1wait posting the solution
I would like to have everybody's idea about the following problem:
suppose we keep two circuits with one loop each(radii x,y) side by side.(resistance r1,r2,, emf e1,e2). not neglecting mutual and self inductance, give a qualitative and quantitative idea about current at any moment through the circuits.
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6 Answers
1let us assume x<y
then for loop one by applying kirchoffs law
μ0πx/2(di1/dt)+μ0πx2/(2y)di2/dt + e1 -i1r1=0
for loop 2
(μ0πx/2)di1/dt+(μ0πy/2)di2/dt + e2 -i2r2=0
solving these i am getting an equation of the form
A(d2i1/dt2)+B(di1/dt)+Ci1+D=0
9Qualitative :
as nothing is mentioned in Q , i suppose both emfs are clockwise
Now inc in i1 will induce backemf in ring one and forward emf for ring2
similar case for i2
Quantitative:
L1 ,L2 - self inductance
K1,K2 - mutual inductance ( i hope u arent asking derivation of exact formula , its tedious )
then for loop one by applying kirchoffs law
L1(di1/dt) - K2(di2/dt) + e1 -i1r1=0
for loop 2
L2(di2/dt) - K1(di1/dt) + e2 -i2r2=0
solving these i got something very interesting!!
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(L2L1 - K2K1)d2i1/dt - (L2r1 + L1r2 )di1/dt + i1(r1r2) = e1r2
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now the abv eq is solvable and has multiple solutions .
but one interesting solution that can be found by observing is i1 = e1/r1 (which is a constant and the other derivatives are 0 )
but nature doesnt permit that solution as for i to be established , it takes a small but definite time ( as speed of EM waves = c )
but after infinite time , i1 = e1/r1 obviously and similar for i2 .
can u solve the abv eq using mathematica and post it ith power ??
1one more Q . r1 r2 are net resistances offered by the entire ckt or reistacnes or coil and source is to be tkaen separately ????
9r1 r2 entire ckt resistance
and can someone solve the DE ?????
