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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 'D'. Can you explain this answer?
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Most Upvoted Answer
Two resistances of 400 Ω and 800 Ω are connected in series with 6 volt...
Ideal volatge across 400ohm=4/(8+4)*6V=2V. But u are using non ideal voltameter with r=10000 ohm thus Req=400*10000/(400+10000)=40000/104 yeah till now i think it's correct.....ok now volatge measured is (40000/104)/(40000/104+800)*6V=1.94982V=1.95V nearly......thus error is 0.05 volts ....done..
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Two resistances of 400 Ω and 800 Ω are connected in series with 6 volt...
To find the error in the measurement of potential difference, we need to calculate the equivalent resistance of the circuit and then determine the current passing through the voltmeter.
Finding the equivalent resistance:
- The resistances are connected in series, so the equivalent resistance (R_eq) is the sum of the individual resistances.
- R_eq = 400 Ω + 800 Ω = 1200 Ω
Calculating the current passing through the voltmeter:
- The total voltage across the circuit is 6 volts.
- According to Ohm's Law, V = I * R, where V is the voltage, I is the current, and R is the resistance.
- Since the voltmeter is connected in parallel with the 400 Ω resistor, the voltage across the 400 Ω resistor is the same as the voltage across the voltmeter.
- Therefore, the current passing through the voltmeter can be calculated using I = V / R, where V is the voltage across the 400 Ω resistor and R is the resistance of the voltmeter.
- Substituting the values, I = 6 volts / 10000 Ω = 0.0006 A
Calculating the error in the measurement:
- The error in the measurement is caused by the resistance of the voltmeter affecting the circuit.
- The voltmeter acts as a parallel resistance, which changes the total resistance of the circuit.
- The error can be calculated using the formula: Error = (R_voltmeter / R_eq) * V, where R_voltmeter is the resistance of the voltmeter, R_eq is the equivalent resistance, and V is the voltage across the 400 Ω resistor.
- Substituting the values, Error = (10000 Ω / 1200 Ω) * 6 volts = 0.05 volts
Therefore, the error in the measurement of potential difference is approximately 0.05 volts, which corresponds to option D.
Finding the equivalent resistance:
- The resistances are connected in series, so the equivalent resistance (R_eq) is the sum of the individual resistances.
- R_eq = 400 Ω + 800 Ω = 1200 Ω
Calculating the current passing through the voltmeter:
- The total voltage across the circuit is 6 volts.
- According to Ohm's Law, V = I * R, where V is the voltage, I is the current, and R is the resistance.
- Since the voltmeter is connected in parallel with the 400 Ω resistor, the voltage across the 400 Ω resistor is the same as the voltage across the voltmeter.
- Therefore, the current passing through the voltmeter can be calculated using I = V / R, where V is the voltage across the 400 Ω resistor and R is the resistance of the voltmeter.
- Substituting the values, I = 6 volts / 10000 Ω = 0.0006 A
Calculating the error in the measurement:
- The error in the measurement is caused by the resistance of the voltmeter affecting the circuit.
- The voltmeter acts as a parallel resistance, which changes the total resistance of the circuit.
- The error can be calculated using the formula: Error = (R_voltmeter / R_eq) * V, where R_voltmeter is the resistance of the voltmeter, R_eq is the equivalent resistance, and V is the voltage across the 400 Ω resistor.
- Substituting the values, Error = (10000 Ω / 1200 Ω) * 6 volts = 0.05 volts
Therefore, the error in the measurement of potential difference is approximately 0.05 volts, which corresponds to option D.
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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 isa)0.01b)0.02c)0.03d)0.05Correct answer is 'D'. Can you explain this answer?
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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 isa)0.01b)0.02c)0.03d)0.05Correct answer is 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 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 isa)0.01b)0.02c)0.03d)0.05Correct answer is 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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 isa)0.01b)0.02c)0.03d)0.05Correct answer is 'D'. Can you explain this answer?.
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 isa)0.01b)0.02c)0.03d)0.05Correct answer is 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 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 isa)0.01b)0.02c)0.03d)0.05Correct answer is 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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 isa)0.01b)0.02c)0.03d)0.05Correct answer is 'D'. Can you explain this answer?.
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