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When a resistance of 2 Ω is connected across the terminals of a cell, the current is 0.5 amperes. When the resistance is increased to 5 ohm, the current is 0.25 amperes. The internal resistance of the cell is
- a)0.5 Ω
- b)1.0 Ω
- c)1.5 Ω
- d)2.0 Ω
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
When a resistance of 2 Ω is connected across the terminals of a cell, ...
V = itV= 5(r+2)=(1/2)(r+2)when resistance increase to 5 then V =25(r+5)= (1/4)(r+5)(1/2)(r+2)= (1/4)(r+5)so , 2r+4= r+5r= 1 ohm that is 1.0 ohm
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When a resistance of 2 Ω is connected across the terminals of a cell, ...
Calculation of Internal Resistance of Cell
Given:
Resistance when current is 0.5 A: R1 = 2 Ω
Current when resistance is 5 Ω: I2 = 0.25 A
Let's assume that the internal resistance of the cell is r.
Using Ohm's law, we can calculate the potential difference across the terminals of the cell:
V = IR
Where I is the current and R is the resistance.
When the resistance is 2 Ω, the potential difference across the terminals of the cell is:
V = IR1 = 0.5 A x 2 Ω = 1 V
Similarly, when the resistance is 5 Ω, the potential difference across the terminals of the cell is:
V = IR2 = 0.25 A x 5 Ω = 1.25 V
Now, we can use the formula for potential difference in a circuit with an internal resistance:
V = E - Ir
Where E is the emf of the cell.
When the resistance is 2 Ω, the potential difference is 1 V, so:
1 V = E - I1r
When the resistance is 5 Ω, the potential difference is 1.25 V, so:
1.25 V = E - I2r
We can solve these two equations simultaneously to find the value of r:
1 V = E - 0.5 Ar
1.25 V = E - 0.25 Ar
Subtracting the second equation from the first, we get:
0.25 V = -0.25 Ar
Solving for r, we get:
r = 1 Ω
Therefore, the internal resistance of the cell is 1 Ω, which is option B.
Given:
Resistance when current is 0.5 A: R1 = 2 Ω
Current when resistance is 5 Ω: I2 = 0.25 A
Let's assume that the internal resistance of the cell is r.
Using Ohm's law, we can calculate the potential difference across the terminals of the cell:
V = IR
Where I is the current and R is the resistance.
When the resistance is 2 Ω, the potential difference across the terminals of the cell is:
V = IR1 = 0.5 A x 2 Ω = 1 V
Similarly, when the resistance is 5 Ω, the potential difference across the terminals of the cell is:
V = IR2 = 0.25 A x 5 Ω = 1.25 V
Now, we can use the formula for potential difference in a circuit with an internal resistance:
V = E - Ir
Where E is the emf of the cell.
When the resistance is 2 Ω, the potential difference is 1 V, so:
1 V = E - I1r
When the resistance is 5 Ω, the potential difference is 1.25 V, so:
1.25 V = E - I2r
We can solve these two equations simultaneously to find the value of r:
1 V = E - 0.5 Ar
1.25 V = E - 0.25 Ar
Subtracting the second equation from the first, we get:
0.25 V = -0.25 Ar
Solving for r, we get:
r = 1 Ω
Therefore, the internal resistance of the cell is 1 Ω, which is option B.
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When a resistance of 2 Ω is connected across the terminals of a cell, the current is 0.5 amperes. When the resistance is increased to 5 ohm, the current is 0.25 amperes. The internal resistance of the cell isa)0.5 Ωb)1.0 Ωc)1.5 Ωd)2.0 ΩCorrect answer is option 'B'. Can you explain this answer?
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When a resistance of 2 Ω is connected across the terminals of a cell, the current is 0.5 amperes. When the resistance is increased to 5 ohm, the current is 0.25 amperes. The internal resistance of the cell isa)0.5 Ωb)1.0 Ωc)1.5 Ωd)2.0 ΩCorrect answer is option 'B'. Can you explain this answer? for NEET 2025 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about When a resistance of 2 Ω is connected across the terminals of a cell, the current is 0.5 amperes. When the resistance is increased to 5 ohm, the current is 0.25 amperes. The internal resistance of the cell isa)0.5 Ωb)1.0 Ωc)1.5 Ωd)2.0 ΩCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for When a resistance of 2 Ω is connected across the terminals of a cell, the current is 0.5 amperes. When the resistance is increased to 5 ohm, the current is 0.25 amperes. The internal resistance of the cell isa)0.5 Ωb)1.0 Ωc)1.5 Ωd)2.0 ΩCorrect answer is option 'B'. Can you explain this answer?.
When a resistance of 2 Ω is connected across the terminals of a cell, the current is 0.5 amperes. When the resistance is increased to 5 ohm, the current is 0.25 amperes. The internal resistance of the cell isa)0.5 Ωb)1.0 Ωc)1.5 Ωd)2.0 ΩCorrect answer is option 'B'. Can you explain this answer? for NEET 2025 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about When a resistance of 2 Ω is connected across the terminals of a cell, the current is 0.5 amperes. When the resistance is increased to 5 ohm, the current is 0.25 amperes. The internal resistance of the cell isa)0.5 Ωb)1.0 Ωc)1.5 Ωd)2.0 ΩCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for When a resistance of 2 Ω is connected across the terminals of a cell, the current is 0.5 amperes. When the resistance is increased to 5 ohm, the current is 0.25 amperes. The internal resistance of the cell isa)0.5 Ωb)1.0 Ωc)1.5 Ωd)2.0 ΩCorrect answer is option 'B'. Can you explain this answer?.
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