Question

2- (4 Points) For the following network, draw the steady state circuit, then calculate the energy stored in the capacitor and in the inductor and the power dissipated in each resistor. .- 2 mH m + MA 30 k12 20k2 6uF V

Answer 1

2 =L: dt The Voltage across inductor is given di 번 by In steady state to dto Ne =0. Hence inductor bechave as shost circuit in steady state. current through the capacitor is given by, i = c d Vc hn steady state tayo dto → .iso Hence in steady state capacitor beehave as open circuit. Hence Circuit for steady state can be drawn just by shost circuiting the inductor and open circuit the Capacitor. dt

+ V 30 KR using current 30 X 5X10-3 VT, I. bl 5mA 320kn Now voltage across Capacitos drop in 20 KR resistor. Iz Can be calculated division rule 30x103 ie In = x 5x10-3 A 30x103+ 20x103 Iz = 50 3m A 5mA-12=( 5-3) mA =2m A stored in inductor = {li ::is EL = ** 2x10-3x (3x10-3)2 9x10-9 EL =9nJ (nano Joules) Energy stored in capacitor = E(= £cr? V= drop across 20 KR resister V= 1X20X103 v » Vs '3x10-3x20x103 v = ) - Iz i. I - Hence, energy > EL -

:: v= 60V . = £ cu ¿ x 6x10-6x (60)2 » Ec = 0.0108 3: Ę = 10-8 m 3 (milli Joules) Power dissipated in 30 Kh resistor = Ego = » 'Ego = (2x10-3); 3oxie | 20 X10-3 I, X 30x103 = Ezo 3 E30 = 120 m power dissipated in 20kr Leesistor = = 1,2x20x103 Eze = (3x10-3)*x 20x105 1808 10-3 J. Exo = 180 m] =) Ezo 1]