Answer
$R$ $=$ $10$ $kohms$
$v$ = $40$ $volts$
$v_{1}$ $=$ $10$ $volts$
$i$ $=$ $I_{T}$ $=$ $1$ $mA$
Work Step by Step
This is a series circuit with missing values for one resistor and the voltage source.
A series circuit has only one current value passing through all its resistors. Let's focus on $R_{2}$ as this resistor has both the value of resistance and the voltage across it.
$I_{T}$ $=$ $I_{R_{2}}$ $=$ $\frac{V_{R_{2}}}{R_{2}}$ $=$ $\frac{v_{1}}{R_{2}}$ $=$ $\frac{\frac{v}{4}}{10\times10^{3}}$ = $\frac{v}{40\times10^{3}}$
Now, given the total power of the circuit $P_{T}$ and the equation of the total current $I_{T}$ we can now determine the value of source $v$:
$P_{T}$ $=$ $I_{T}\times{v}$
$40\times10^{-3}$ $=$ $\frac{v}{40\times10^{3}}\times{v}$
$40\times10^{-3}\times40\times10^{3}$ = $v\times{v}$
$1,600$ $=$ $v^{2}$
$v$ = $40$ $volts$
Therefore, $I_{T}$ is:
$I_{T}$ $=$ $\frac{v}{40\times10^{3}}$
$I_{T}$ $=$ $\frac{40}{40\times10^{3}}$
$I_{T}$ $=$ $1$ $mA$
and $v_{1}$ s:
$v_{1}$ $=$ $\frac{v}{4}$
$v_{1}$ $=$ $\frac{40}{4}$
$v_{1}$ $=$ $10$ $volts$
We can now compute for the total resistance $R_{T}$ and $R$ using ohm's law:
$I_{T}$ $=$ $\frac{v}{R_{T}}$
Arranging to get $R_{T}$
$R_{T}$ $=$ $\frac{v}{I_{T}}$
$R_{T}$ $=$ $\frac{40}{1\times10^{-3}}$
$R_{T}$ $=$ $40$ $kohms$
$R_{T}$ $=$ $R_{1}$ + $R_{2}$ + $R$ + $R_{3}$
$40$ $kohms$ $=$ $8$ $kohms$ + $10$ $kohms$ + $R$ + $12$ $kohms$
$R$ $=$ $10$ $kohms$
