Question

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Answer 1

Answer:

A and B are terminals having half wire of parallel connection

hope it helps you

Here 1:1 is ratio of capacitance so ! the potential difference becomes half that is 2 volt

Answer 2

Answer:

(4) 8μF

Explanation:

8 & 8 are in series .....hence potentials are in (1/8 : 1/8 ) 1 : 1 ratio & resultant capacity 8×8/(2×8) = 4

now this 4& other 4 are in // hence...the volatage is same but capacity is 4 + 4 = 8

then this 8 & other 2 are in series ....hence voltages are in (2:8) 1: 4 ratio...

now since total V = 20V then the voltage in sub circuit is 20 × 1/5 = 4V

this 4V is now split into 1:1 ratio giving 2V at the 8μF capacitor