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Two resistances of 400 Ω and 800 Ω are connected in series with 6 volt battery of negligible internal resistance. A voltmeter of resistance 10,000 Ω is used to measure the potential difference across 400 Ω . The error in the measurement of potential difference in volts approximately is
- a)0.01
- b)0.02
- c)0.03
- d)0.05
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
Two resistances of 400 and 800 are connected in series with 6 volt b...
Calculation of Error in Voltage Measurement
1. Calculate Total Resistance:
- Resistances in series add up: R_total = R1 + R2 = 400 + 800 = 1200 Ω
2. Calculate Total Current:
- Using Ohm's Law: I = V_total / R_total = 6 / 1200 = 0.005 A
3. Calculate Voltage Drop across 400 Ω resistor:
- Using Ohm's Law: V_400 = I * R_400 = 0.005 * 400 = 2 V
4. Calculate Voltage Reading Error:
- When voltmeter is connected, it forms a voltage divider with the resistor:
- V_actual = V_400 * (R_voltmeter / (R_400 + R_voltmeter))
- V_actual = 2 * (10000 / (400 + 10000)) ≈ 1.98 V
5. Calculate Error in Voltage Measurement:
- Error = |V_actual - V_400| = |1.98 - 2| = 0.02 V
Therefore, the error in the measurement of potential difference in volts is approximately 0.02 V, which corresponds to option 'D'.
1. Calculate Total Resistance:
- Resistances in series add up: R_total = R1 + R2 = 400 + 800 = 1200 Ω
2. Calculate Total Current:
- Using Ohm's Law: I = V_total / R_total = 6 / 1200 = 0.005 A
3. Calculate Voltage Drop across 400 Ω resistor:
- Using Ohm's Law: V_400 = I * R_400 = 0.005 * 400 = 2 V
4. Calculate Voltage Reading Error:
- When voltmeter is connected, it forms a voltage divider with the resistor:
- V_actual = V_400 * (R_voltmeter / (R_400 + R_voltmeter))
- V_actual = 2 * (10000 / (400 + 10000)) ≈ 1.98 V
5. Calculate Error in Voltage Measurement:
- Error = |V_actual - V_400| = |1.98 - 2| = 0.02 V
Therefore, the error in the measurement of potential difference in volts is approximately 0.02 V, which corresponds to option 'D'.
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Community Answer
Two resistances of 400 and 800 are connected in series with 6 volt b...
The correct option is D 0.05 V
Before connecting voltmeter potential difference across 400W resistance is
Vi=400(400+800)×6
=2V
After connecting voltmeter equivalent resistance between A and B
=400×10,000(400+10,000)
=384.6Ω
Hence, potential difference measured by voltmeter
Vf=384.6(384.6+800)×6
=1.95V
Error in measurement =
Vi−Vf=2−1.95
= 0.05V.
Before connecting voltmeter potential difference across 400W resistance is
Vi=400(400+800)×6
=2V
After connecting voltmeter equivalent resistance between A and B
=400×10,000(400+10,000)
=384.6Ω
Hence, potential difference measured by voltmeter
Vf=384.6(384.6+800)×6
=1.95V
Error in measurement =
Vi−Vf=2−1.95
= 0.05V.
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Two resistances of 400 and 800 are connected in series with 6 volt battery of negligible internal resistance. A voltmeter of resistance 10,000 is used to measure the potential difference across 400 . The error in the measurement of potential difference in volts approximately isa)0.01b)0.02c)0.03d)0.05Correct answer is option 'D'. Can you explain this answer?
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Two resistances of 400 and 800 are connected in series with 6 volt battery of negligible internal resistance. A voltmeter of resistance 10,000 is used to measure the potential difference across 400 . The error in the measurement of potential difference in volts approximately isa)0.01b)0.02c)0.03d)0.05Correct answer is option 'D'. 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 Two resistances of 400 and 800 are connected in series with 6 volt battery of negligible internal resistance. A voltmeter of resistance 10,000 is used to measure the potential difference across 400 . The error in the measurement of potential difference in volts approximately isa)0.01b)0.02c)0.03d)0.05Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two resistances of 400 and 800 are connected in series with 6 volt battery of negligible internal resistance. A voltmeter of resistance 10,000 is used to measure the potential difference across 400 . The error in the measurement of potential difference in volts approximately isa)0.01b)0.02c)0.03d)0.05Correct answer is option 'D'. Can you explain this answer?.
Two resistances of 400 and 800 are connected in series with 6 volt battery of negligible internal resistance. A voltmeter of resistance 10,000 is used to measure the potential difference across 400 . The error in the measurement of potential difference in volts approximately isa)0.01b)0.02c)0.03d)0.05Correct answer is option 'D'. 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 Two resistances of 400 and 800 are connected in series with 6 volt battery of negligible internal resistance. A voltmeter of resistance 10,000 is used to measure the potential difference across 400 . The error in the measurement of potential difference in volts approximately isa)0.01b)0.02c)0.03d)0.05Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two resistances of 400 and 800 are connected in series with 6 volt battery of negligible internal resistance. A voltmeter of resistance 10,000 is used to measure the potential difference across 400 . The error in the measurement of potential difference in volts approximately isa)0.01b)0.02c)0.03d)0.05Correct answer is option 'D'. Can you explain this answer?.
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Two resistances of 400 and 800 are connected in series with 6 volt battery of negligible internal resistance. A voltmeter of resistance 10,000 is used to measure the potential difference across 400 . The error in the measurement of potential difference in volts approximately isa)0.01b)0.02c)0.03d)0.05Correct answer is option 'D'. Can you explain this answer?, a detailed solution for Two resistances of 400 and 800 are connected in series with 6 volt battery of negligible internal resistance. A voltmeter of resistance 10,000 is used to measure the potential difference across 400 . The error in the measurement of potential difference in volts approximately isa)0.01b)0.02c)0.03d)0.05Correct answer is option 'D'. Can you explain this answer? has been provided alongside types of Two resistances of 400 and 800 are connected in series with 6 volt battery of negligible internal resistance. A voltmeter of resistance 10,000 is used to measure the potential difference across 400 . The error in the measurement of potential difference in volts approximately isa)0.01b)0.02c)0.03d)0.05Correct answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
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