For the circuit s

The resistance of 6 ohm and 3 ohm are connected in series. Therefore, their net resistance can be calculated as

R= R₁+R₂

Here, R₁= 6ohm

R₂= 3ohm

so R= 6ohm +3ohm= 9 ohm

The current through 6 ohm resistor= current through line 1 = I= V/R =4/9=0.44V

(ii) The resistance of 12ohm and 3 ohm are connected in series. Therefore, their net resistance can be calculated as

R= R₁+R₂

Here, R₁= 12ohm

R₂= 3ohm

so R= 12ohm +3ohm= 15 ohm

The current through them = I= V/R =4/15

Potential difference across 12 ohm resistor= 4/15 ×12 =3.2