1sure ..
what amount of heat is generated in a coil of resistance r due to chaarge q passing through it if the current in coil decreases to 0 uniformly during a time interval Δt
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19 Answers
62q is the total charge
i= a - 1/a t (a is a constant!)
dq/dt=a-1/a t
now can you solve it?
you will have to find a
and then you will have to find the integral of
i^2(t) . r dt
62q is the total charge
i= a - b t (a is a constant!)
dq/dt=a-b t
also at t=t1, i=0
so t1=a/b
q=at1-bt12
and then you will have to find the integral of
(a - b t)^2 . r dt
(a^2 + t^2 b^2 -2abt).r . dt
(a^2t1 + t1^3/3a^2 -t1^2ab).r
Havent dirtied my hands totally but does this not give the final answer?
1i = i° - i°t/Δt
dq/dt = i° - i°t/Δt
∫dq =0Δt∫ ( i° - i°t/Δt )dt = i°Δt - i°Δt/2 = i°Δt/2 = q
therefore i° = 2q/Δt
i =2q/Δt( 1 - t/Δt )
heat = 0Δt∫ i2r dt = ∫(2q/Δt( 1 - t/Δt ))2 r dt = 4q2r/Δt2∫(1 + (t/Δt)2 - 2t/Δt)dt
=4q2r/Δt2(Δt + Δt/3 - Δt )
= 4q2r/3Δt