In the circuit shown. The potential drop across each capacitor is (assuming the two diodes are ideal).
.

Options

A. 12 V,12 V
B. 16 V,8 V
C. zero, 24 V
D. 8 V, zero

Answer

B

Explanation

The diode connected parallel to the battery is reverse bias. So current will not pass through it. So total emf divided among `C_(1)` and `C_(2)` and in the inverse ratio of their capacities
`V=(q)/(C),(V_(1))/(V_(2)) =(C_(2))/(C_(1))`
`V_(1)=(VC_(2))/(C_(1)+C_(2)),V_(2)=(VC_(1))/(C_(1)+C_(2))`.