In the figure below, what is the value ($$kOmega$$) of R needed to make $$i_1=1.25"mA"$$?

 

We first want to simplify the circuit to find the equivalent resistance ($R_{eq}$).

Resistors in series add together, so we can add the two 4k$\Omega$ resistors together.

$R_{eq1}=4k\Omega +4k\Omega =8k\Omega$

Next, we want to get an expression for the combined parallel resistance of $R_{eq1}$ and $R$. We can use the equation for finding the equivalent resistance of two resistors in parallel to do so:

$$\begin{gathered} R_{eq}=\frac{R_1{R}_2}{R_1+R_2} \\ {R}_{eq2}=\frac{\left(R\right)\left(8k\Omega \right)}{R+8k\Omega } \end{gathered}$$

Finally, we add this to the 4k$\Omega$ resistor in series to find the equivalent resistance of the circuit.

$$\begin{gathered} R_{eq}=4k\Omega +R_{eq2} \\ {R}_{eq}=4k\Omega +\frac{\left(R\right)\left(8k\Omega \right)}{R+8k\Omega } \\ {R}_{eq}=\frac{\left(4k\Omega \right)\left(R\right)+\left(32000k\Omega \right)}{R+8k\Omega }+\frac{\left(R\right)\left(8k\Omega \right)}{R+8k\Omega } \\ {R}_{eq}=\frac{\left(12k\Omega \right)\left(R\right)+\left(32000k\Omega \right)}{R+8k\Omega } \end{gathered}$$

Now that we have the equivalent resistance, we use Ohm's law to solve for R:

$$\begin{gathered} V = IR \\ 10V= 1.25mA *\left(\frac{\left(12k\Omega \right)\left(R\right)+\left(32000k\Omega \right)}{R+8k\Omega }\right) \\ 8000 = \frac{\left(12k\right)\left(R\right)+32000k}{R+8k} \\ 8kR + 64000k = 12kR + 32000k \\ 4kR = 32000k \\ R = 8k\Omega \end{gathered}$$

Therefore, R must be equal to 8k$\Omega$.