I have to set the battery voltage to 14.5V from the first divider, then from that I have to attenuate it to 1V to feed to the input of an op amp. Now this answer is inspiring but doesn't solve my purpose. Maximum I can assume the R2 & R4. R3 & R5 still unknown.
Please help.
Well, we are trying to analyze the following circuit:
simulate this circuit – Schematic created using CircuitLab
When we use and apply KCL, we can write the following set of equations:
$$\text{I}_1=\text{I}_2+\text{I}_3\tag1$$
When we use and apply Ohm's law, we can write the following set of equations:
$$ \begin{cases} \text{I}_1=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_1}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_1-\text{V}_2}{\text{R}_3}\\ \\ \text{I}_3=\frac{\text{V}_2}{\text{R}_4} \end{cases}\tag2 $$
Substitute \$(2)\$ into \$(1)\$, in order to get:
$$ \begin{cases} \frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}=\frac{\text{V}_1}{\text{R}_2}+\frac{\text{V}_1-\text{V}_2}{\text{R}_3}\\ \\ \frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}=\frac{\text{V}_1}{\text{R}_2}+\frac{\text{V}_2}{\text{R}_4} \end{cases}\tag3 $$
Now, we know the values of \$\text{V}_\text{i}\$, \$\text{R}_1\$, \$\text{R}_3\$, \$\text{V}_1\$, and \$\text{V}_2\$. So coloring the things we know red, we see:
$$ \begin{cases} \frac{\color{red}{\text{V}_\text{i}}-\color{red}{\text{V}_1}}{\color{red}{\text{R}_1}}=\frac{\color{red}{\text{V}_1}}{\text{R}_2}+\frac{\color{red}{\text{V}_1}-\color{red}{\text{V}_2}}{\color{red}{\text{R}_3}}\\ \\ \frac{\color{red}{\text{V}_\text{i}}-\color{red}{\text{V}_1}}{\color{red}{\text{R}_1}}=\frac{\color{red}{\text{V}_1}}{\text{R}_2}+\frac{\color{red}{\text{V}_2}}{\text{R}_4} \end{cases}\tag4 $$
So, we have two equations and two unknowns, namely \$\text{R}_2\$ and \$\text{R}_4\$. Using your values, we see that:
$$ \begin{cases} \frac{16-\frac{29}{2}}{10000}=\frac{\frac{29}{2}}{\text{R}_2}+\frac{\frac{29}{2}-1}{1000}\\ \\ \frac{16-\frac{29}{2}}{10000}=\frac{\frac{29}{2}}{\text{R}_2}+\frac{1}{\text{R}_4} \end{cases}\tag5 $$
It is not hard to solve this (so I let you do that). I used a simple Mathematica code to solve this:
In[1]:=Vi = 16;
V1 = 29/2;
V2 = 1;
R1 = 10000;
R3 = 1000;
FullSimplify[
Solve[{I1 == I2 + I3, I1 == (Vi - V1)/R1, I2 == V1/R2,
I3 == (V1 - V2)/R3, I3 == V2/R4}, {I1, I2, I3, R2, R4}]]
Out[1]={{I1 -> 3/20000, I2 -> -(267/20000), I3 -> 27/2000,
R2 -> -(290000/267), R4 -> 2000/27}}
So, we get:
$$\text{R}_2=-\frac{290000}{267}\approx-1086.14\space\Omega\space\space\space\wedge\space\space\space\text{R}_4=\frac{2000}{27}\approx74.0741\space\Omega\tag6$$
Which is not possible because the resistance must be bigger than zero.
