
Five cells each of emf \[{\rm{E}}\] and internal resistance \[{\rm{r}}\] send the same amount of current through an external resistance \[{\rm{R}}\] whether the cells are connected in parallel or in series. Then the ratio \[\left( {\dfrac{R}{r}} \right)\] is
A. \[2\]
B. \[\dfrac{1}{2}\]
C. \[\dfrac{1}{5}\]
D. \[1\]
Answer
218.4k+ views
Hint: To solve such questions of combinations of cells, one should know the formulas of current and voltage for combinations of cells. By knowing the formulas we can solve the question directly.
Formula Used:
For a cell of emf \[E\] and internal resistance \[r\],
\[I{\rm{ }} = \]Total resistance (net emf)
\[E - V = Ir\]
Where \[I = \]Current, and \[V = \]potential difference across external resistance.
Complete answer:
If the number of cells is connected end to end and the positive terminal of one cell is connected to the negative terminal of the succeeding cell, then it is called a series arrangement of cells.
Given : Number of cells, \[n = 5\], emf of each cell\[ = E\] Internal resistance of each cell \[ = r\] In series, current through resistance \[R\],
\[I = \dfrac{{nE}}{{nr + R}} = \dfrac{{5E}}{{5r + R}}\]
In parallel, current through resistance R \[{I^\prime } = \dfrac{{E}}{{\dfrac{r}{n} + R}} = \dfrac{{nE}}{{r + nR}} = \dfrac{{5E}}{{r + 5R}}\]
According to question, \[I = {I^\prime }\]
Therefore\[\dfrac{{5E}}{{5r + R}} = \dfrac{{5E}}{{r + 5R}}{\rm{ }}\] \[ \Rightarrow \] \[\;5{\rm{ }}r + R = r + 5{\rm{ }}R\]
or \[\ R = r\]
\[\dfrac{R}{r} = 1\]
Therefore, the correct option is D.
Note:The terminal voltage across the cell when the cell is not in use is the equivalent EMF. In a series circuit, every component is joined end to end to create a single path for electricity to flow. In a parallel circuit, each component is connected across every other component, creating two sets of electrically connected points.
Formula Used:
For a cell of emf \[E\] and internal resistance \[r\],
\[I{\rm{ }} = \]Total resistance (net emf)
\[E - V = Ir\]
Where \[I = \]Current, and \[V = \]potential difference across external resistance.
Complete answer:
If the number of cells is connected end to end and the positive terminal of one cell is connected to the negative terminal of the succeeding cell, then it is called a series arrangement of cells.
Given : Number of cells, \[n = 5\], emf of each cell\[ = E\] Internal resistance of each cell \[ = r\] In series, current through resistance \[R\],
\[I = \dfrac{{nE}}{{nr + R}} = \dfrac{{5E}}{{5r + R}}\]
In parallel, current through resistance R \[{I^\prime } = \dfrac{{E}}{{\dfrac{r}{n} + R}} = \dfrac{{nE}}{{r + nR}} = \dfrac{{5E}}{{r + 5R}}\]
According to question, \[I = {I^\prime }\]
Therefore\[\dfrac{{5E}}{{5r + R}} = \dfrac{{5E}}{{r + 5R}}{\rm{ }}\] \[ \Rightarrow \] \[\;5{\rm{ }}r + R = r + 5{\rm{ }}R\]
or \[\ R = r\]
\[\dfrac{R}{r} = 1\]
Therefore, the correct option is D.
Note:The terminal voltage across the cell when the cell is not in use is the equivalent EMF. In a series circuit, every component is joined end to end to create a single path for electricity to flow. In a parallel circuit, each component is connected across every other component, creating two sets of electrically connected points.
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