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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

A wire \(50 \mathrm{~cm}\) long and \(1 \mathrm{~mm}^{2}\) in cross-section carries a current of \(4 \mathrm{~A}\) when connected to a \(2 \mathrm{~V}\) battery. The resistivity of the wire is: (A) \(2 \times 10^{-7} \Omega \mathrm{m}\) (B) \(5 \times 10^{-7} \Omega \mathrm{m}\) (C) \(4 \times 10^{-6} \Omega \mathrm{m}\) (D) \(1 \times 10^{-6} \Omega \mathrm{m}\)

If \(\sigma_{1}, \sigma_{2}\), and \(\sigma_{3}\) are the conductance's of three conductor then equivalent conductance when they are joined in series, will be. (A) \(\sigma_{1}+\sigma_{2}+\sigma_{3}\) (B) $\left(1 / \sigma_{1}\right)+\left(1 / \sigma_{2}\right)+\left(1 / \sigma_{3}\right)$ (C) $\left\\{\left(\sigma_{1} \sigma_{2} \sigma_{3}\right) /\left(\sigma_{1}+\sigma_{2}+\sigma_{3}\right)\right\\}$ (D) None of these.

The tungsten filament of bulb has resistance equal to \(18 \Omega\) at \(27^{\circ} \mathrm{C}\) temperature \(0.25 \mathrm{~A}\) of current flows, when \(45 \mathrm{~V}\) is connected to it If $\alpha=4.5 \times 10^{-3} \mathrm{~K}^{-1}$ for a tungsten then find the temperature of the filament. (A) \(2160 \mathrm{~K}\) (B) \(1800 \mathrm{~K}\) (C) \(2070 \mathrm{~K}\) (D) \(2300 \mathrm{~K}\)

An infinite sequence of resistances is shown in the figure. The resultant resistance between \(\mathrm{A}\) and \(\mathrm{B}\) will be, when \(\mathrm{R}_{1}=1 \mathrm{ohm}\) and \(\mathrm{R}_{2}=2 \mathrm{ohm}\) (A) \(3 \Omega\) (B) \(2 \Omega\) (C) \(1 \Omega\) (D) \(1.5 \Omega\)

The effective resistance of a n number of resistors connected in parallel in \(\mathrm{x} \mathrm{ohm}\). When one of the resistors is removed, the effective resistance becomes y ohm. The resistance of the resistor that is removed is.... (A) \(\\{(\mathrm{xy}) /(\mathrm{x}+\mathrm{y})\\}\) (B) \(\\{(\mathrm{xy}) /(\mathrm{y}-\mathrm{x})\\}\) (C) \((\mathrm{y}-\mathrm{x})\) (D) \(\sqrt{x y}\)
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