Physics, asked by aaditi3845, 1 year ago

two capacitors of capacitance 6 micro F and 12 micro F are connected in series with battery. the voltage across the 6 micro F capacitor is 2 V . Compute the total batter voltage.

Answers

Answered by Anonymous
83

Given,

C 1 = 6 μ F

C 2 = 12 μ F

Potential Difference across C 1 = 2 V

let the potential Difference across C 2 be V volt.

As the two capacitors are connected in series , the charge on each capacitor must be same.

∴ Charge on 6 μ F capacitor = Charge on 12μ F capacitor.

on putting the values,

6 F× 2 V = 12 F× V volt

on solving,

\frac{6*2}{12} = 1V

We know,

The emf ( E) is equal to the sum of potential difference across all the components of a circuit.

Battery voltage (E)  = V1 + V2

                                 = 2 V + 1V

                                 = 3 V


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Answered by Anonymous
47

we know

Q = C ×V

in 6uf capacitor v = 2 given

then Q= 6×2= 12 micro coloumb

in series we know that Q is same.

in series c equivalent

c1×c2/(c1+c2) = 4

then Q = cequivalent × net potential

V= 12/4 = 3

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