Question

A capacitor (C=250 mF) and resistor (R= 450 ohms) are connected in series to a 9.00 V power supply. a.) How many seconds after their connection is the charge on the capacitor 80% of its maximum? b.) What is the current at t= 50.0 seconds?

Answer 1

capacitance, c = 2'50 mg = 250 X10-3F Resistance, R = 450 s2 For a CR series circuit (for charging of the capacitor ), charge on the capacitor at time t is, q=@o Cl- e-ticR) q. = Maximum charge on capacitor Given , q = 80 % = 0.8% - 0.890 = qo ci-e-T/CR): = 0.8 = l-e-ticR => e-tICR = 0.2 > en e-tier = ln (0.2) * - = -1.609 et (250 x10-3)x(450) = 1.609 = t=181.01S L b) current in the circuit, I = ţe-ther voltage of the power supply ,v=gu Given, t=50s . -I = e - 7250x10–3) *(450) = 12,82 X10-3 A = 112.89 may so