Question

Four resistors of resistance $15\Omega ,12\Omega ,4\Omega ,$ and $10\Omega$ respectively in cyclic order to form a Wheatstone’s network. The resistance that is to be connected in parallel with the resistance of $10\Omega$ to balance the network is

Answer 1

Step 1. Given data,

Four resistance, $15\Omega ,12\Omega ,4\Omega ,$ and$10\Omega$,

From the diagram,

Let, Resistance connected on $AB$ be, $R_{AB}=15\Omega$

Resistance connected on $BC$ be, $R_{BC}=12\Omega$

Resistance connected on $CD$ be, $R_{CD}=4\Omega$

Resistance connected on $DA$ be, $R_{DA}=R_{eq}$

Step 2. Finding the resistance, $R$ that is to be connected with $10\Omega$ resistance,

To calculate the Resistances connected on $DA$ we have to use the formula of equivalent resistance connected in parallel,

By using the formula of equivalent resistance connected in parallel, $R_{eq}$

$\frac{1}{R_{eq}}=\frac{1}{R}+\frac{1}{10}$

$\Rightarrow$$\frac{1}{R_{eq}}=\frac{10+R}{10R}$

$\Rightarrow$ $R_{eq}=\frac{10R}{10+R}$

By balancing the Wheatstone bridge,

$\frac{R_{AB}}{R_{DA}}=\frac{R_{BC}}{R_{CD}}$

$\Rightarrow$ $\frac{15}{R_{eq}}=\frac{12}{4}$

$\Rightarrow$ $\frac{15}{{\displaystyle \frac{10R}{10+R}}}=\frac{12}{4}$

$\Rightarrow$ $\frac{15\left(10+R\right)}{{\displaystyle 10R}}=\frac{12}{4}$

$\Rightarrow$ $120R=600+60R$

$\Rightarrow$ $120R-60R=600$

$\Rightarrow$ $60R=600$

$\Rightarrow$ $R=10\Omega$

Therefore, the resistance of $10\Omega$ is to be connected to balance the network.