Repeat of Question (2) info: ou have two capacitors: CA 470. nF and C3 285 nF, a
ID: 1659418 • Letter: R
Question
Repeat of Question (2) info: ou have two capacitors: CA 470. nF and C3 285 nF, and a DC battery (Vis resistance in the battery.) 9.00 V). (Ignore any internal Now, the two capacitors are arranged in SERIES and connected to the battery. Enough time passes that the capacitors become fully charged. e. (1 pt.) Draw a schematic diagram for the circuit, showing the two capacitors and the DC battery. Use proper schematic symbols for each. Be neat - use a straightedge or ruler! r.(2 pts.) What is the equivalent capacitance of the capacitor network?C470nF 26n F Show your work. 4 g. (2 pts.) Find the amount of charge on each capacitor, Q and Q- Show your work h. (2 pts.) Find the potential across each capacitor, VA and Vg. Show your work. i. (2 pts.) Find the total energy stored in the capacitor network. Show your work.Explanation / Answer
a) C1=470nF C2=285nF
The equivalent capacitance is given by:
C=(C1*C2)/(C1+C2)= 177.42 nF
b) Total charge = Q = CV = 177.42*9 nC = 1596.76 nC
which is the charge stored in each capacitor as the same current flows
c) Va = Qa/C1 = 3.4V
Vb = Qb/C2 = 5.6V
d) Total energy = .5C1V1^2 + .5C2v2^2 = 2716.6 + 4668.8 nJ = 7185.4 nJ