Answer
64.8k+ views
Hint: Here the variable resistance is connected in series with the cell having an internal resistance. Then the voltage across the variable resistance will be the difference between the emf of the cell and the potential drop across the internal resistance of the cell. Using the given meter readings for two different values of resistances, we can form two equations for the potential difference across the variable resistor.
Formula used:
The potential difference across a resistor is given by, $V = IR$ where $I$ is the current through the resistor and $R$ is the resistance of the resistor.
Complete step by step answer:
Step 1: Sketch a circuit diagram of the given arrangement.
![](https://www.vedantu.com/question-sets/cee78381-8807-4307-bd14-2b9f33e333307673266183167244825.png)
Here the resistance of the variable resistor is denoted by $R$ and the internal resistance of the cell is $r$.
Let $E$ be the emf of the cell and let $I$ be the current through the circuit.
Also, let $V$ be the potential difference across the variable resistor.
Step 2: Express the potential drop across the variable resistor for the different meter readings.
The potential difference across the variable resistor can be expressed as $V = E - \left( {I \times r} \right)$
$ \Rightarrow E = V + \left( {I \times r} \right)$ ------ (1)
Substituting for $I = 0.3{\text{A}}$ and $V = 0.9{\text{V}}$ in equation (1) we get, $E = 0.9 + 0.3r$ ------- (2)
Substituting for $I = 0.25{\text{A}}$ and $V = 1{\text{V}}$ in equation (1) we get, $E = 1 + 0.25r$ ------- (3)
Step 3: Using equations (2) and (3) obtain the internal resistance of the cell.
We can equate the R.H.S of equations (2) and (3) to obtain the internal resistance of the cell.
i.e., $0.9 + 0.3r = 1 + 0.25r$
$ \Rightarrow r = \dfrac{{0.1}}{{0.05}} = 2\Omega $
Thus the internal resistance of the cell is $r = 2\Omega $ .
So the correct option is (B).
Note: Here the internal resistance of the cell and the variable resistor are essentially connected in series. In a series connection between two resistors, the current through each resistor will be the same. However, the potential drop across the internal resistance and the variable resistance are different. The term $I \times r$ in equation (1) is the potential drop across the internal resistance of the cell. Equation (1) can also be referred to as the voltage equation of the given circuit.
Formula used:
The potential difference across a resistor is given by, $V = IR$ where $I$ is the current through the resistor and $R$ is the resistance of the resistor.
Complete step by step answer:
Step 1: Sketch a circuit diagram of the given arrangement.
![](https://www.vedantu.com/question-sets/cee78381-8807-4307-bd14-2b9f33e333307673266183167244825.png)
Here the resistance of the variable resistor is denoted by $R$ and the internal resistance of the cell is $r$.
Let $E$ be the emf of the cell and let $I$ be the current through the circuit.
Also, let $V$ be the potential difference across the variable resistor.
Step 2: Express the potential drop across the variable resistor for the different meter readings.
The potential difference across the variable resistor can be expressed as $V = E - \left( {I \times r} \right)$
$ \Rightarrow E = V + \left( {I \times r} \right)$ ------ (1)
Substituting for $I = 0.3{\text{A}}$ and $V = 0.9{\text{V}}$ in equation (1) we get, $E = 0.9 + 0.3r$ ------- (2)
Substituting for $I = 0.25{\text{A}}$ and $V = 1{\text{V}}$ in equation (1) we get, $E = 1 + 0.25r$ ------- (3)
Step 3: Using equations (2) and (3) obtain the internal resistance of the cell.
We can equate the R.H.S of equations (2) and (3) to obtain the internal resistance of the cell.
i.e., $0.9 + 0.3r = 1 + 0.25r$
$ \Rightarrow r = \dfrac{{0.1}}{{0.05}} = 2\Omega $
Thus the internal resistance of the cell is $r = 2\Omega $ .
So the correct option is (B).
Note: Here the internal resistance of the cell and the variable resistor are essentially connected in series. In a series connection between two resistors, the current through each resistor will be the same. However, the potential drop across the internal resistance and the variable resistance are different. The term $I \times r$ in equation (1) is the potential drop across the internal resistance of the cell. Equation (1) can also be referred to as the voltage equation of the given circuit.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
According to classical free electron theory A There class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)