A small heating coil has a constant resistance of 25 W. a) How much power would the coil consume if the voltage drop across it is 96 V?
回答 (9)
2020-08-06 3:40 pm
a)
The resistance should "25 Ω" instead of "25 W".
Power of the coil consumed
= v²/R
= 96²/25
= 370 W (to 2 sig. fig.)
The resistance should "25 Ω" instead of "25 W".
Power of the coil consumed
= v²/R
= 96²/25
= 370 W (to 2 sig. fig.)
2020-08-06 11:10 pm
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2020-10-26 12:33 pm
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2020-08-11 6:57 pm
25 W is not a resistance.
You need to learn some electrical terminology before coming here to ask a question.
You need to learn some electrical terminology before coming here to ask a question.
2020-08-09 12:14 am
The "units" are extremely important. Did you mean 'ohms'??
IF ohms, then:
I = V/R and Watts = I*V So Watts = V²/R
W = (96)²/25 = 368.64 w, but I would round to 370 w
IF ohms, then:
I = V/R and Watts = I*V So Watts = V²/R
W = (96)²/25 = 368.64 w, but I would round to 370 w
2020-08-07 10:50 am
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2020-12-07 2:28 am
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2020-08-09 5:16 am
resistance is measured in Ohms,
Not Watts.If you meant Ohms, IE the coil has 25 Ohms resistancethem power expended in the coil is given by Power = V x Ior V^2 / R= 368.64 Watts
Not Watts.If you meant Ohms, IE the coil has 25 Ohms resistancethem power expended in the coil is given by Power = V x Ior V^2 / R= 368.64 Watts
2020-08-07 3:30 am
The resistance is 25w/96v=.260Ω
It will use 25W.
If the coil is 25Ω the the power will be 96^2/25= 387 W.
It will use 25W.
If the coil is 25Ω the the power will be 96^2/25= 387 W.
收錄日期: 2021-04-18 18:35:49
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