance of \(50 \Omega\) the balance point is moved through $10 \mathrm{~cm}\(. What are the resistance of \)\mathrm{P}\( and \)\mathrm{Q}$ ? (A) \([(100) / 3] \Omega,[(100) / 2] \Omega\) respectively (B) \([(50) / 3] \Omega,[(50) / 2] \Omega\) respectively (C) \([(25) / 3] \Omega,[(25) / 2] \Omega\) respectively (D) \([(75) / 3] \Omega,[(75) / 2] \Omega\) respectively

What is the resistance of an open key? (A) \(\infty\) (B) Can't be determined (C) 0 (D) depends on the other resistance in the circuit

In meter bridge experiment, A thin uniform wire \(\mathrm{AB}\) of length $1 \mathrm{~m}\( and unknown resistance \)\mathrm{x}\( and a resistance of \)12 \Omega$ are connected. In the above question, after appropriate conditions are made, it is found that no deflection takes places in the galvanometer when the sliding jockey touches the wire at a distance of \(60 \mathrm{~cm}\) from \(\mathrm{A}\). What is the value of the resistance \(\mathrm{X}\) ? (A) \(18 \Omega\) (B) \(8 \Omega\) (C) \(16 \Omega\) (D) \(4 \Omega\)

What is the unit of temperature coefficient of resistance? (A) \(\Omega^{-1}{ }^{\circ} \mathrm{C}\) (B) \(\Omega^{1}{ }^{\circ} \mathrm{C}^{-1}\) (C) \({ }^{\circ} \mathrm{C}^{-1}\) (D) \(\Omega^{0}{ }^{\circ} \mathrm{C}^{-1}\)

See all solutions

Recommended explanations on English 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