Question

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.

Answer 1

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

Answer 2

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