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12. Physics | Current Electricity | Leaky Capacitor due to Resistivity in Dielectric | (GA)

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in this illustration, we’ll study about
the leaky capacitor due to resistivity in dielectric. here we are given that a parallel
capacitor is filled with a faulty dielectric of dielectric constant k. and it also has
some resistivity ro. and if a charge q is given to the capacitor. it discharges through
the medium between the plates and we are required to find the time constant of discharging of
this capacitor and the leakage current. in capacitor. here, you can see in solution if
we are given. with a capacitor a parallel plate capacitor. and, a dielectric is filled
in it. say the terminals of capacitor are x and y. if we consider its plate area is,
ay. and plates separation, as d. and we know in this situation. the capacitance, of capacitor
can be written as. c is k epsilon not, ay by d, where this k is the dielectric constant
of the medium which is filled between the plates. here in this situation as we are given
that this medium also has some resistivity ro due to. some conductivity in the medium
we can also write. the resistance. across plates, of capacitor, this can be written
as ro d by ay. where d is the path length along which the charge carriers will flow
and ay is the cross sectional area of the medium through which it is flowing. so its
resistances ro d by ay. here if a charge plus q and minus q is given to the plates. here
the charge will be discharge through the conductivity of the medium or through this resistance between
the plates. so here we can consider this, capacitor which is having a charge q is being
discharged wire this resistance. so we can, calculate the time constant. of discharging.
which is given here as tau, this is r c and if you multiply the 2 we are getting the result
as ro k epsilon not. that is a time constant of discharging. so in this situation if we
calculate the leakage current in case of discharging of a charge capacitor. then we know for. a
given. r c circuit. discharge current. is given as. i is equal to we already studied
this in concept videos this q by r c e to power minus t by, r c. so here we can substitute
the value so this current is q by, r c we can substitute as ro k epsilon not e to power
minus t by ro k. epsilon not. so this is the discharge current which can also be written
as, leakage current. through the dielectric and this is the final result of this problem.

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