$A$ network of resistors is connected to a $16\; V$ battery with internal resistance of $1\; \Omega$ as shown in Figure:
$(a)$ Compute the equivalent resistance of the network.
$(b)$ Obtain the current in each resistor.
$(c)$ Obtain the voltage drops $V_{A B}, V_{B C}$ and $V_{C D}$.

(N/A) The network is a simple series and parallel combination of resistors. First,the two $4\; \Omega$ resistors in parallel are equivalent to a resistor $= [(4 \times 4) / (4 + 4)]\; \Omega = 2\; \Omega$.
In the same way,the $12\; \Omega$ and $6\; \Omega$ resistors in parallel are equivalent to a resistor of $[(12 \times 6) / (12 + 6)]\; \Omega = 4\; \Omega$.
The equivalent resistance $R$ of the network is obtained by combining these resistors ($2\; \Omega$ and $4\; \Omega$) with the $1\; \Omega$ resistor in series,that is,$R = 2\; \Omega + 4\; \Omega + 1\; \Omega = 7\; \Omega$.
$(b)$ The total current $I$ in the circuit is:
$I = \frac{\varepsilon}{R + r} = \frac{16\; V}{(7 + 1)\; \Omega} = 2\; A$.
Consider the resistors between $A$ and $B$. If $I_{1}$ is the current in one of the $4\; \Omega$ resistors and $I_{2}$ the current in the other,$I_{1} \times 4 = I_{2} \times 4$,that is,$I_{1} = I_{2}$,which is obvious from the symmetry. Since $I_{1} + I_{2} = I = 2\; A$,we have $I_{1} = I_{2} = 1\; A$.
Thus,the current in each $4\; \Omega$ resistor is $1\; A$. The current in the $1\; \Omega$ resistor between $B$ and $C$ is $2\; A$.
Now,consider the resistances between $C$ and $D$. If $I_{3}$ is the current in the $12\; \Omega$ resistor and $I_{4}$ is the current in the $6\; \Omega$ resistor,$I_{3} \times 12 = I_{4} \times 6$,i.e.,$I_{4} = 2 I_{3}$. Since $I_{3} + I_{4} = I = 2\; A$,we have $I_{3} = (2/3)\; A$ and $I_{4} = (4/3)\; A$.
$(c)$ The voltage drop across $AB$ is $V_{A B} = I_{1} \times 4 = 1\; A \times 4\; \Omega = 4\; V$.
The voltage drop across $BC$ is $V_{B C} = I \times 1\; \Omega = 2\; A \times 1\; \Omega = 2\; V$.
The voltage drop across $CD$ is $V_{C D} = I_{3} \times 12\; \Omega = (2/3)\; A \times 12\; \Omega = 8\; V$.