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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

Area of cross-section of a copper wire is equal to area of a square of $2 \mathrm{~mm}\( length It carries a current of \)8 \mathrm{~A}$ Find drift velocity of electrons (Density of free electrons in copper $\left.=8 \times 10^{28} \mathrm{~m}^{-3}\right)$ (A) \(1.56 \times 10^{-2} \mathrm{~ms}^{-1}\) (B) \(1.56 \times 10^{-4} \mathrm{~ms}^{-1}\) (C) \(3.12 \times 10^{-2} \mathrm{~ms}^{-1}\) (D) \(3.12 \times 10^{-3} \mathrm{~ms}^{-1}\)

Assertion and reason are given in following questions each question has four options one of them is correct select it. (a) Both assertion and reason are true and the reason is correct reclamation of the assertion. (b) Both assertion and reason are true, but reason is not correct explanation of the assertion. (c) Assertion is true, but the reason is false. (d) Both, assertion and reason are false. Assertion: There is no current in the metals in the absence of electric field. Reason: Motion of free electrons is random. (A) a (B) \(b\) (C) \(\mathrm{c}\) (D) \(\mathrm{d}\)

In each of the following questions, match column \(\mathrm{I}\) and column II and select the correct match out of the four given choices Column I \(\quad\) Column II (a) The unit of electrical resistivity is (p) \(\mathrm{m}^{2} \mathrm{~S}^{-1} \mathrm{~V}^{-1}\) (b) The unit of current density is (q) \(\Omega^{-1} \mathrm{~m}^{-1}\) (c) The unit of electrical conductivity is (r) \(\mathrm{Am}^{-2}\) (d) The unit of electric mobility is (s) \(\Omega \mathrm{m}\) (A) \(a-p, b-q, c-r, d-s\) (B) \(a-s, b-r, c-q, d-p\) (C) \(a-r, b-q, c-p, d-s\) (D) \(a-q, b-r, c-s, d-p\)

Two resistances \(\mathrm{R}_{1}\) and \(\mathrm{R}_{2}\) have effective resistance \(\mathrm{R}_{\mathrm{s}}\) when connected in sires combination and \(R_{p}\) when connected in parallel combination if $\mathrm{R}_{8} \mathrm{R}_{\mathrm{p}}=16\( and \)\left(\mathrm{R}_{1} / \mathrm{R}_{2}\right)=4\( the values of \)\mathrm{R}_{1}\( and \)\mathrm{R}_{2}$ are (A) \(2 \Omega\) and \(0.5 \Omega\) (B) \(1 \Omega\) and \(0.25 \Omega\) (C) \(8 \Omega\) and \(2 \Omega\) (D) \(4 \Omega\) and \(1 \Omega\)

The drift velocity of free electrons through a conducting wire of radius \(\mathrm{r}\), carrying current \(\mathrm{I}\), is \(\mathrm{V}_{\mathrm{d}}\) if the same current is passed through a conductor of radius \(2 \mathrm{r}\) what will be the drift velocity? (A) \(\left(\mathrm{V}_{\mathrm{d}} / 4\right)\) (B) \(\mathrm{V}_{\mathrm{d}}\) (C) \(2 \mathrm{~V}_{\mathrm{d}}\) (D) \(24 \mathrm{~V}_{\mathrm{d}}\)
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