Answer
$R_{eq}$ $=$ $10$ $ohms$
$i$ $=$ $0.9$ $A$
$i_{1}$ $=$ $0.8$ $A$
$v$ $=$ $7.2$ $volts$
Work Step by Step
The equivalent resistance of the parallel circuit can be solved by paralleling $72$ $ohms$ and $9$ $ohms$ resistors then series with $2$ $ohms$ resistors as shown below:
$R_{eq}$ $=$ $\frac{72\times{9}}{72+9}$ + $2$
$R_{eq}$ $=$ $10$ $ohms$
Since the $2$ $ohms$ resistor is in series with the source, current $i$ is equal to $I_{T}$:
$i$ $=$ $I_{T}$ $=$ $\frac{V}{R_{eq}}$
$i$ $=$ $\frac{9}{10}$
$i$ $=$ $0.9$ $A$
and therefore, the voltage across the $2$ $ohms$ resistor is:
$v$ $=$ $0.9\times{2}$
$v$ $=$ $1.8$ $volts$
Using KVL, we will have:
$V$ - $v$ - $v_{1}$ $=$ $0$
$9$ - $1.8$ - $v_{1}$ $=$ $0$
$v_{1}$ $=$ $7.2$ $volts$
The current $i_{1}$ can now be solved:
$i_{1}$ $=$ $\frac{7.2}{9}$
$i_{1}$ $=$ $0.8$ $A$
Since the $9$ $ohms$ and $72$ $ohms$ resistors are parallel, they have the same voltage value.
$v_{1}$ $=$ $v$ $=$ $7.2$ $volts$
