A network of resistor is connected to a 16 V battery with internal resistance of `1 Omega`, a shown in fig.
Obtain the voltage drops `V_(AB), V_(BC)` and `V_(CD)`

Explanation

The voltage drop across AB is
`V_(AB) = I_(1) xx 4 = 1 A xx 4 Omega = 4V`
This can also be obtained by multiplying 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) = 2A xx 1 Omega = 2V`
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) = 2A xx 4 Omega = 8V`
Note that the total voltage drop across AD is `4V + 2V + 8V = 14 V`. Thus the terminal voltage of the battery is 14V, while its emf is 16V. The loss of the voltage (=2V) is accounted for by the internal resistance `1 Omega` of the battery `[2A + 1 Omega = 2V]`