Question

Figure shows a capacitor connected to a voltage source. There are two dielectric bricks in the capacitor. d = 0.2mm V = 5cos (10°t)v &rl = 5 &r2 = 8 I= 5mm w = 2mm W, = 3mm Problem [5] <10 points> Calculate the magnitude of the total current on the capacitor. Problem [6] <10 points> Calculate the magnitude of the displacement current on dielectric brick 1.

Answer 1

Given 11d = 0.2mm Lons lewe e-mm v=Sws(109) -2mm wn- 3mm let Capacitance with Eri be, ca & tor bela C.=. A to try d A = 2mm x mm - 10mm² to 8.88x1012 Elm d= 0.2mm x 5 10x10 6 x 8.8501012 0.24103 2. 2125X10 F Ar to be AL = 3mmx 5 mm = 15 mm² d 8.85x10-12 flm d= 0.2mm 15x10 by 8.85 X 10-12 x 8 0.2x10-3 5.310 x 10 18 F

And these two capacitances are parallel to each other The circuit becomes G 5 cs(104) 2 V x = xutxa w=109 Total current Xc jwa 109 2.2128 x 1018 -j 0.45 19 4109 on j Ұсу juca 109x 5-310x1018 jo. 1883 yola V - 5 cos (107) -j[6.4519 +0:1883] x 10 7.809 cos/109t +90°) x100A 90] [ Li 7-8094109 A

dreledne brick Ma Displacement Current through ix [. By current division rule Gta 7.809x101 costat +90) x 2.2186 Xiare (2.2125+ 5.80) xial с 2.2967 X109 cos Cat +90°). Magnitude : 2.29674109 A C