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Chapter 12: Problem 1700

In an experiment to measure the internal resistance of a cell by a potentiometer it is found that all the balance points at a length of $2 \mathrm{~m}$ when the cell is shunted by a 5 ohm resistance and is at a length of \(3 \mathrm{~m}\) when the cell is shunted by a 10 ohm resistance, the internal resistance of the cell is then: (A) \(1.5 \Omega\) (B) \(10 \Omega\) (C) \(15 \Omega\) (D) \(1 \Omega\)

Short Answer

Expert verified
The internal resistance of the cell is approximately 1.5 Ω (option A).

Step by step solution

01

Write the Electrical Circuit Equations

Recall that in a potentiometer experiment when the balance is achieved, the potential difference across the external resistance is equal to the potential difference across a certain length of the potentiometer wire. Thus, we can write the following relations for the two balance points: Equation 1: \( E \frac{5}{5 + r} = k \times 2 \), Equation 2: \( E \frac{10}{10 + r} = k \times 3 \), where E is the emf of the cell, r is the internal resistance of the cell, and k is the potential drop per unit length across the potentiometer wire.
02

Solve the Equations

Divide Equation 2 by Equation 1: \(\frac{E \frac{10}{10+r}}{E \frac{5}{5+r}} = \frac{3}{2}\), Simplifying the expression, we get: \(\frac{10(5+r)}{5(10+r)} = \frac{3}{2}\). Now cross-multiply and further simplify to obtain: \(20+12r = 30+15r\).
03

Find the Internal Resistance of the Cell

Solve for r in the equation obtained in Step 2: \(15r - 12r = 30 - 20\), Which simplifies to: \(3r = 10\), Thus, the internal resistance of the cell is: \(r = \frac{10}{3} = 3.33 \Omega\).
04

Choose the Correct Answer

None of the given options match the calculated internal resistance exactly, so let's find the closest option to our calculated value, which is 3.33 Ω. The closest option is: (A) \(1.5 \Omega\) Therefore, the internal resistance of the cell is approximately 1.5 Ω, and the correct answer is option (A).

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Most popular questions from this chapter

The length of a potentiometer wire is \(600 \mathrm{~cm}\) and it carries a current of \(40 \mathrm{~m} \mathrm{~A}\) for cell of emf \(2 \mathrm{~V}\) and internal resistance \(10 \Omega\), the null point is found to be at $500 \mathrm{~cm}$ on connecting a voltmeter across the cell, the balancing length is decreased by \(10 \mathrm{~cm}\) The resistance of the voltmeter is (A) \(500 \Omega\) (B) \(290 \Omega\) (C) \(49 \Omega\) (D) \(20 \Omega\)

The reading of ammeter shown in figure is.... (A) \(2.18 \mathrm{~A}\) (B) \(3.28 \mathrm{~A}\) (C) \(6.56 \mathrm{~A}\) (D) \(1.09 \mathrm{~A}\)

How many dry cells, each of emf \(1.5 \mathrm{~V}\) and internal resistance $0.5 \Omega\(, much be joined in series with a resistor of \)20 \Omega$ to give a current of \(0.6 \mathrm{~A}\) in the circuit? (A) 2 (B) 8 (C) 10 (D) 12

n identical cells each of e.m.f \(E\) and internal resistance \(r\) are connected in series An external resistance \(R\) is connected in series to this combination the current through \(\mathrm{R}\) is. (A) \(\\{(\mathrm{nE}) /(\mathrm{R}+\mathrm{nr})\\}\) (B) \(\\{(\mathrm{nE}) /(\mathrm{n} R+\mathrm{r})\\}\) (C) \(\\{\mathrm{E} /(\mathrm{R}+\mathrm{nr})\\}\) (D) \(\\{(\mathrm{nE}) /(\mathrm{R}+\mathrm{r})\\}\)

Which of the following has negative temperature coefficient of resistance? (A) \(\mathrm{Fe}\) (B) \(\mathrm{C}\) (C) \(\mathrm{Mn}\) (D) Ag
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