Question

capacitors are connected as shown to a potential difference of V=14.0 volts. CisC,-Cr=C,-C,-20 μ1: The charge on C5 is A) 40 pC B) 60 HC C) 80 uC D) 120 HC E) 160 μC C1 C3 C2 29·Four identical 1 mF capacitors are connected together electrically what is the greatest possible capacitance of the combination? A) 8 mF B) 4 mF 30 Five capacitors are connected as shown. C2 40 μ F, C4# 10 μ1, Ci=C,sCF 20 μ F The charge on capacitor Cs is 40 μ(. The potential difference between point A and B, VA-VB- Cz Cs

Answer 1

Series c_12 = c_1 c_2/c_1 + c_2 = 20 times 20/2 0 + 20 = 10 mu F parallel c_1245 = c_12 + c_4 + c_5 = 10 + 20 + 20 = 50mu F Series c_eq = 20 times 50/20 + 50 = 100mu F/70 Q = c_eq times V = 100/7 times 10^-6 times 200muc potential across 50 mu F is V = Q/c = 200 times 10^-6/50 times 10^-6 = 4v charge on c_5 in Q_5 = c_5 v_5 = 20 times 10^-6 times 4v Q = 80muC Greater possible capacitance is possible when they joined in parallel c_eq = c_1 + c_2 + c_3 + c_5 rightarrow 1 + 1 + 1 + 1 = 4mF