
In a mixed grouping of identical cells, $5$ rows are connected in parallel by each row contains $10$ cells. This combination sends a current ${{i}}$ through an external resistance of ${{20 \Omega }}$. If the emf and internal resistance of each cell is ${{1}}{{.5 V}}$ and $1{{ }}\Omega $, respectively then the value of ${{i}}$ is:
A) 0.14
B) 0.25
C) 0.75
D) 0.68
Answer
123.3k+ views
Hint: When n cells of same emf are connected in series combination the value of current, ${{I = }}\dfrac{{{E}}}{{{r}}}$. When n cells of same emf are connected in parallel combination the value of current, ${{I = }}\dfrac{{{E}}}{{{{R + }}\dfrac{{{r}}}{{{n}}}}}$. Parallel combination of mix grouping of cells, the value of current is given by
${{I = }}\dfrac{{{{nE}}}}{{{{R + }}\dfrac{{{{nr}}}}{{{m}}}}}$.
Complete step by step solution:
Given: The number of cells in a row, ${{n = 10}}$
The number of rows, ${{m = 5}}$
Emf of each cell, ${{E = 1}}{{.5 V}}$
Internal resistance of each cell, ${{r = 1 \Omega }}$
Electric current is defined as the rate of flow of charge through a conductor in any cross sectional area in a definite direction. SI unit of electric current is ampere represented by A.
Emf is defined as the maximum potential difference between the terminal of the cell when no current is drawn from the cell or the circuit is open. SI unit of emf is voltage represented by V.
Internal resistance is the resistance offered by the electrolyte of the cells to the flow of charge. SI unit of internal resistance is ohms represented by $\Omega $.
Formula for current in mix grouping of cells is given by
$\Rightarrow {{i = }}\dfrac{{{{nE}}}}{{{{R + }}\dfrac{{{{nr}}}}{{{m}}}}}...{{(i)}}$
Where total resistance ${{ = R + }}\dfrac{{{{nr}}}}{{{m}}}$
Now substituting the given values in formula (i), we get
$
\Rightarrow {{i = }}\dfrac{{{{10 \times 15}}}}{{{{20 + }}\dfrac{{{{10 \times 1}}}}{{{5}}}}} \\
\Rightarrow {{i = }}\dfrac{{{{55}}}}{{{{22}}}} \\
\therefore {{i = 0}}{{.68 A}} $
Note: When the cells are joined in a mix group, then to get the maximum current the load resistance must be equal to the internal resistance of the cell. The value of emf of the cells connected in parallel combination remains the same i.e. ${{{E}}_{{{eqv}}}}{{ = E}}$.
For example: If 2 cells are connected in parallel or 100 cells connected in parallel combination the value of equivalent emf remains the same in both the cases.
${{I = }}\dfrac{{{{nE}}}}{{{{R + }}\dfrac{{{{nr}}}}{{{m}}}}}$.
Complete step by step solution:
Given: The number of cells in a row, ${{n = 10}}$
The number of rows, ${{m = 5}}$
Emf of each cell, ${{E = 1}}{{.5 V}}$
Internal resistance of each cell, ${{r = 1 \Omega }}$
Electric current is defined as the rate of flow of charge through a conductor in any cross sectional area in a definite direction. SI unit of electric current is ampere represented by A.
Emf is defined as the maximum potential difference between the terminal of the cell when no current is drawn from the cell or the circuit is open. SI unit of emf is voltage represented by V.
Internal resistance is the resistance offered by the electrolyte of the cells to the flow of charge. SI unit of internal resistance is ohms represented by $\Omega $.
Formula for current in mix grouping of cells is given by
$\Rightarrow {{i = }}\dfrac{{{{nE}}}}{{{{R + }}\dfrac{{{{nr}}}}{{{m}}}}}...{{(i)}}$
Where total resistance ${{ = R + }}\dfrac{{{{nr}}}}{{{m}}}$
Now substituting the given values in formula (i), we get
$
\Rightarrow {{i = }}\dfrac{{{{10 \times 15}}}}{{{{20 + }}\dfrac{{{{10 \times 1}}}}{{{5}}}}} \\
\Rightarrow {{i = }}\dfrac{{{{55}}}}{{{{22}}}} \\
\therefore {{i = 0}}{{.68 A}} $
Note: When the cells are joined in a mix group, then to get the maximum current the load resistance must be equal to the internal resistance of the cell. The value of emf of the cells connected in parallel combination remains the same i.e. ${{{E}}_{{{eqv}}}}{{ = E}}$.
For example: If 2 cells are connected in parallel or 100 cells connected in parallel combination the value of equivalent emf remains the same in both the cases.
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