Question

A network of resistances is constructed with $R_1$ and $R_2$ as shown in the figure. The potential at the points 1, 2, 3 … n are $V_1,V_2,V_3,\dots V_n$ respectively, each having a potential k times smaller than the previous one.

The ratio $\frac{R_2}{R_3}$ is

• A. $\frac{{(k-1)}^2}{k}$
• B. $k^2-\frac{1}{k}$
• C. $\frac{k}{k-1}$
• D. $k-\frac{1}{k^2}$

Answer 1

The correct option is C $\frac{k}{k-1}$

Since each point have potential $k$ times smaller than previous one, thus $V_1=\frac{V_o}{k}$, $V_2=\frac{V_1}{k}$, $V_3=\frac{V_2}{k}$ and so on. Let current $I$ flow through $R_1$ and gets distributed as shown.

Current ,$\begin{array}{}I=I_1+I_2& \text{(1)}\end{array}$
Since current, $I=\frac{\text{Potential difference}}{\text{resistance}}$
Therefore equation (1) can be written as,
$\begin{array}{}\frac{V_o-V_1}{R_1}=\frac{V_1-V_2}{R_1}+\frac{V_o-0}{R_2}& \text{(2)}\end{array}$
Substituting $V_1=\frac{V_o}{k}$, $V_2=\frac{V_1}{k}=\frac{V_o}{k^2}$ in (2),
$\frac{V_2{k}^2-V_{2}k}{R_1}=\frac{V_{2}k-V_2}{R_1}+\frac{V_{2}k}{R_2}$
$\frac{k^2-k}{R_1}=\frac{k-1}{R_1}+\frac{k}{R_2}$
On rearranging the above relation, we get
$\frac{R_1}{R_2}=\frac{(k-1{)}^2}{k}$
$\begin{array}{}\Rightarrow R_1=R_2\frac{(k-1{)}^2}{k}& \text{(3)}\end{array}$
Consider the last branch of the circuit and let current flow through $R_1$ as shown.
Now current through resistor $R_1$ of the last branch in the circuit and $R_3$ is same.
Therefore equating currents in form of potential difference and resistance,
$\frac{V_{n-1}-V_n}{R_1}=\frac{V_n-0}{R_3}$
$\frac{R_1}{R_3}=\frac{V_{n-1}}{V_n}-1$
Substituting the value of $R_1$ from (3),
$\begin{array}{}\frac{R_2(k-1{)}^2}{k{R}_3}=\frac{V_{n-1}}{V_n}-1& \text{(4)}\end{array}$
Since it is given that potential at a point is $k$ times smaller than previous one, therefore $V_n=\frac{V_{n-1}}{k}$
Therefore (4) becomes,
$\frac{R_2(k-1{)}^2}{k{R}_3}=k-1$
$\Rightarrow \frac{R_2}{R_3}=\frac{k}{k-1}$