Answer
396.9k+ views
Hint: Express the current in the potentiometer wire in terms of emf of the cell and internal resistance. Recall the balance condition for the potentiometer. Substitute the emf of the cell and potential across the wire of the potentiometer in the balance condition to get the value of internal resistance.
Formula used:
Balance condition, \[\dfrac{E}{V} = \dfrac{{{l_1}}}{{{l_2}}}\]
Here, E is the emf of the cell, V is the potential across the wire of the potentiometer, \[{l_1}\] and \[{l_2}\] are the points on the wire where the potentiometer is balanced.
Complete step by step answer:
Let a cell of emf E is connected across the length of the potentiometer wire and the internal resistance of this potentiometer is r. when we switch on the circuit, the current I flows through the wire of the potentiometer. We can express the current in the potentiometer wire as follows,
\[I = \dfrac{E}{{R + r}}\]
\[ \Rightarrow E = I\left( {R + r} \right)\] …… (1)
This is the emf of the cell.
Now, the potential difference across the length of the potentiometer wire is can be expressed using Ohm’s law as,
\[V = IR\] …… (2)
Here, R is the resistance.
We know the balance condition for potentiometer,
\[\dfrac{E}{V} = \dfrac{{{l_1}}}{{{l_2}}}\]
Here, \[{l_1}\] and \[{l_2}\] are the points on the wire where the potentiometer is balanced.
Using equation (1) and (2) in the above equation, we get,
\[\dfrac{{I\left( {R + r} \right)}}{{IR}} = \dfrac{{{l_1}}}{{{l_2}}}\]
\[ \Rightarrow \dfrac{{R + r}}{R} = \dfrac{{{l_1}}}{{{l_2}}}\]
\[ \Rightarrow 1 + \dfrac{r}{R} = \dfrac{{{l_1}}}{{{l_2}}}\]
Substituting 110 cm for \[{l_1}\], 100 cm for \[{l_2}\] and \[10\,\Omega \] for R in the above equation, we get,
\[1 + \dfrac{r}{{10}} = \dfrac{{110}}{{100}}\]
\[ \Rightarrow \dfrac{r}{{10}} = 1.1 - 1\]
\[ \Rightarrow \dfrac{r}{{10}} = 0.1\]
\[ \therefore r = 1\,\Omega \]
Therefore, the internal resistance of the cell is \[1\,\Omega \].
So, the correct answer is option A.
Note:The internal resistance of the cell is negligible and in often cases we neglect the internal resistance of the cell if the cell provides a high supply. The internal resistance is the resistance provided by the electrolytic solution in the cell. In the closed circuit while determining the emf of the cell, the Ohm’s law should be expressed such that it has resistance which is the sum of resistance of the circuit and internal resistance.
Formula used:
Balance condition, \[\dfrac{E}{V} = \dfrac{{{l_1}}}{{{l_2}}}\]
Here, E is the emf of the cell, V is the potential across the wire of the potentiometer, \[{l_1}\] and \[{l_2}\] are the points on the wire where the potentiometer is balanced.
Complete step by step answer:
Let a cell of emf E is connected across the length of the potentiometer wire and the internal resistance of this potentiometer is r. when we switch on the circuit, the current I flows through the wire of the potentiometer. We can express the current in the potentiometer wire as follows,
\[I = \dfrac{E}{{R + r}}\]
\[ \Rightarrow E = I\left( {R + r} \right)\] …… (1)
This is the emf of the cell.
Now, the potential difference across the length of the potentiometer wire is can be expressed using Ohm’s law as,
\[V = IR\] …… (2)
Here, R is the resistance.
We know the balance condition for potentiometer,
\[\dfrac{E}{V} = \dfrac{{{l_1}}}{{{l_2}}}\]
Here, \[{l_1}\] and \[{l_2}\] are the points on the wire where the potentiometer is balanced.
Using equation (1) and (2) in the above equation, we get,
\[\dfrac{{I\left( {R + r} \right)}}{{IR}} = \dfrac{{{l_1}}}{{{l_2}}}\]
\[ \Rightarrow \dfrac{{R + r}}{R} = \dfrac{{{l_1}}}{{{l_2}}}\]
\[ \Rightarrow 1 + \dfrac{r}{R} = \dfrac{{{l_1}}}{{{l_2}}}\]
Substituting 110 cm for \[{l_1}\], 100 cm for \[{l_2}\] and \[10\,\Omega \] for R in the above equation, we get,
\[1 + \dfrac{r}{{10}} = \dfrac{{110}}{{100}}\]
\[ \Rightarrow \dfrac{r}{{10}} = 1.1 - 1\]
\[ \Rightarrow \dfrac{r}{{10}} = 0.1\]
\[ \therefore r = 1\,\Omega \]
Therefore, the internal resistance of the cell is \[1\,\Omega \].
So, the correct answer is option A.
Note:The internal resistance of the cell is negligible and in often cases we neglect the internal resistance of the cell if the cell provides a high supply. The internal resistance is the resistance provided by the electrolytic solution in the cell. In the closed circuit while determining the emf of the cell, the Ohm’s law should be expressed such that it has resistance which is the sum of resistance of the circuit and internal resistance.
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)