Find (a) Equivalent resistance
(b) and current in each resistance

Explanation

(a) `R_(eq)=(1/16+1/8+1/16)^-1+1=5Omega`
`i=60/(4+1)=12 A`
Hence 12 A will flow through the cell
By using current distribution law.
Current in resistance `10Omega` and `6Omega` = A
Current in resistance `5Omega` and `3Omega` = 6 A
Current in rrsistance `20Omega` = 0
Current in resistance `16Omega` = 3 A