Question

ID: See __ Group: - Q1. Consider a parallel-plate capacitor connected to a battery, a) How does the charge in this capacitor change if Potential difference is tripled (by varying the voltage of the battery). (1 mark) b) How the energy does stored change if the capacitance of the capacitor is doubled? Q2. Answer to the following questions by using the above capacitor combination circuit. The values of capacitors are C=18 uF, C2-6 uF C;=36 uF C=18 uF and Supply voltage, V=72v a) What is the total charge stored in the equivalent capacitor? (2 mark) PHY2011 Final Exam Fall 19-20 Couse Instructor-Sajeena Harees q=0(1-e%c) += RC 1= 1,8*% c=Y U- v2

Answer 1

a) Q = CV V 3V - Q -3CV + 30 1. The change is tripled 3) E22cu2 - caza - Ey zacv² = 2 (cu?) RE - Energy stored in capacitor also deables 2) a) Cz 2 (3 are ipso in parallel - Ceq : Carlz = 6+36= 42 u F - į , cep, 2 4 are in series to E c t test ap 5 tetot + to cey = 4 8 7.412 uF Any Qe leave 71412 * 72 · 533.65 cc b) Total Electric Pe stored a & Ceq uzo 0.5% 7.412 x1067222 0.0/ 925 - Any (e ci same a cey, 2 cq are in series a will have charge Q = 70 533165 uL vez do = V3 V4 = - v, ce 533.65 - 29.65 V 29.65 V. Bry 533 65. 12.711 2 = Ay A V2 = V₂ = V = 18 3 Przy staner = 29.65V As