A network of Resistors is Connected to a 16V battery with internal resistance `1Omega` as shown in Figure below.
(a) Compute the equivalent resistance of the network
(b) Calculate the total current in the circuit

Explanation

`R_(P)= (R_(1)R_(2))/(R_(1)+ R_(2))`
4ohm and 4ohm in parallel: `R_(P^(1))= (4 xx 4)/(4 + 4)=2` ohm
12ohm and 6ohm in parallel: `R_(P^(2)) = (12 xx 6)/(12 + 6) = 4` ohm
`R_(P^(1))`, 1 ohm and `R_(P^(2))` in series: Equivalent resistance of network `R_(eq)= R_(P^(1)) + 1 + R_(P^(2)) =2 + 1 + 4= 7` ohm
Total current in the circuit: `I= (epsi)/((R_(eq) + r))= (16)/(7+1)= 2A`