Question 723: Current Electricity

Question

36 cells each of internal resistance $0.5 \Omega$ and emf 1.5 V each are used to send current through an external circuit of $2 \Omega$ resistor. For getting maximum current, $n$ cells are combined in series in each row. If total number of such rows are $m$, then the value of $(n / m)$ will be

Accepted Answer

Given:
emf, $E = 1.5 \, 	ext{V}$
internal resistance per cell, $r = 0.5 \, \Omega$
external resistance, $R = 2 \, \Omega$
total number of cells, $mn = 36$
For optimal current flow using a mixed arrangement of cells:
The relationship between the internal resistance of each series connection and the total external resistance is described by:
$    \frac{nr}{m} = R    $
Plugging in the given values:
$    \frac{n 	imes 0.5}{m} = 2    $
Multiplying the equations for the total number of cells and the condition for maximum current gives:
$    0.5n^2 = 72    $
Solving for $n$:
$    n^2 = 144      $
$    n = 12    $
$    m = \frac{36}{12} = 3    $
Therefore:
$ \frac{n}{m} = \frac{12}{3} = 4 $

Answer

4

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6

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8

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3

Current Electricity

36 cells each of internal resistance 0.5 \Omega and emf 1.5 V each are used to send current through an external circuit of 2 \Omega resistor. For getting maximum current, n cells are combined in series in each row. If total number of such rows are m, then the value of (n / m) will be

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