Question

Problem 2: For the circuit shown below, find the following: The expression of i(t) fort > 0. The voltage vc (t) fort > 0 Calculate the peak energy stored in the capacitor Calculate the real power dissipated in the load formed by R and C. b) d) 40 UF + 0 -V-24 Ve0 400 Hz

Answer 1

given circuit is i(t),40MF 152 VsCU) 24sinct = 24600 م رے f=400H2 W=214 = 800 T = 2513.274 capacitive reactance , xc a hoc j (801) (40xwor) :- 39.94712 a) current ilt) = . V(+) SRTEZ 2460° 15-j9.947 = 1.334 L33.549º. = 1.334 sincwt + 33.549) -1.334_sin ( 80T7+33.549A b) voltage across capacitors ve(t) = fit) at y (t) = abrio 1-334 sin ( 80078 33.599)

Let u = 80077t + 33.549. sin(8oor + + 33.544) de = f sinuat u=8007t + 33.549 da = 80077 - 800 - 800 SooT I sinudt af sinu do not - I sinudu - (-cosu) Jsinc 800 Ft +33.549) dt = goot (rosu) - Lot (= cos(Botrz+33.5an)) - equation ☺ substitute in equation , we get ve (t) = 1.334 1-1 sin cos (80077433-549) 40X166 ( 80071 ve(t) = -13.2695 cos (B06TTt +33.549) v

c) Energy stored in a capacitor, E= { cv According to this equation, to calculate Peak energy stored in the capacitor first we have to calculate the peak voltage across the capacitor. voltage across Capacitoy, z VC+) = -13.2695 cos(800Tt+33.549) Peak voltage across capacitor. creak) = 13.2695 V Peak energy stored in the capacitor, Ecpeat = crear? = { 2408106(13.2695) = 0.003525 = 3.52 mJ is dissipated in capacitor Real power equal to zero Reve power dissipated in a Resis tog is P = (Syms]xr = G) IS - 13.35W x-