Answer
$R_{eq}$ $=$ $10$ $ohms$
$I_{2}$ $=$ $0.05$ $A$
Work Step by Step
To get the equivalent resistance, we must start simplifying the resistor values beginning from the one farthest from the source:
$R_{5}$ and $R_{6}$ **parallel**
$R_{56}$ $=$ $\frac{R_{5}\times{R_{6}}}{R_{5}+R_{6}}$ $=$ $\frac{4\times{4}}{4+4}$ $=$ $2$ $ohms$
$R_{4}$ and $R_{56}$ **series**
$R_{456}$ $=$ $R_{4}$ + $R_{56}$ $=$ $22$ + $2$ $=$ $24$ $ohms$
$R_{456}$ and $R_{7}$ **parallel**
$R_{4567}$ $=$ $\frac{R_{456}\times{R_{7}}}{R_{456}+R_{7}}$ $=$ $\frac{24\times{8}}{24+8}$ $=$ $6$ $ohms$
$R_{4567}$ and $R_{3}$ **series**
$R_{34567}$ $=$ $R_{3}$ + $R_{4567}$ $=$ $4$ + $6$ $=$ $10$ $ohms$
$R_{34567}$ and $R_{2}$ **parallel**
$R_{234567}$ $=$ $\frac{R_{34567}\times{R_{2}}}{R_{34567}+R_{2}}$ $=$ $\frac{10\times{90}}{10+90}$ $=$ $9$ $ohms$
$R_{234567}$ and $R_{1}$ **series**
$R_{1234567}$ $=$ $R_{eq}$ $=$ $R_{1}$ + $R_{234567}$ $=$ $1$ + $9$ $=$ $10$ $ohms$
Refer to the attached figure below for the current directions.
Since $R_{1}$ is in series with the voltage source, we can say that $I_{1}$ $=$ $I_{T}$.
$I_{1}$ $=$ $\frac{V}{R_{eq}}$ $=$ $\frac{50}{10}$ $=$ $5$ $A$
and
$V_{1}$ $=$ $I_{1}\times{R_{1}}$ $=$ $5\times{1}$ $=$ $5$ $volts$
To get the value of $i$ or $I_{2}$, we will need to get the value of $V_{2}$ using the voltage divider method:
$V_{2}$ $=$ $V_{1}\times{\frac{R_{2}}{R_{2}+{R_34567}}}$ $=$ $5\times{\frac{90}{90+10}}$ $=$ $4.5$ $volts$
and
$I_{2}$ $=$ $\frac{V_{2}}{R_{2}}$ $=$ $\frac{4.5}{90}$ $=$ $0.05$ $A$
