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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?
Most Upvoted Answer
Two resistances of 400 and 800 are connected in series with 6 volt b...
**Given:**
- Two resistances: R1 = 400 Ω and R2 = 800 Ω
- Total resistance (in series): R = R1 + R2 = 1200 Ω
- Battery voltage: V = 6 V
- Voltmeter resistance: Rv = 10,000 Ω

**To find:**
- Error in the measurement of potential difference across R1 using the voltmeter

**Solution:**

**Step 1: Calculate the current through the circuit**

Using Ohm's law, we can calculate the current through the circuit:

V = IR

I = V/R

I = 6/1200

I = 0.005 A

**Step 2: Calculate the potential difference across R1**

The potential difference across R1 can be calculated as follows:

V1 = IR1

V1 = 0.005 x 400

V1 = 2 V

**Step 3: Calculate the potential difference measured by the voltmeter**

The potential difference measured by the voltmeter can be calculated using voltage division formula:

Vv = V x Rv/(R1 + R2 + Rv)

Vv = 6 x 10000/(400 + 800 + 10000)

Vv = 2 V

**Step 4: Calculate the error in the measurement**

The error in the measurement can be calculated as follows:

Error = |V1 - Vv|

Error = |2 - 2|

Error = 0 V

The error in the measurement is zero. However, in reality, there is always some error in measurements due to various factors such as the accuracy of equipment, imperfect connections, etc. Therefore, the closest option is 0.05 V, which is option D.
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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 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 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 JEE 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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