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Sea water at frequency v=4xx10^(8) Hz has permittivity in ~~ 80 in_(0), permeability mu = mu_(0) and resistivity rho = 0.25 Omega, Imagine a parallel plate capacitor immersed in sea water and driven by an alternating voltage source V(t)=V_(0)sin 2pi vt.What fraction of the conduction current density is the displacement current current density ? |
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Answer» SOLUTION :Let distance between two plates of capacitor be .d. and voltage applied is `V(t)=V_(0)sin 2pi vt` Hence electric FIELD, `E=(V_(0))/(d)sin 2pi v t ""`….(1) By Ohm.s law, `I_(c )=sigma E=(E )/(rho)` `therefore I_(c ) = (V_(0))/(rho d) sin 2pi v t` Let`(V_(0))/(rho d) = J_(0)^(c )""`....(2) `therefore J_(c ) = J_(0)^(c )sin 2pi vt ""`.....(3) Now displacement current density, `J_(d)=in(delta E)/(dt)=(in delta)/(dt)[(V_(0))/(d)sin 2pi vt]""`[`because`From equation (1)] `=(in 2piv V_(0))/(d)cos 2pi vt` Let`(2pi v V_(0))/(d)=J_(0)^(d)""`....(4) `therefore J_(d) = J_(0)^(d)cos pi vt ""`...(5) By taking ratio of (4) and (2), `(J_(0)^(d))/(J_(0)^(2))=(2pi v in V_(0))/(d)xx(rho d)/(V_(0))` `= 2 pi v in rho` `= 2piv xx80 in_(0) xx 0.25 [because in = 80 in_(0), rho = 0.25 Omega m]` `= 4pi in_(0) v xx 10` `=(10 v)/(9xx 10^(9)) "" [because4pi in_(0)=(1)/(9xx10^(9))]` `= (10xx4xx10^(8))/(9xx10^(9))[because v=4xx10^(8)Hz]` `=(4)/(9)` |
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