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The resistances of the four arms P, Q, R and S in a Wheatstone’s bridge are 10 ohm, 30 ohm, 30 ohm and 90 ohm, respectively. The e.m.f. and internal resistance of the cell are 7 volt and 5 ohm respectively. If the galvanometer resistance is 50 ohm, the current drawn from the cell will be -
- a)1.0 A
- b)0.2 A
- c)0.1 A
- d)2.0 A
Correct answer is option 'B'. Can you explain this answer?
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Calculation of Current Drawn from the Cell:
- The formula to calculate the current drawn from the cell in a Wheatstone's bridge is given by:
\[ I = \frac{E}{(r + R_{eq})} \]
Where,
- I is the current drawn from the cell
- E is the e.m.f. of the cell (7 V)
- r is the internal resistance of the cell (5 ohm)
- \( R_{eq} \) is the equivalent resistance of the bridge
Calculating Equivalent Resistance:
- The equivalent resistance of the Wheatstone's bridge can be calculated using the formula:
\[ R_{eq} = \frac{PR}{Q + R} \]
Where,
- P, Q, R, and S are the resistances in the arms of the bridge
- Substituting the given values:
\[ R_{eq} = \frac{(10)(90)}{30 + 30} = \frac{900}{60} = 15 ohm \]
Calculating Current Drawn:
- Substituting the values of E, r, and \( R_{eq} \) in the formula:
\[ I = \frac{7}{(5 + 15)} = \frac{7}{20} = 0.35 A \]
- However, since the galvanometer resistance is 50 ohm, the effective resistance in the circuit would be \( 15 + 50 = 65 ohm \)
- Recalculating the current drawn using the corrected resistance:
\[ I = \frac{7}{(5 + 65)} = \frac{7}{70} = 0.1 A \]
Therefore, the current drawn from the cell is 0.1 A, which corresponds to option C.
- The formula to calculate the current drawn from the cell in a Wheatstone's bridge is given by:
\[ I = \frac{E}{(r + R_{eq})} \]
Where,
- I is the current drawn from the cell
- E is the e.m.f. of the cell (7 V)
- r is the internal resistance of the cell (5 ohm)
- \( R_{eq} \) is the equivalent resistance of the bridge
Calculating Equivalent Resistance:
- The equivalent resistance of the Wheatstone's bridge can be calculated using the formula:
\[ R_{eq} = \frac{PR}{Q + R} \]
Where,
- P, Q, R, and S are the resistances in the arms of the bridge
- Substituting the given values:
\[ R_{eq} = \frac{(10)(90)}{30 + 30} = \frac{900}{60} = 15 ohm \]
Calculating Current Drawn:
- Substituting the values of E, r, and \( R_{eq} \) in the formula:
\[ I = \frac{7}{(5 + 15)} = \frac{7}{20} = 0.35 A \]
- However, since the galvanometer resistance is 50 ohm, the effective resistance in the circuit would be \( 15 + 50 = 65 ohm \)
- Recalculating the current drawn using the corrected resistance:
\[ I = \frac{7}{(5 + 65)} = \frac{7}{70} = 0.1 A \]
Therefore, the current drawn from the cell is 0.1 A, which corresponds to option C.
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The resistances of the four arms P, Q, R and S in a Wheatstone’s bridge are 10 ohm, 30 ohm, 30 ohm and 90 ohm, respectively. The e.m.f. and internal resistance of the cell are 7 volt and 5 ohm respectively. If the galvanometer resistance is 50 ohm, the current drawn from the cell will be -a)1.0 Ab)0.2 Ac)0.1 Ad)2.0 ACorrect answer is option 'B'. Can you explain this answer?
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The resistances of the four arms P, Q, R and S in a Wheatstone’s bridge are 10 ohm, 30 ohm, 30 ohm and 90 ohm, respectively. The e.m.f. and internal resistance of the cell are 7 volt and 5 ohm respectively. If the galvanometer resistance is 50 ohm, the current drawn from the cell will be -a)1.0 Ab)0.2 Ac)0.1 Ad)2.0 ACorrect answer is option 'B'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The resistances of the four arms P, Q, R and S in a Wheatstone’s bridge are 10 ohm, 30 ohm, 30 ohm and 90 ohm, respectively. The e.m.f. and internal resistance of the cell are 7 volt and 5 ohm respectively. If the galvanometer resistance is 50 ohm, the current drawn from the cell will be -a)1.0 Ab)0.2 Ac)0.1 Ad)2.0 ACorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The resistances of the four arms P, Q, R and S in a Wheatstone’s bridge are 10 ohm, 30 ohm, 30 ohm and 90 ohm, respectively. The e.m.f. and internal resistance of the cell are 7 volt and 5 ohm respectively. If the galvanometer resistance is 50 ohm, the current drawn from the cell will be -a)1.0 Ab)0.2 Ac)0.1 Ad)2.0 ACorrect answer is option 'B'. Can you explain this answer?.
The resistances of the four arms P, Q, R and S in a Wheatstone’s bridge are 10 ohm, 30 ohm, 30 ohm and 90 ohm, respectively. The e.m.f. and internal resistance of the cell are 7 volt and 5 ohm respectively. If the galvanometer resistance is 50 ohm, the current drawn from the cell will be -a)1.0 Ab)0.2 Ac)0.1 Ad)2.0 ACorrect answer is option 'B'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The resistances of the four arms P, Q, R and S in a Wheatstone’s bridge are 10 ohm, 30 ohm, 30 ohm and 90 ohm, respectively. The e.m.f. and internal resistance of the cell are 7 volt and 5 ohm respectively. If the galvanometer resistance is 50 ohm, the current drawn from the cell will be -a)1.0 Ab)0.2 Ac)0.1 Ad)2.0 ACorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The resistances of the four arms P, Q, R and S in a Wheatstone’s bridge are 10 ohm, 30 ohm, 30 ohm and 90 ohm, respectively. The e.m.f. and internal resistance of the cell are 7 volt and 5 ohm respectively. If the galvanometer resistance is 50 ohm, the current drawn from the cell will be -a)1.0 Ab)0.2 Ac)0.1 Ad)2.0 ACorrect answer is option 'B'. Can you explain this answer?.
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