An ideal task.

Relevant wiki: Kirchoff's Voltage Law (closed loops)

This scheme can be simplified like so. On the simplified picture below, we have also assumed the directions of currents $I_1$ , $I_2$ , and $I_a$ .

From the picture, we see that the voltage the voltmeter measures is $U_v = R \times I_a$ .

We can now calculate $R$ as $R = \frac{U_v}{I_a} = 1kΩ.$

From the task, we have the power of the resistor $4R$ and that power is $P = 4R \times I_2^2$ . From that, we calculate $I_2$ as $I_2 = \sqrt {\dfrac{P}{4R}} = 5 mA.$

Using Kirchoff's current law (conservation at nodes) , we can say that $I_a = I_1 + I_2$ , and we can calculate $I_1$ as $I_a-I_2$ , which is equal to $5 mA$ .

Because the resistors are in a parallel connection, their voltages are the same, and we can write $R_x \times I_1 = 4R \times I_2$ , from where we calculate $R_x$ as $4R$ , which is equal to $4kΩ$ .

On the even simpler picture above, I have marked points $A$ and $B$ . The voltage between them is $U_{ab} = 4R \times I_a = 40V$ . That voltage is also equal to $R \times I$ , from where we calculate $I$ as $I = \frac{U_ab}{R} = 40mA.$ The desired current Is is calculated as $I+I_a = 50mA$ .

The power of the generator is calculated as $U_{ab} \times I_s = 2W$ .

The text states that the inputed answer should be $1 + 4 + 50 + 2 = \boxed{57}.$

1a. $R = x \text{ k}\Omega$ : We note that the current through $2R||2R$ together is $I_a=10 \text{ mA}$ . Therefore,

$\begin{aligned} \quad I_a(2R||2R) & = U_v \\ 0.01 R & = 10 \\ \implies R & = \frac {10}{0.01} = 1000 = 1 \text{ k}\Omega \\ \implies x & = 1 \end{aligned}$

1b. $R_x = y \text{ k}\Omega$ : Let the current through $4R$ be $I_{4R}$ , then we have:

$\begin{aligned} \quad I_{4R}^2(4R) & = 0.1 \text{ W} \\ 4000 I_{4R}^2 & = 0.1 \\ I_{4R}^2 & = \frac {0.1}{4000} \\ I_{4R} & = \frac 1{200} = 5 \text{ mA} \end{aligned}$

$\quad \implies$ The current through $R_x$ , $I_x = I_a - I_{4R} = 10 - 5 = 5 \text{ mA} = I_{4R}$ $\implies R_x = 4R = 4 \text{ k}\Omega$ $\implies y = 4$ .

2. $P_{cg} = z \text{ W}$ : Let the right branch of resistors of current generator $I_s$ be $R_a$ then the left branch $R_b = R$ . We have $R_a = Rx||4R + R + 2R||2R = 2R + R + R = 4R = 4R_b$ . Now, the current through $R_a$ is $I_a$ and by current division, the current through (R_b), $I_b = 4I_a$ . Therefore,

$\begin{aligned} \quad P_{cg} & = I_a^2R_a + I_b^2R_b \\ & = 4I_a^2R + (4I_a)^2R \\ & = 20 I_a^2R = 20 (0.01^2) (1000) = 2 \text{ W} \\ \implies z & = 2 \end{aligned}$

3. $I_s = q \text{ mA}$ : $I_s = I_a + I_b = I_a + 4I_a = 5I_a = 50 \text{ mA}$ $\implies q = 50$ .

Therefore, $x+y+z+q = 1 + 4 + 2 + 50 = \boxed{57}$