Two capacitors 60 farad and 40 farad are connected in series. If an external battery is applied which is 60 V, find the charge stored on each capacitor plate, and the total electrical energy stored in the system of two capacitors. Along with finding the potential difference across each capacitor.

Asked by Topperlearning User | 22nd Apr, 2015, 11:43: AM

Expert Answer:

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Applying formula for series capacitance, we have equivalent capacitance for two capacitors = begin mathsize 11px style fraction numerator left parenthesis straight C 1 space cross times space straight C 2 right parenthesis over denominator left parenthesis straight C 1 space plus space straight C 2 right parenthesis end fraction space equals space fraction numerator 60 space cross times space 40 over denominator 60 space plus space 40 end fraction space equals space 24 space straight F end style

 Now, the electrical energy stored in a capacitor ,

 It is easier to use the formulabegin mathsize 11px style 1 half straight Q squared over straight C end style , because capacitors in series have the same charge stored on their plates.

Let’s first find the charge on each capacitor

Charge on each capacitor = 24 × 60 C = 1440 C.                                                

The energy stored in the system of two capacitors,

=

Now, the capacitors in series with different capacitances will not have the same potential difference across them but the potential differences should add up to the total applied voltage of 60 V.

The potential difference across the first capacitor = V=q/C= 1440/60 = 24 V

and the potential difference across the second capacitor = 60 – 24 V = 36 V.

Answered by  | 22nd Apr, 2015, 01:43: PM