Question

A pair of 60 cm² area plates are separated by a 2.0-mm-thick layer of ideal dielectric characterized by er = 3.0. If a voltage v(t) = 1.0 sin(41 x 10ºt) V is placed across the plates, determine the displacement current.

Answer 1

solution - Given that: of 11 cm = 10²m3 Area of plates A= 60 cm2 A = 60x107 m2 - A = 6x 103 m2 Separation between plates da dio mm - (d = 2.0 X103m] L. 1 mm= 15m? constant for = 3 .0 The efectic permittinity Dielectric 7 E = 3 to of medium & Voltage vects = 1.0 sin (ATX 103 t) v The capacitance of plate is given as: с. СА (= đ C-300 A (= I 3x8, 86x1072 X 6X103 = c = 2x10s sto=8.86410 Sto=8.86x1024 - - C = 79.74 x 1012 - C = 7.974 X 10ll farad is capacitance of plate. This

Now dult at = a dt 21.0 fin (ATX 103 t) { I dult > duch) 40 411X103 cos (47 X10't) Vls. 1-0 at The How displacement current (ID) is given as ID = c d ult) at Id= 7.974 x10" x 411 X 103 cos (40X10 +) A {using 0 40% - ID = 1.00 x 106 cos (41 X 103 t) A ID = 1.0 COS (411X1094) MA {IMA= 100A The displacement current is Jp = 1.0cog (401x1039) MA
Answer: displacement current =1.0 Cos(4 π x10³t)  μ A