$8.3.15^*$ The attenuator is a voltage divider, the circuit of which is shown in the figure. What should be the resistances $R_1$ and $R_2$ so that at each subsequent resistance $R_1$ the voltage is ten times less than at the previous one?
Let's consider the following figure
Applying Kirchhoff Second Law (Loop's law). For loop $I$ $$-i_{11}R_1-i_{22}R_2+i_{21}R_2=0$$ $$-i_{11}R_1+R_2(i_{21}-i_{22})=0 \;(1)$$ As $i_{11}R_1 = 10 i_{12}R_1$, that is, $i_{1n} = 10 i_{1(n+1)}$, hence $$i_{1n} = \frac{i_{11}}{10^{n-1}} \;\forall~n\geq2 \;(2)$$ For loop $II$ and according $(2)$ $$-i_{N+1}(R_1+r)+i_{2N}R_2=0$$ $$-\frac{i_{11}}{10^{N-1}}(R_1+r)+i_{2N}R_2=0 \;(3)$$ For subsequent bifurcations (applying Kirchhoff First Law) $$i_{11} = i_{12}+i_{22}$$ $$i_{22} = i_{11}-i_{12}$$ So, $$i_{2n} = i_{1(n-1)}-i_{1n} \;\forall~n\geq2 \;(4)$$ According $(2)$, let's modify $(4)$ $$i_{2n} = \frac{i_{11}}{10^{n-2}}-\frac{i_{11}}{10^{n-1}}$$ $$i_{2n} = \frac{9i_{11}}{10^{n-1}} \;(5)$$ According $(5)$ into $(3)$ $$9R_2 = R_1+r \;(6)$$ According $(5)$, from $(1)$ $$R_1 = \frac{81}{10} R_2 \;(7)$$ Finally, solving the equation system formed by $(6)$ and $(7)$, it is obtained $$\boxed{R_1 = 9r}$$ and, $$\boxed{R_2 = \frac{10}{9}r}$$
BSc. Luis Daniel Fernández Quintana
Physics Department (FCNE)
Universidad de Oriente, Cuba