A 0-200 V voltmeter has an accuracy of 0.75% of full-scale reading. If the voltage measured is 100 V, the error is
1.5%
This problem involves calculating the error in a voltmeter reading based on its accuracy specification. Voltmeter accuracy is often stated as a percentage of the full-scale reading, which represents the maximum possible value the instrument can measure.
The accuracy specification tells us the maximum possible deviation from the true value. Since it's given as a percentage of the full-scale reading, we first calculate the absolute error in volts:
Absolute Error = Accuracy Percentage $\times$ Full-Scale Reading
Using the given values:
Absolute Error = $0.75\% \times 200 \, \text{V}$
To calculate this, we convert the percentage to a decimal:
Absolute Error = $\frac{0.75}{100} \times 200 \, \text{V}$
Absolute Error = $0.0075 \times 200 \, \text{V}$
Absolute Error = $1.5 \, \text{V}$
So, the maximum possible error for this voltmeter, regardless of the actual reading, is 1.5 Volts.
The options provided are percentages. Typically, when asked for the error in this format, it refers to the percentage error relative to the *actual measured value*. Let's calculate this:
Percentage Error = $\frac{\text{Absolute Error}}{\text{Measured Voltage}} \times 100\%$
Using the calculated absolute error and the measured voltage:
Percentage Error = $\frac{1.5 \, \text{V}}{100 \, \text{V}} \times 100\%$
Percentage Error = $0.015 \times 100\%$
Percentage Error = $1.5\%$
The calculated error, expressed as a percentage of the measured voltage (100 V), is 1.5%. This matches one of the options.
An ammeter of 0-50 A range has an accuracy of ± 1% of full scale reading. The current measured is 10 A. The uncertainty in measured value is
The radius of a Sphere is given as 40.0 ± 0.5 mm. The estimated error in its mass is :
Which of the following types of error CANNOT be produced in bridge measurement system?
The difference between the indicated value and the true value of a quantity is: