Answer
425.4k+ views
Hint: For n cells with internal resistance r connected in series,
${E_{eq}} = nE$ and ${r_{eq}} = nr$
For n cells with internal resistance r connected in parallel,
${E_{eq}} = E$ and ${r_{eq}} = \dfrac{r}{n}$
Here ${E_{eq}}$ is the equivalent emf of all the battery sources and ${r_{eq}}$ is the equivalent internal resistance.
Current for both the connection is given by $I = \dfrac{{{E_{eq}}}}{{{R_{eq}}}}$ where ${R_{eq}}$ is the total equivalent resistance including internal and external resistance.
Complete step by step solution:
First, we have to calculate the equivalent EMF and total equivalent resistance for the type of connections (series and parallel).
For n cells with internal resistance r connected in series,
${E_{eq}}$ is the algebraic sum of all the EMFs i.e. ${E_{eq}} = nE$
As all the internal resistances are also in series, so the equivalent internal resistance will be ${r_{eq}} = nr$
Now as the equivalent internal resistance and the external resistance R are in series connection, so the total equivalent resistance for the circuit will be given as, ${R_{eq}} = R + {r_{eq}} = R + nr$
Current for both the connection is given by $I = \dfrac{{{E_{eq}}}}{{{R_{eq}}}}$ where ${R_{eq}}$ is the total equivalent resistance including internal and external resistance.
Let current through this series connection be ${I_s}$
So, ${I_s} = \dfrac{{nE}}{{R + nr}}$
Now, for n cells with internal resistance r connected in parallel,
${E_{eq}}$ in the parallel connection will be ${E_{eq}} = E$ as the EMF will remain the same.
As all the internal resistances are also in parallel, so the equivalent internal resistance will be ${r_{eq}} = \dfrac{r}{n}$
Now as the equivalent internal resistance and the external resistance R are in series connection, so the total equivalent resistance for the circuit will be given as, ${R_{eq}} = R + {r_{eq}} = R + \dfrac{r}{n}$
Current for both the connection is given by $I = \dfrac{{{E_{eq}}}}{{{R_{eq}}}}$ where ${R_{eq}}$ is the total equivalent resistance including internal and external resistance.
Let current through this series connection be ${I_p}$
So, ${I_p} = \dfrac{E}{{\left( {R + \dfrac{r}{n}} \right)}}$
Now, as given in the question that the current in the circuit is the same whether the cells are connected in series or in parallel which means ${I_s} = {I_p}$
So, $\dfrac{{nE}}{{R + nr}} = \dfrac{E}{{\left( {R + \dfrac{r}{n}} \right)}}$
On further solving we have,
$\dfrac{{nE}}{{R + nr}} = \dfrac{{nE}}{{nR + r}}$
Or we can say, $R + nr = nR + r$
On simplifying we get,
$r = R$
$\therefore$The internal resistance r is equal to $R$. Hence, option (C) is the correct answer.
Note:
While calculating the overall equivalent resistance after calculating equivalent internal resistance, remember that ${r_{eq}}$ and the external resistance R will be in series.
Remember that the equivalent EMF for the parallel connection will remain as original.
${E_{eq}} = nE$ and ${r_{eq}} = nr$
For n cells with internal resistance r connected in parallel,
${E_{eq}} = E$ and ${r_{eq}} = \dfrac{r}{n}$
Here ${E_{eq}}$ is the equivalent emf of all the battery sources and ${r_{eq}}$ is the equivalent internal resistance.
Current for both the connection is given by $I = \dfrac{{{E_{eq}}}}{{{R_{eq}}}}$ where ${R_{eq}}$ is the total equivalent resistance including internal and external resistance.
Complete step by step solution:
First, we have to calculate the equivalent EMF and total equivalent resistance for the type of connections (series and parallel).
For n cells with internal resistance r connected in series,
${E_{eq}}$ is the algebraic sum of all the EMFs i.e. ${E_{eq}} = nE$
As all the internal resistances are also in series, so the equivalent internal resistance will be ${r_{eq}} = nr$
Now as the equivalent internal resistance and the external resistance R are in series connection, so the total equivalent resistance for the circuit will be given as, ${R_{eq}} = R + {r_{eq}} = R + nr$
Current for both the connection is given by $I = \dfrac{{{E_{eq}}}}{{{R_{eq}}}}$ where ${R_{eq}}$ is the total equivalent resistance including internal and external resistance.
Let current through this series connection be ${I_s}$
So, ${I_s} = \dfrac{{nE}}{{R + nr}}$
Now, for n cells with internal resistance r connected in parallel,
${E_{eq}}$ in the parallel connection will be ${E_{eq}} = E$ as the EMF will remain the same.
As all the internal resistances are also in parallel, so the equivalent internal resistance will be ${r_{eq}} = \dfrac{r}{n}$
Now as the equivalent internal resistance and the external resistance R are in series connection, so the total equivalent resistance for the circuit will be given as, ${R_{eq}} = R + {r_{eq}} = R + \dfrac{r}{n}$
Current for both the connection is given by $I = \dfrac{{{E_{eq}}}}{{{R_{eq}}}}$ where ${R_{eq}}$ is the total equivalent resistance including internal and external resistance.
Let current through this series connection be ${I_p}$
So, ${I_p} = \dfrac{E}{{\left( {R + \dfrac{r}{n}} \right)}}$
Now, as given in the question that the current in the circuit is the same whether the cells are connected in series or in parallel which means ${I_s} = {I_p}$
So, $\dfrac{{nE}}{{R + nr}} = \dfrac{E}{{\left( {R + \dfrac{r}{n}} \right)}}$
On further solving we have,
$\dfrac{{nE}}{{R + nr}} = \dfrac{{nE}}{{nR + r}}$
Or we can say, $R + nr = nR + r$
On simplifying we get,
$r = R$
$\therefore$The internal resistance r is equal to $R$. Hence, option (C) is the correct answer.
Note:
While calculating the overall equivalent resistance after calculating equivalent internal resistance, remember that ${r_{eq}}$ and the external resistance R will be in series.
Remember that the equivalent EMF for the parallel connection will remain as original.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)