Two currents from different sources flow in opposite directions through a resistor. I1 is
measured as 79 mA on a 100 mA analog instrument with an accuracy of ±3% of full scale.
I2, determined as 31 mA, is measured on a digital instrument with a ±100 µA accuracy.
Calculate the maximum and minimum levels of the current in Rl
Imax = I1max - I2min.
Imax = (79+0.03*100) - (31-0.1) = 82 - 30.9 = 51.1 mA.
Imin = I1min - I2max.
Imin = (79-0.03*100) - (31+0.1) = 76 - 31.1 = 44.9 mA.
What is (31-0.1)
How to get "0.1"
هندسه
Well, let's calculate the maximum and minimum levels of the current in Rl.
For I1 measured with the analog instrument, the accuracy is ±3% of full scale, which means the actual value could be 3% higher or lower. So we have:
Maximum value of I1 = 79 mA + (3% x 100 mA) = 79 mA + 3 mA = 82 mA
Minimum value of I1 = 79 mA - (3% x 100 mA) = 79 mA - 3 mA = 76 mA
For I2 measured with the digital instrument, the accuracy is ±100 µA, which means the actual value could be 100 µA higher or lower. So we have:
Maximum value of I2 = 31 mA + 100 µA = 31.1 mA
Minimum value of I2 = 31 mA - 100 µA = 30.9 mA
Now, since the two currents flow in opposite directions through the resistor, their combined effect would result in the maximum and minimum levels of the current in Rl.
Maximum level of current in Rl = I1 (maximum) + I2 (maximum) = 82 mA + 31.1 mA = 113.1 mA
Minimum level of current in Rl = I1 (minimum) + I2 (minimum) = 76 mA + 30.9 mA = 106.9 mA
So, the maximum and minimum levels of the current in Rl are approximately 113.1 mA and 106.9 mA, respectively. Keep in mind that these values are subject to the accuracy limitations of the instruments used for measurement!
To calculate the maximum and minimum levels of the current in Rl, we need to consider the accuracy of the instruments used to measure I1 and I2. Let's start by calculating the maximum and minimum values for I1.
Given:
I1 = 79 mA (measured on a 100 mA analog instrument)
Accuracy of the analog instrument = ±3% of full scale
To calculate the maximum value of I1, we add the maximum percentage error to the measured value:
Max I1 = 79 mA + (3% of 100 mA) = 79 mA + 3 mA = 82 mA
To calculate the minimum value of I1, we subtract the maximum percentage error from the measured value:
Min I1 = 79 mA - (3% of 100 mA) = 79 mA - 3 mA = 76 mA
Next, let's compute the maximum and minimum values for I2.
Given:
I2 = 31 mA (measured on a digital instrument)
Accuracy of the digital instrument = ±100 µA
To calculate the maximum value of I2, we add the maximum accuracy value to the measured value:
Max I2 = 31 mA + 100 µA = 31 mA + 0.1 mA = 31.1 mA
To calculate the minimum value of I2, we subtract the maximum accuracy value from the measured value:
Min I2 = 31 mA - 100 µA = 31 mA - 0.1 mA = 30.9 mA
Finally, to find the maximum and minimum levels of the current in Rl, we add and subtract the maximum and minimum values of I1 and I2, respectively:
Maximum level of current in Rl = Max I1 + Max I2 = 82 mA + 31.1 mA = 113.1 mA
Minimum level of current in Rl = Min I1 + Min I2 = 76 mA + 30.9 mA = 106.9 mA
Therefore, the maximum and minimum levels of the current in Rl are 113.1 mA and 106.9 mA, respectively.