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OpenStudy (anonymous):
How do I find the current if the problem only gives me resistance and power?
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OpenStudy (joannablackwelder):
P=IV and V=IR so P=I(IR)
OpenStudy (joannablackwelder):
Thus I=sqrt(P/R)
OpenStudy (anonymous):
So an example being, A device has a resistance of 67 ohms and a Power rating of 43 watts what is the current? How would you do this?
OpenStudy (anonymous):
I am still confused about how you got to that formula.
OpenStudy (joannablackwelder):
Are you familiar with the two formluas I mentioned before substituting?
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OpenStudy (anonymous):
Yes
OpenStudy (joannablackwelder):
43 W=I^2 * 67 ohms, solve for I.
OpenStudy (joannablackwelder):
Using dimensional analysis, I get the unit of A.
OpenStudy (anonymous):
You are going wayyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyy beyond the 9th grade level with vocab.
OpenStudy (joannablackwelder):
Do you mean dimensional analysis? That is just ignoring the numbers and looking only at the units.
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OpenStudy (anonymous):
OH. I am so stupid. o. o
OpenStudy (anonymous):
What is the unit "A"
OpenStudy (joannablackwelder):
No worries. A is the unit of current.
OpenStudy (anonymous):
Power consumed by
heating = P = I^2 R = current squared times resistance
so
I = sqrt(P/R), as noted by J-B-.
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