Question

The circuit in the figure below contains a 90.0 V battery and four capacitors. In the top parallel branch, there are two capacitors, one with a capacitance of C = 1.00 pF and another with a capacitance of 6.00 pF. In the bottom parallel branch, there are two more capacitors, one with a capacitance of 2.00 pF and another with a capacitance of C2 = 3.00 uF. C 6.00 uF 2.00 uF 90.0 V (a) What is the equivalent capacitance (in F) of the entire circuit? 1.03 X Which capacitors are in series, and which are in parallel? Find the equivalent capacitance for the two capacitors in series across each branch first. Then find the equivalent capacitance of the circuit from the two parallel equivalent capacitances. UF (b) What is the charge (in C) on each capacitor? on 1 on C₂ on the 6.00 uF capacitor on the 2.00 pF capacitor (c) What is the potential difference in V) across each capacitor? across C1 across C2

Answer 1

Sexies 띠 6Mf L HP series corcuit I 2씨 덜 900 한한 구상= Ceq = e ut 할 =능동 Ceq = 8 ut Equivalent circuit : 특서 Parallel -circuit I 은빠 Ceg =월용 =끓 35 = 20057Uf ! Equivalent Capacitante = 2.057Mf Equivalent circut 끓f - Circuit O 90V

In series connection, charge across capacitors remains same * Potential varies parallel connection, & charge varies * Potential across capacitors remains same. In For circuit II c = 72 uf V= 90V coa q=cv - 72 x 90 = 185.14 Mc For circuit II, potential at 3 uf = 90v potential at & df = 900 charge at 3 mf = cv = 6 x 90 = 97.14 MC Charge at & ut = cv = x 90 = 108 Mc

for circuit I, [series connection) charge charge at at a 6uf = 77. 14 MC = 77.14 uc charge charge at at ruf = 108mC a = 108 mc (series connection] potential at c = a = 77.14 = 77.14 1 potential at 6lf a = 77.14 = 12.864 potential at allf = a C = 108 = 540 poteritial at es = a 2 = 103 = 364