Consider an infinte ladder network shown in figure. A voltage $V$ is applied between the points $A$ and $B$ . This applied value of voltage is halved after each section. Then:

\frac{R_{1}}{R_{2}} = 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{R_{1}}{R_{2}} = \frac{1}{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{R_{1}}{R_{2}} = 2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{R_{1}}{R_{2}} = 3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{R_{1}}{R_{2}} = \frac{1}{2}$
Let the potential at $A$ be $V$ , Potential at next two junctions can be considered as $\frac{V}{2}, \frac{V}{4}$

Using Kirchhoff's Junction Law
$i = i_{1} + i_{2}$
$\frac{\frac{V}{2} - V}{R_{1}} = \frac{\frac{V}{4} - \frac{V}{2}}{R_{1}} + \frac{0 - \frac{V}{2}}{R_{2}}$
$- \frac{V}{2 R_{1}} = - \frac{V}{2 R_{2}} - \frac{V}{4 R_{1}}$
$\frac{V}{2 R_{1}} = \frac{V}{2 R_{2}} + \frac{V}{4 R_{1}}$
$\frac{1}{R_{1}} = \frac{1}{R_{2}} + \frac{1}{2 R_{1}}$
$\Rightarrow \frac{R_{2}}{R_{1}} = 2$

Suggest Corrections

Similar questions

\frac{R_{1}}{R_{2}}

R_{1} / R_{2}

A

B

R_{1} / R_{2}

r_{1}

r_{2}

R_{1}

R_{2}

R_{1}

R_{2}

Join BYJU'S Learning Program