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(dw)=q(dv) v=ir by ohms law dv=r*(di) q=it (dw)=rti(di) integrating, w=rt*i^2/2 w/t=P(power)=i^2*r/2 but actually its i^2*r only wheres the mistake please point out

(dw)=q(dv)
v=ir by ohms law
dv=r*(di)
q=it
(dw)=rti(di)
integrating,
w=rt*i^2/2
w/t=P(power)=i^2*r/2
but actually its i^2*r only
wheres the mistake
please point out

Grade:12

2 Answers

Sher Mohammad IIT Delhi
askIITians Faculty 174 Points
7 years ago
In actual, the i is the peak current, Imax , which sqrt(2)times the rms quantity, just take care of rms quantity and peak value.
Nikhil Kumar
18 Points
7 years ago
No dought you are doing integration to find power and your ans is half. Actually if you study integration in math, it will area between wrt to which you are doing integration. And it will provide avg value. But actually peak value is double

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