Question

A 10.0-uF capacitor is plugged into a 110 V-rms 60.0-Hz voltage source, with an ammeter in series. Show that the rms current through the capacitor is 0.415 A.

step by step plase

Answer 1

la) c = Wc 376.992 x 10-5 (6) The instantaneous current in ce is given by soin Here Vrms = lov frequency: f= 60 Hz capacitor = 10.0uF = 10x 10-6f Now first we find the angular frequency w = 2 T f W = 2x 3.1416 X 60 376.992 rad / sec. since 1 X 10-5 F Xo Reactance will be 2 65.2576 using the coloumbs low rms current is given by Irms = Vrms хс 265.2576 0.415 Amp Hence Proved. LC when, is closed is I = wo Qmax cos (wt + II) where w= Q = initial charge across Sic the capacitor Now il 6 こ max

first we will find w=1 Snc putting value of 10.20H) (4x10-6)F L and c. 1118 rad Is Qmax CV (4x10-6F) (200) 800 x 10-6c = 8 x 10-4c we get. Now putting the value of all in 0.70A = (1118 rad/s) (800 x10-6 c) Co 3 (1118 radis)t +T 2 cos ((1118 rad is) t + IT -0.7826 24 2 t = Lost (-0.78 264) - 11/2 1118 radis 8.047 810-4 S Instantaneous charge across capacitor is Qmax Coswt. (800x10-6 I cos 61118 rad,s) (8.047810-45) = 497.5 x 10-6 c Thus energy is 497.5x10-6 30.98 10-3 ] 2 x 4.0x 10-6 u = Q2 2c 31 ms Hince Proued