Circuit, circuit, and circuit

From the three circuits, we have:

$\begin{cases} R_1||R_2 + R_3 = \dfrac {R_1R_2}{R_1+R_2}+R_3 = 40 & \implies {\color{#3D99F6}R_1R_2+R_2R_3+R_3R_1} = 40 (R_1+R_2) & ...(1) \\ R_3||R_1 + R_2 = \dfrac {R_3R_1}{R_3+R_1}+R_2 = 45 & \implies {\color{#3D99F6}R_1R_2+R_2R_3+R_3R_1} = 45 (R_3+R_1) & ...(2) \\ R_2||R_3 + R_1 = \dfrac {R_2R_3}{R_2+R_3}+R_1 = 72 & \implies {\color{#3D99F6}R_1R_2+R_2R_3+R_3R_1} = 72 (R_2+R_3) & ...(3) \end{cases}$

$\begin{aligned} \implies 40 (R_1+R_2) & = 45 (R_3+R_1) = 72 (R_2+R_3) \end{aligned}$

$\begin{cases} (1)=(2): & R_1 = 8R_2-9R_3 & ...(4) \\ (2)=(3): & 5R_1 = 8R_2+3R_3 & ...(5) \end{cases}$

$(4)+3(5): \quad 16R_1 = 32R_2 \implies R_1=2R_2$

$(4): \quad 2R_2 = 8R_2-9R_3 \implies R_3 = \dfrac 23 R_2$

$\begin{aligned} (1): \quad \frac {{\color{#3D99F6}R_1}R_2}{{\color{#3D99F6}R_1}+R_2}+R_3 & = 40 & \small \color{#3D99F6} \text{Note that } R_1 = 2 R_2 \\ \frac {2R_2^2}{3R_2} + R_3 & = 40 \\ {\color{#3D99F6}\frac 23 R_2} + R_3 & = 40 & \small \color{#3D99F6} \text{Note that } R_3 = \frac 23 R_2 \\ {\color{#3D99F6}R_3} + R_3 & = 40 \\ \implies R_3 & = 20 \ \Omega \\ R_2 & = \frac 32 R_3 = 30 \ \Omega \\ R_1 & = 2 R_2 = 60 \ \Omega \end{aligned}$

Therefore, the difference between the highest and lowest resistance among the three resistors: $R_1-R_3 = 60 - 20 = \boxed{40}$ .

Let $x=R_1 R_2+R_1 R_3+R_2 R_3$ .

For every kind of circuit above apply,

Those,

If sum them all,

So

With the same way, we can get,

If substituted to the initial form becomes,

So

So the difference between the maximum resistance and the minimum resistance of the three resistors above is 60Ω - 20Ω = 40Ω.