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

Explanation

(a) The network is a simple series and parallel combination of resistors. First the two `4Omega` resistors in parallel are equivalent to a resistor `= [(4xx 4) //(4 +4) Omegta = 2 alpha`
In the same way. The `12 Omega and 6 Omega` resistores in parallel are equivalent ot a resistor of
` [ (12 xx 6) //(12+ 6) ]Omega = 4 Omega`
The equivalent resistance R of the networn is obtained by conmbining these resistore `(2 Omega and 4 Omega)` with `1 Omega` in series. that is.
`R = 2 Omega + 4 Omega + 1 Omega = 7 Omega`
(b) The total current In in the circuit is
`I = (in )/(R +r) = (16 v)/((7 +1) Omega) = 2A`
Consider the restors between A and B. If `I _(1)` is the current in one of the `4Omega` resistors and ` I (2)` the current in the other.
`I _(1) xx 4 = I _(2) xx 4`
then is, ` I _(1) = I _(2) ,` which is otherwise obvious from the symmetry of the two arma. But `I _(1) + I _(2) = I =2 A.` Thus.
`I _(1) = I _(2) =1A`
that , is current in each `4Omega` resistor is 1A. Current in `1Omega` resistor between B and C would be 2 A.
Now, consider the resistance between C and D. If `I _(3)` is the current in the `12Omega` resistor, and `I_(4)` in the `6 Omega` rsistor.
`I _(3) xx 12 =I _(4) xx 6,i.e., I _(4) = 2I _(3)`
But, `I_(3) + I _(4) =I =2A`
Thus, ` I _(3) = ((2)/(3)) A, I _(4) = ((4)/(3)) A`
that is, the current in the `12Omega` resistor is (2/3) A, whithe the current in the `6 Omega` resistor is (4/3) is
`V _(AB) = I _(A) xx 4 = 1A xx 4 Omega = 4V.`
This can also be obtained by multtplying the total current between A and B by the equivalent resistance between A and B, that is.
`V_(AB) =2A xx 2 Omega =4V`
The voltage drop across BC is
`V _(BC)=2 A xx 1 Omega =2 V`
Finally, the voltage drop across CD is
`V _(CD) =12 Omega xx I _(3) =12 Omega xx ((2)/(3)) A =8V.`
This can alternately be obtained by multiplying total current between C and D by the equivalent resistance between C and D. that is
`V _(CD) =2 A xx 4 Omega =8V`
Note that the total voltage drop across AD is `4 V + 2V +8V =14V.` Thus, the terminal voltage of the bettery is 14 V, while its emf is 16 V. The loss of the voltege `(=2V)` is accounted for by the internal resistance `1 Omega` of the bettery `[2 A xx 1 Omega = 2V].`