/*! 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 15 A cell of emf \(E\) is connected... [FREE SOLUTION] | 91影视

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Chapter 14: Problem 15

A cell of emf \(E\) is connected across a resistance \(R\). The potential difference between the terminals of the cell is found to be \(V\). The internal resistance of the cell must be (A) \(\frac{2(E-V) V}{R}\) (B) \(\frac{2(E-V) R}{E}\) (C) \(\frac{(E-V) R}{V}\) (D) \((E-V) R\)

Short Answer

Expert verified
The internal resistance of the cell is given by option (C) \(r = \frac{(E - V) R}{V}\).

Step by step solution

01

Use Kirchhoff's law and Ohm's law

: According to Kirchhoff's law, the sum of the emf and potential difference across the internal resistance of the cell is equal to the potential difference across the external resistance R. Let's denote the current passing through the circuit as I and the internal resistance of the cell as r. From Ohm's law, we know that \(I = \frac{V}{R}\), where V is the potential difference across the resistance R. Using Kirchhoff's law: \[E - Ir = V\]
02

Substitute the value of current (I) from Ohm's law

: As \(I = \frac{V}{R}\), we can substitute it back into the equation from the first step: \(E - \frac{V}{R}r = V\) Now we will solve for r, the internal resistance of the cell.
03

Rearrange the equation to solve for r

: We need to isolate r. First, multiply both sides by R to get rid of the fraction: \(ER - Vr = VR\) Now, we move all the terms containing r to one side and everything else to the other side: \(Vr = E - V)R\) Finally, we divide both sides by V to solve for r: \(r = \frac{(E - V)R}{V}\) Therefore, the internal resistance of the cell is given by the option (C) \(\frac{(E - V) R}{V}\).

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Most popular questions from this chapter

In the circuit shown, the current in the resistor is (A) \(0 \mathrm{~A}\) (B) \(0.13 \mathrm{~A}\), from \(Q\) to \(P\) (C) \(0.13 \mathrm{~A}\), from \(P\) to \(Q\) (D) \(1.3 \mathrm{~A}\), from \(P\) to \(Q\)

The resistance of a bulb filament is \(100 \Omega\) at a temperature \(100^{\circ} \mathrm{C}\). If its temperature coefficient of resistance be \(0.005 /{ }^{\circ} \mathrm{C}\), its resistance will become \(200 \Omega\) at a temperature of (A) \(300^{\circ} \mathrm{C}\) (B) \(400^{\circ} \mathrm{C}\) (C) \(500^{\circ} \mathrm{C}\) (D) \(200^{\circ} \mathrm{C}\)c

Figure \(14.47\) shows the circuit of a potentiometer. The length of the potentiometer wire \(A B\) is \(50 \mathrm{~cm}\). The EMF \(E_{1}\) of the battery is \(4 \mathrm{~V}\), having negligible internal resistance. Value of \(R_{1}\) and \(R_{2}\) are \(15 \Omega\) and \(5 \Omega\), respectively. When both the keys are open, the null point is obtained at a distance of \(31.25 \mathrm{~cm}\) from \(A\), but when both the keys are closed, the balance length reduces to \(5 \mathrm{~cm}\) only. Given \(R_{A B}=10 \Omega\) The internal resistance of the cell \(E_{2}\) is (A) \(4.5 \Omega\) (B) \(5.5 \Omega\) (C) \(6.5 \Omega\) (D) \(7.5 \Omega\)

The temperature dependence of resistances of \(\mathrm{Cu}\) and Si (not doped) in the temperature range \(300-400 \mathrm{~K}\), is best described by (A) linear increase for Cu, exponential increase for Si. (B) linear increase for Cu, exponential decrease for Si. (C) linear decrease for Cu, linear decrease for Si. (D) linear increase for Cu, linear increase of Si.

The resistance of the series combination of two resistances is \(S\). When they are joined in parallel, the total resistance is \(P\). If \(S=n P\), then the minimum possible value of \(n\) is (A) 4 (B) 3 (C) 2 (D) 1
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