Question

A 19.0-μF capacitor and a 44.0-μF capacitor are charged by being connected across separate 45.0-V batteries.

Answer 1

Answer:

Explanation:

C1 = 19 micro farad

C2 = 44 micro farad

V = 45 V

A.

The relation between the charge and the potential is

q = C x V

where, c is the capacitance and V be the potential difference

q1 = C1 x V = 19 x 45 = 855 micro Coulomb

q2 = C2 x V = 44 x 45 = 1980 micro Coulomb

B.

Now the charge is shared, the charge on both the capacitors = (q1 + q2)/ 2

q = ( 855 + 1980) / 2 = 1417.5 micro Coulomb

C.

Potentia across 44 micro farad

V' = q/C2 = 1417.5 / 44 = 32.2 V