Question

114a-three capacitors C1,C2,C3 connected in series across DC voltage source of 15V. the corresponding values of the capacitances are C1=8UF,C2=1UF, C3=3UF

draw the equivalent circuit and calculate the a-equivalent capacitance of the circuit b-charge of each capacitor

114b-two capacitors are connected in series giving an equivalent capacitance of 60uF, the total charge for the equivalent capacitance is 15x10^-3 calculate the stored energy in the capacitor

Answer 1

Answers:

114a) The circuit described in the problem is as shown below:

In series the capacitances add up in inverse fashion, that is

1/Ceq = 1/C1 + 1/C2 + 1/C3

Therefore

1/Ceq = 1/8 + 1 + 1/3 = 12/24

Therefore Ceq = 2 microfarad

Therefore the equivalent circuit is:

Charge on each capacitor:

In a series circuit all the capacitors have the same charge. Therefore each capacitor has Charge Q = Ceq*V = 2*15 = 30 micro-coloumbs

Answer 114b:

Energy Stored in a capacitor is given by E = 0.5 * C * V²

We also know that V = Q/C

Therefore E = 0.5 * C * Q²/C²

or E = 0.5*Q²/C

E = 0.5**225*10^-6/60*10^-6

E = 0.5*3.75 = 1.875 J

Hope this helps.