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 *
V ^{2}**

We also know that V = Q/C

Therefore E = 0.5 * C * Q^{2}/C^{2}

or E = 0.5*Q^{2}/C

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

**E = 0.5*3.75 = 1.875 J**

Hope this helps.

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