
When terminals of a cell of emf \[1.5V\] are connected to an ammeter of resistance $4\Omega $, the ammeter reads $0.30A$. Which of the following statements are correct?
A) The internal resistance of the cell is zero.
B) If a $4\Omega $ resistance is also across the terminals of the cell, the ammeter will read $0.50A$ .
C) If a $4\Omega $ resistance is also across the terminals of the cell, the ammeter will read $0.40A$ .
D) Suppose a voltmeter of resistance $4\Omega $ is used to measure the potential difference between terminals of a cell, it will read $1.0V$.
Answer
135k+ views
Hint: The ammeter is used to measure the current in the circuit and always connected in series with the external resistance, whereas voltmeter is used to measure the potential difference across an external resistance and always connected in parallel.
Complete step by step answer:
A cell is a single electrical energy source which uses chemical reaction to produce current.
When a current produce in a cell due to chemical reactions then a factor that’s opposing to the flow of current in cell is called internal resistance
The EMF (Electromotive force) of cell
EMF of cell $=1.5V$
Resistance of ammeter $=4\Omega $
Resistance of circuit $=\dfrac{V}{i}=\dfrac{1.5}{0.3}=5\Omega $
Now the ammeter provides $4\Omega $ resistance only, so the surplus is coming from the cell.
The internal resistance of the cell is $=5-4=1\Omega $
If a $4\Omega $ resistance is connected across cell (assuming parallel connection) resistance of the circuit will be$=\dfrac{4\times 4}{4+4}+1=3$
The final Emf across the cell will be
$E-ir=1.5-\left( \dfrac{1.5}{3} \right)\times 1=1V$
Current through ammeter $=\dfrac{1}{4}=0.25A$
When a voltmeter is connected to across the cell, the final resistance will be
$\dfrac{4\times 4}{4+4}$ (resistance of circuit)$+1$ (internal resistance of circuit)
$\begin{align}
& =\dfrac{16}{8}+1 \\
& =2+1 \\
& =3\Omega \\
\end{align}$
The reading of the voltmeter
$\begin{align}
& =E-ir \\
& =1.5-\left( \dfrac{1.5}{3} \right)\times 1 \\
& =1volt \\
\end{align}$
Voltmeter resistance $4\Omega $ is used to measure the potential difference between terminals of the cell, it will be read $1volt$.
$\therefore$ D is the correct option.
Note: The voltage provided/emf of an ideal battery and the one with internal resistance are different. Always connect the external resistance across the battery in parallel as otherwise the circuit would be incomplete.
Complete step by step answer:
A cell is a single electrical energy source which uses chemical reaction to produce current.
When a current produce in a cell due to chemical reactions then a factor that’s opposing to the flow of current in cell is called internal resistance
The EMF (Electromotive force) of cell
EMF of cell $=1.5V$
Resistance of ammeter $=4\Omega $
Resistance of circuit $=\dfrac{V}{i}=\dfrac{1.5}{0.3}=5\Omega $
Now the ammeter provides $4\Omega $ resistance only, so the surplus is coming from the cell.
The internal resistance of the cell is $=5-4=1\Omega $
If a $4\Omega $ resistance is connected across cell (assuming parallel connection) resistance of the circuit will be$=\dfrac{4\times 4}{4+4}+1=3$
The final Emf across the cell will be
$E-ir=1.5-\left( \dfrac{1.5}{3} \right)\times 1=1V$
Current through ammeter $=\dfrac{1}{4}=0.25A$
When a voltmeter is connected to across the cell, the final resistance will be
$\dfrac{4\times 4}{4+4}$ (resistance of circuit)$+1$ (internal resistance of circuit)
$\begin{align}
& =\dfrac{16}{8}+1 \\
& =2+1 \\
& =3\Omega \\
\end{align}$
The reading of the voltmeter
$\begin{align}
& =E-ir \\
& =1.5-\left( \dfrac{1.5}{3} \right)\times 1 \\
& =1volt \\
\end{align}$
Voltmeter resistance $4\Omega $ is used to measure the potential difference between terminals of the cell, it will be read $1volt$.
$\therefore$ D is the correct option.
Note: The voltage provided/emf of an ideal battery and the one with internal resistance are different. Always connect the external resistance across the battery in parallel as otherwise the circuit would be incomplete.
Recently Updated Pages
JEE Main 2021 July 25 Shift 2 Question Paper with Answer Key

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 20 Shift 2 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

How to find Oxidation Number - Important Concepts for JEE

Half-Life of Order Reactions - Important Concepts and Tips for JEE

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Elastic Collisions in One Dimension - JEE Important Topic

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Dual Nature of Radiation and Matter Class 12 Notes: CBSE Physics Chapter 11

Displacement-Time Graph and Velocity-Time Graph for JEE

Formula for number of images formed by two plane mirrors class 12 physics JEE_Main

JEE Advanced 2024 Syllabus Weightage

JEE Main Chemistry Question Paper with Answer Keys and Solutions
