charge

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

19 Answers

62
Lokesh Verma ·

is delta t a small time?

then it might be of some use!

1
ANKIT MAHATO ·

thankx bhaiya .. :)

62
Lokesh Verma ·

ankit i want to give u two pinks ;)

1
ANKIT MAHATO ·

sure ..

1
vector ·

check out my other q too

1
ANKIT MAHATO ·

welcome ...

1
vector ·

thanks

1
ANKIT MAHATO ·

[50] ..

1
ANKIT MAHATO ·

done check my original post #3..

1
vector ·

ve not mentioned

62
Lokesh Verma ·

q 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

1
vector ·

still not getting the ans

1
ANKIT MAHATO ·

i am also trying this out once agaen ..

62
Lokesh Verma ·

q 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?

62
Lokesh Verma ·

ok trying this one..

1
vector ·

@nishant bhaiya i tried it bt my ans does nt match wid crct ans

1
vector ·

ans is 4q2r/3Δt

1
ANKIT MAHATO ·

is the answer correct .. .. :)

1
ANKIT MAHATO ·

i = 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

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