Physics, asked by jkjackal2346, 1 year ago

To get maximum current in a resistance of 3 ohms, one can use n rows of m cells (connected in series) connected in parallel. If the total number of cells is 24 and the internal resistance of a cell is 0.5 ohms then(a) m = 12, n = 2(b) m = 8, n = 3(c) m = 2, n = 12(d) m = 6, n = 4

Answers

Answered by Anonymous
11
Hey mate.....

here's the answer....

To get maximum current in a resistance of 3 ohms, one can use n rows of m cells (connected in series) connected in parallel. If the total number of cells is 24 and the internal resistance of a cell is 0.5 ohms then(a) m = 12, n = 2(b) m = 8, n = 3(c) m = 2, n = 12(d) m = 6, n = 4

\huge{Option \: B}

Hope this helps❤

sarajsingh: hello
Answered by Shazia055
0

Given:

Resistance \[ = 3\Omega \]

Number of rows =n

Number of cells in one row =m

Total number of cells =24

Internal resistance of a cell \[ = 0.5\,\Omega \]

To Find: The value of m and n

Solution:

The equivalent potential for 'm' number of cells connected in series combination is given as:

\[V = {E_1}{r_1} + {E_2}{r_2} + {E_3}{r_3} + {E_4}{r_4} + \,... + {E_m}{r_m}\]

Where, \[{E_m} = \] potential of each cell, \[{r_n} = \] internal resistance of each cell

If the total potential of the cells in the series is E and total internal resistance is R, then,

\[V = mER\]                ... (i)

The potential for a number of cells connected in parallel combination is given as:

\[V = \frac{{{E_1}}}{{{r_1}}} + \frac{{{E_2}}}{{{r_2}}} + \frac{{{E_3}}}{{{r_3}}} + \frac{{{E_4}}}{{{r_4}}} + \,... + \frac{{{E_n}}}{{{r_n}}}\]

If the total potential is E and total internal resistance is r, then,

\[V = n\frac{E}{r}\]                  ... (ii)

As 24 cells are to be connected in n rows of m cells

From Ohm's law, we have,

\[\begin{gathered}  \frac{{mr}}{n} + R = \frac{{mE}}{I} \hfill \\  I = \frac{{mE}}{{\frac{{mr}}{n} + R}} \hfill \\  I = \frac{{mnE}}{{mr + nR}} \hfill \\ \end{gathered} \]               ... (iii)

In order to get maximum current, the value of \[{mr + nR}\] should be minimum.

It will be minimum for \[R = \frac{{mr}}{n}\]          ... (iv)

Therefore, the maximum current is given by

\[I = \frac{{mE}}{{2R}}\]

Putting the values of R and r in equation (iv), we have,

\[\begin{gathered}  \frac{m}{n} = \frac{R}{r} \hfill \\  \frac{m}{n} = \frac{3}{{0.5}} = 6 \hfill \\ \end{gathered} \]

Thus, \[m = 12\] and \[n = 2\]

Hence, the value of m and n is 12 and 2 respectively. Thus, the correct option is (a) m = 12, n = 2.

#SPJ2

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