A network of resistances is connected to a 16 V battery with internal resistance of `1 Omega` as shown in figure. (a) Compute the equivlaent resistance of the network. (b) Obtain the current in each resistor and (c ) obtain the voltage drop`V_(AB), V_(BC)` and `V_(CD)`.

Explanation

(a) In the circuit shown, `4 Omega` and `4 Omega` are in parallel , `12 Omega` and `6Omega` are in parallel. These two combination of resistances are in series with `1 Omega` resistance in the circuit . Total resistance of circuit between A and D is
`R = ( 4 xx 4)/(4+4) + 1 +(12 xx 6)/(12+6) = 2 +1 + 4 =7 Omega`
(b) Total current, `I= (epsilon)/((R +r)) = (16)/(7+1)= 2A`
Current at A is divided equally in each of `4 Omega` resistance in parallel. So,
`I_(1) = I_(2)=2/2 = 1A`
pot. diff. across C and D,
`V_(CD) =I xx R_(CD)= 2 xx 4 = 8 V`
`I_(3) = V_(CD)/12 = 8/12 =2/3A , I_(4) - V_(CD)/6 = 8/6 = 4/3A`
(c ) `V_(AB) = I xx R_(AB) = 2 xx 2= 4V`,
`V_(BC)= I xx 1 = 2 xx 1= 2V` ,
`V_(CD) = I xx R_(CD) = 2 xx 4= 8V`