Question

INSTRUCTIONS: Read the questions carefully and answer the following items accordingly. Show your solution properly. Q6 An Amplitude-versus-time characteristic curve is given in Figure 2. It is the graphic response of a DC supplied series RC circuit. Determine: a) Draw the circuit, name the parts of a Series RC DC circuit and show the mathematical relations of initial current, voltage across the capacitor at charging state and stored energy by the capacitor. (2 Mark) b) Describe in detail about the item (a) according to the graphic response in Figure 2. (5 Mark) c) The capacitance is 75uF while the resistance is 60k 2, calculate: i) time constant, ii) the voltage source E, iii) the capacitor at t=1000ms, and iv) the voltage across the capacitor at t3 Sec. (3 Mark) Time-domain Graph of a DC series RC Circuit Amplitude M time (m5) Charging State Steady Stote -100 State - v Figure 2

Answer 1

page- (1) Source Free first Order RC Circuit: + Vco) TC R only teoo - This capacitor will have initial condition. This is initial voltage, Vc (0) Given, Vc (0)= V(o) os elemente Vc (0)=100v. loov st I=Rc state -t/Rc V=Vclo) e - t/Rc V=100 e volt discharging steady E = – cv² (11-) step response of first order RC circuit: t=0 R X- Here. Vc(o)=ov. V ст. at (as graph is starting O and rises.) Vc (00) = loov

page-2 V(t) = Vs + (vo - Vs) e the Amplitude Y() + ( VOO) - V(o)) loov = 100+ (0-100) & the éthi) Holt t(m) V(t) = 100 100(1-e charging time steady state -t/Rc V=100 e V = (coli-etiz) 턀글2 E = 1/2 cve Amplitude -HRC » V(t)=100e tims) Amplitude 100 * V(t)= 100/1-ette) volt

page-3 c) (i.) I =Rc Go X10 X 75X106 = 4500X11 x 10 3 I= 4.5 sec * + - VR = 1008 VC(O)=ov to E By KVL: -E + VR=0 E=VR E = 100V (iii) t=1000ms * In V = 100 - ERC (IV.) V = 100 e - 3/4.5 as to ie, as t becomes very large, circuit enters into steady state steady state, "Capacitor is open! 100 (1-2 Hz) =10011-€ 345) 1V=51-34 Volt = 100 e IV=48.658 Volt