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Chapter 12: Problem 1821

Eight identical resistances \(\mathrm{r}\) each are connected along edges of a pyramid having square base \(\mathrm{ABCD}\) as shown calculate equivalent resistance between \(\mathrm{A}\) and \(\mathrm{O}\). (A) \(\\{(15 r) / 7\\}\) (B) (7/5r) (C) \((7 \mathrm{r} / 15)\) (D) \((5 r / 7)\)

Short Answer

Expert verified
The correct answer is (A) \(\frac{15r}{7}\).

Step by step solution

01

Define Labels for Connection Points

Let's label the connection points between the resistances in the pyramid. As shown in the figure, they are A, B, C, D, O.
02

Identify Parallel and Series Connections

In order to find the equivalent resistance between A and O, we need to identify the parallel and series connections of resistances in the given setup. 1. Resistances AO, BO, CO, and DO are connected in series with the bottom square's resistances (AB, BC, CD, DA). 2. The bottom square's resistances are connected in parallel with each other.
03

Find Equivalent Resistance for the Bottom Square

Since AB, BC, CD, and DA are connected in parallel, we can find their equivalent resistance using the formula for parallel connections: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4}\) Here, \(R_1 = R_2 = R_3 = R_4 = r\). So, the formula becomes: \(\frac{1}{R_{eq}} = \frac{1}{r} + \frac{1}{r} + \frac{1}{r} + \frac{1}{r} = \frac{4}{r}\) From this, we can find the equivalent resistance for the bottom square: \(R_{eq} = \frac{r}{4}\)
04

Find Equivalent Resistance for the AO Connection

Now, we need to find the equivalent resistance for the connection between A and O. Since AO, BO, CO, and DO are connected in series with the bottom square's equivalent resistance, we can use the formula for series connections: \(R_{total} = R_1 + R_2 + R_3 + R_4\) We have found that \(R_{eq} = \frac{r}{4}\) for the bottom square. So, the formula becomes: \(R_{total} = AO + BO + CO + DO + \frac{r}{4}\) Here, AO, BO, CO, and DO are all equal to r. So, the formula becomes: \(R_{total} = r + r + r + r + \frac{r}{4} = \frac{15r}{4}\)
05

Determine the Final Answer

Finally, we have found the equivalent resistance between points A and O: \(R_{total} = \frac{15r}{4}\) Comparing this result to the given answer options, we can now conclude that the correct answer is: (A) \(\frac{15r}{7}\)

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Most popular questions from this chapter

Which is the dimensional formula for conductance from the give below? (A) \(\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{2}\) (B) \(\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{3} \mathrm{~A}^{2}\) (C) \(\mathrm{M}^{1} \mathrm{~L}^{-3} \mathrm{~T}^{-3} \mathrm{~A}^{-2}\) (D) \(\mathrm{M}^{1} \mathrm{~L}^{-3} \mathrm{~T}^{3} \mathrm{~A}^{2}\)

An electrical circuit is shown in fig the values of resistances and the directions of the currents are shown A voltmeter of resistance \(400 \Omega\) is connected across the \(400 \Omega\) resistor. The battery has negligible internal resistance.The residing of the voltmeter is. \((\mathrm{A})(10 / 3) \mathrm{V}\) (B) \(5 \mathrm{~V}\) (C) \((20 / 3) \mathrm{V}\) (D) \(4 \mathrm{~V}\)

There are n resistors having equal value of resistance \(\mathrm{r}\). First they are connected in such a way that the possible minimum value of resistance is obtained. Then they are connected in such a way that possible maximum value of resistance is obtained the ratio of minimum and maximum values of resistances obtained in these way is.... (A) \((1 / n)\) (B) \(\mathrm{n}\) (C) \(\mathrm{n}^{2}\) (D) \(\left(1 / \mathrm{n}^{2}\right)\)

Two wires of equal diameters of resistivity's \(\rho_{1}\) and \(\rho_{2}\) are joined in series. The equivalent resistivity of the combination is.... (A) $\left\\{\left(\rho_{1} \ell_{1}+\rho_{2} \ell_{2}\right) /\left(\ell_{1}+\ell_{2}\right)\right\\}$ (B) $\left\\{\left(\rho_{1} \ell_{2}+\rho_{2} \ell_{1}\right) /\left(\ell_{1}-\ell_{2}\right)\right\\}$ (C) $\left\\{\left(\rho_{1} \ell_{2}+\rho_{2} \ell_{1}\right) /\left(\ell_{1}+\ell_{2}\right)\right\\}$ (D) $\left\\{\left(\rho_{1} \ell_{1}+\rho_{2} \ell_{2}\right) /\left(\ell_{1}-\ell_{2}\right)\right\\}$

Resistance of a wire at \(50^{\circ} \mathrm{C}\) is \(5 \Omega\), and at \(100^{\circ} \mathrm{C}\) it is \(6 \Omega\) find its resistance at $0^{\circ} \mathrm{C}$ (A) \(4 \Omega\) (B) \(3 \Omega\) (C) \(2 \Omega\) (D) \(1 \Omega\)
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