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Chapter 12: Problem 1792

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) \(\\{(\mathrm{nE}) /(\mathrm{R}+\mathrm{r})\\}\)

Short Answer

Expert verified
The short answer to the question is: The current through the external resistance R is given by (A) \(\\{(\mathrm{nE}) /(\mathrm{R}+\mathrm{nr})\\}\).

Step by step solution

01

Understand Series Connection of Cells

In a series connection, the cells are connected end-to-end such that the positive terminal of one cell is connected to the negative terminal of the adjacent cell. The total e.m.f of the combination is the sum of the e.m.f of each of the cells. Similarly, the total internal resistance of the combination is the sum of the internal resistances of each cell. Total e.m.f of the combination (E_total) = nE Total internal resistance (r_total) = nr
02

Setup the Circuit with the External Resistance

When the external resistance R is connected in series to the combination of the cells, the total resistance of the circuit becomes the sum of the total internal resistance and the external resistance: Total resistance (R_total) = r_total + R = nr + R
03

Apply Ohm's Law to Find the Current

Ohm's Law states that the current passing through a conductor between two points is directly proportional to the voltage across the two points and inversely proportional to the resistance: Current (I) = Voltage (V) / Resistance (R) For this circuit, the total voltage is given by the total e.m.f of the combination E_total, and the total resistance is given by R_total. Therefore, the current passing through the external resistance R can be determined as: I = E_total / R_total = nE / (nr + R) Comparing our answer to the given options, we can see that it matches option (A). So, the current through R is given by: (A) \(\\{(\mathrm{nE}) /(\mathrm{R}+\mathrm{nr})\\}\)

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Most popular questions from this chapter

In the circuit shown in fig the potential difference across \(3 \Omega\) is. (A) \(2 \mathrm{~V}\) (B) \(4 \mathrm{~V}\) (C) \(8 \mathrm{~V}\) (D) \(16 \mathrm{~V}\)

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

At what temperature will the resistance of a copper wire be three times its value at \(0^{\circ} \mathrm{C}\) ? (Given: temperature coefficient of resistance for copper \(=4 \times 10^{-3}{ }^{\circ} \mathrm{C}^{-1}\) ) (A) \(400^{\circ} \mathrm{C}\) (B) \(450^{\circ} \mathrm{C}\) (C) \(500^{\circ} \mathrm{C}\) (D) \(550^{\circ} \mathrm{C}\)

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) \(\left(1 / \mathrm{n}^{2}\right)\)

For a cell of e.m.f \(2 \mathrm{~V}\), a balance is obtained for $50 \mathrm{~cm}\( of the potentiometer wire If the cell is shunted by a \)2 \Omega$ resistor and the balance is obtained across \(40 \mathrm{~cm}\) of the wire, then the internal resistance of the cell is. (A) \(1 \Omega\) (B) \(0.5 \Omega\) (C) \(1.2 \Omega\) (D) \(2.5 \Omega\)
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