Sep 9, 2018
Current against Farad capacitance. If the load is 90 Farad, the current is 30A against 1E-9C, which is 0.03mA. 0.02mA, 0.039mA.
Please refer to this question.
A:
0.03mA is about 10% of the normal current.
On the simple side, because tiny 1W resistors don't cost much.
On the more complex side, we can engineer 10% resistors to be smaller than they would have to be.
On the simple side, because tiny 1W resistors don't cost much.
On the complex side, the resistor value is chosen from experience in that 0.03mA works in that application.
I have seen systems where all the resistors are carefully matched, but it has always been to save money. For example, when the load was 120V, so a 1W 10% resistor was used, for 120/1W = 12. That was really expensive.
A:
As already mentioned, most electronic components are chosen to be cheap or to be of such a small size that they make the component worthwile to be used. For electronic components in the mega-ohm range, they often have such a small size that the only way to reduce the price by a large amount is to make a lot of them. So they have to be cheap.
If you had a load with a thousand times higher current requirement, then you could choose a component with a higher current rating and have it cost more. But if you have a requirement of something that's close to a small wire, then you don't have a lot of other options.
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b) start over
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a) repair
b) start over
i chose repair, and it showed me this message again "A problem has been detected, and windows could not be installed/uninstalled properly." the drive was C:. it gave me 2 options :
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b) start over
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a) repair
b) start over
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