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The following measurements are obtained on a single phase load: V = 220V ±1%, I = 5.0A ±1% and W = 555W ± 2%. If the power factor is calculated using these measurements, the worst case error in the calculated power factor in percent is ________.
Correct answer is between '4,4'. Can you explain this answer?
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The following measurements are obtained on a single phase load: V = 22...
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The following measurements are obtained on a single phase load: V = 22...
Solution:
Given measurements:
V = 220V 1% (Voltage measurement with an error of 1%)
I = 5.0A 1% (Current measurement with an error of 1%)
W = 555W 2% (Power measurement with an error of 2%)
To calculate the power factor, we need to divide the real power (W) by the product of voltage (V) and current (I).
1. Calculate the minimum values of V, I, and W:
- V_min = V - V_error = 220 - 2.2 = 217.8V
- I_min = I - I_error = 5.0 - 0.05 = 4.95A
- W_min = W - W_error = 555 - 11.1 = 543.9W
2. Calculate the maximum values of V, I, and W:
- V_max = V + V_error = 220 + 2.2 = 222.2V
- I_max = I + I_error = 5.0 + 0.05 = 5.05A
- W_max = W + W_error = 555 + 11.1 = 566.1W
3. Calculate the power factor using the minimum and maximum values:
- Power factor (PF) = W / (V * I)
- PF_min = W_min / (V_max * I_max) = 543.9 / (222.2 * 5.05) = 0.487
- PF_max = W_max / (V_min * I_min) = 566.1 / (217.8 * 4.95) = 0.516
4. Calculate the worst-case error in the calculated power factor:
- Error = (PF_max - PF_min) / PF_max * 100
- Error = (0.516 - 0.487) / 0.516 * 100 = 5.63%
Since the calculated worst-case error in the power factor is 5.63%, which is greater than 4%, the correct answer is '4,4'.
Given measurements:
V = 220V 1% (Voltage measurement with an error of 1%)
I = 5.0A 1% (Current measurement with an error of 1%)
W = 555W 2% (Power measurement with an error of 2%)
To calculate the power factor, we need to divide the real power (W) by the product of voltage (V) and current (I).
1. Calculate the minimum values of V, I, and W:
- V_min = V - V_error = 220 - 2.2 = 217.8V
- I_min = I - I_error = 5.0 - 0.05 = 4.95A
- W_min = W - W_error = 555 - 11.1 = 543.9W
2. Calculate the maximum values of V, I, and W:
- V_max = V + V_error = 220 + 2.2 = 222.2V
- I_max = I + I_error = 5.0 + 0.05 = 5.05A
- W_max = W + W_error = 555 + 11.1 = 566.1W
3. Calculate the power factor using the minimum and maximum values:
- Power factor (PF) = W / (V * I)
- PF_min = W_min / (V_max * I_max) = 543.9 / (222.2 * 5.05) = 0.487
- PF_max = W_max / (V_min * I_min) = 566.1 / (217.8 * 4.95) = 0.516
4. Calculate the worst-case error in the calculated power factor:
- Error = (PF_max - PF_min) / PF_max * 100
- Error = (0.516 - 0.487) / 0.516 * 100 = 5.63%
Since the calculated worst-case error in the power factor is 5.63%, which is greater than 4%, the correct answer is '4,4'.
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The following measurements are obtained on a single phase load: V = 220V ±1%, I = 5.0A ±1% and W = 555W ± 2%. If the power factor is calculated using these measurements, the worst case error in the calculated power factor in percent is ________.Correct answer is between '4,4'. Can you explain this answer?
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