1.

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»

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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