Math Problem Statement

Figure 1 shows a circuit diagram for measurement of resistance. The meter reads 10A, on a 100A range, and the voltmeter reads 125V on a 150V range. The instrument scales are such that 0.1 of a scale can be distinguished. The constructional error of the ammeter is ±0.3% and that of the voltmeter is ±0.4%. The resistance of the ammeter is 0.25 Ω. Determine the total systematic error in instrument readings, measured value of resistor R, true value of resistor R, and percentage error in the value of R.

Solution

Let's go through the problem one by one.

Question 2(b)

The question refers to Figure 1, a circuit diagram used for measuring resistance RR, using an ammeter and a voltmeter. Here's a breakdown of the requested items:

Given:

  • Ammeter reading I=10AI = 10 \, \text{A} (on a 100A100 \, \text{A} range)
  • Voltmeter reading V=125VV = 125 \, \text{V} (on a 150V150 \, \text{V} range)
  • Ammeter error: ±0.3%\pm 0.3\%
  • Voltmeter error: ±0.4%\pm 0.4\%
  • Ammeter resistance Ra=0.25ΩR_a = 0.25 \, \Omega

To Find:

(i) Total systematic error in instrument readings
(ii) Measured value of resistor RR
(iii) True value of resistor RR
(iv) Percentage error in the value of RR

Step-by-step solutions:

  1. Systematic error
    For the ammeter:

    • Instrument range Irange=100AI_{\text{range}} = 100 \, \text{A}
    • Error due to construction: 0.3%0.3\%
    • Scale error: 0.1 of a scale Total error for the ammeter ΔI=(0.3%×Irange)+0.1\Delta I = (0.3\% \times I_{\text{range}}) + 0.1.

    For the voltmeter:

    • Instrument range Vrange=150VV_{\text{range}} = 150 \, \text{V}
    • Error due to construction: 0.4%0.4\%
    • Scale error: 0.1 of a scale Total error for the voltmeter ΔV=(0.4%×Vrange)+0.1\Delta V = (0.4\% \times V_{\text{range}}) + 0.1.
  2. Measured value of the resistor RR
    From Ohm's law: Rmeasured=VmeasuredImeasured=125V10A=12.5ΩR_{\text{measured}} = \frac{V_{\text{measured}}}{I_{\text{measured}}} = \frac{125 \, \text{V}}{10 \, \text{A}} = 12.5 \, \Omega But we also need to correct for the ammeter resistance.

  3. True value of resistor RR
    The true value RtrueR_{\text{true}} should take into account the internal resistance of the ammeter: Rtrue=RmeasuredRa=12.5Ω0.25Ω=12.25ΩR_{\text{true}} = R_{\text{measured}} - R_a = 12.5 \, \Omega - 0.25 \, \Omega = 12.25 \, \Omega

  4. Percentage error in RR
    The percentage error is calculated as: Percentage error=RmeasuredRtrueRtrue×100\text{Percentage error} = \frac{R_{\text{measured}} - R_{\text{true}}}{R_{\text{true}}} \times 100

    Substituting values: Percentage error=12.512.2512.25×100=2.04%\text{Percentage error} = \frac{12.5 - 12.25}{12.25} \times 100 = 2.04\%

Would you like me to calculate the errors more precisely?


Additional Questions:

  1. How do you calculate the total uncertainty in measurements with multiple sources of error?
  2. What is the significance of ammeter internal resistance in resistance measurements?
  3. How can measurement errors be minimized in electrical circuits?
  4. How does percentage error change with increasing voltmeter or ammeter range?
  5. Why is it important to consider both systematic and random errors in an experiment?

Tip:

When calculating measurement errors, always check if the instrument’s internal resistances (like that of an ammeter or voltmeter) need to be included in your calculations for more accurate results.

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Math Problem Analysis

Mathematical Concepts

Error Analysis
Ohm's Law
Percentage Error

Formulas

Ohm's Law: V = IR
Systematic Error: ΔV = (0.4% × V_range) + 0.1
Systematic Error: ΔI = (0.3% × I_range) + 0.1
Percentage Error: ((Measured Value - True Value) / True Value) × 100

Theorems

Ohm's Law

Suitable Grade Level

College Level (Engineering/Physics)