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A cell whose e.m.f. is 2V and internal resistance is 0.1Ω, is connected with a resistance of 3.9Ω. The voltage across the cell terminal will be
- a)0.50 V
- b)1.90 V
- c)1.95 V
- d)2.00 V
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
A cell whose e.m.f. is 2V and internal resistance is 0.1Ω, is connecte...
Answer :- c
Solution :- EMF of the cell(E) =2 V
internal resistance(r)= 0.1 V
External resistance(R)=3.9 V
V=IR
=(E÷(R+r))×R
=(2÷(3.9+0.1))×3.9
=1.95 V
Therefore potential difference across the terminals of the cell = 1.95V
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A cell whose e.m.f. is 2V and internal resistance is 0.1Ω, is connecte...
Given data:
e.m.f. (E) = 2V
Internal resistance (r) = 0.1
Resistance (R) = 3.9
To find: Voltage across the cell terminal
Formula used:
As per Ohm’s law, V = IR
The voltage across the cell terminal can be found by applying Kirchhoff's voltage law (KVL) to the circuit which states that the algebraic sum of the potential differences in any closed loop of a circuit must be zero.
Calculation:
The circuit diagram is as follows:
By KVL,
E - Ir - IR = 0
2 - 0.1I - 3.9I = 0
-1.9I = -2
I = 1.05 A
Voltage across the 3.9-ohm resistor,
V = IR = 1.05 x 3.9 = 4.095 V
Voltage across the cell terminal,
V = E - Ir = 2 - 0.1 x 1.05 = 1.95 V
Therefore, the voltage across the cell terminal is 1.95 V.
Hence, the correct option is (c) 1.95 V.
e.m.f. (E) = 2V
Internal resistance (r) = 0.1
Resistance (R) = 3.9
To find: Voltage across the cell terminal
Formula used:
As per Ohm’s law, V = IR
The voltage across the cell terminal can be found by applying Kirchhoff's voltage law (KVL) to the circuit which states that the algebraic sum of the potential differences in any closed loop of a circuit must be zero.
Calculation:
The circuit diagram is as follows:
By KVL,
E - Ir - IR = 0
2 - 0.1I - 3.9I = 0
-1.9I = -2
I = 1.05 A
Voltage across the 3.9-ohm resistor,
V = IR = 1.05 x 3.9 = 4.095 V
Voltage across the cell terminal,
V = E - Ir = 2 - 0.1 x 1.05 = 1.95 V
Therefore, the voltage across the cell terminal is 1.95 V.
Hence, the correct option is (c) 1.95 V.
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A cell whose e.m.f. is 2V and internal resistance is 0.1Ω, is connected with a resistance of 3.9Ω. The voltage across the cell terminal will bea) 0.50 V b) 1.90 V c) 1.95 V d) 2.00 V Correct answer is option 'C'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A cell whose e.m.f. is 2V and internal resistance is 0.1Ω, is connected with a resistance of 3.9Ω. The voltage across the cell terminal will bea) 0.50 V b) 1.90 V c) 1.95 V d) 2.00 V Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A cell whose e.m.f. is 2V and internal resistance is 0.1Ω, is connected with a resistance of 3.9Ω. The voltage across the cell terminal will bea) 0.50 V b) 1.90 V c) 1.95 V d) 2.00 V Correct answer is option 'C'. Can you explain this answer?.
A cell whose e.m.f. is 2V and internal resistance is 0.1Ω, is connected with a resistance of 3.9Ω. The voltage across the cell terminal will bea) 0.50 V b) 1.90 V c) 1.95 V d) 2.00 V Correct answer is option 'C'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A cell whose e.m.f. is 2V and internal resistance is 0.1Ω, is connected with a resistance of 3.9Ω. The voltage across the cell terminal will bea) 0.50 V b) 1.90 V c) 1.95 V d) 2.00 V Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A cell whose e.m.f. is 2V and internal resistance is 0.1Ω, is connected with a resistance of 3.9Ω. The voltage across the cell terminal will bea) 0.50 V b) 1.90 V c) 1.95 V d) 2.00 V Correct answer is option 'C'. Can you explain this answer?.
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