Question

When ${}^{\prime }{n}^{\prime }$ wires which are identical are connected in series, the effective resistance exceeds that when they are in parallel by $\frac{X}{Y}\Omega$. Then the resistance of each wire is:-

• A. $\frac{xn}{y(n^2-1)}$
• B. $\frac{yn}{x(n^2-1)}$
• C. $\frac{xn}{y(n-1)}$
• D. $\frac{yn}{x(n-1)}$

Answer 1

The correct option is A $\frac{xn}{y(n^2-1)}$

Let the resistance of each wire is $R$. $n$ identical wires of resistance $R$ is connected in series combination,

So, $R_{eq}=R+R+R+R+R+.........ntimes$

$R_{eq}=nR$

Similarly, $n$ same identical wires of resistance $R$ is connected in parallel combination,

So, $\frac{1}{R_{eq}^{\prime }}=\frac{1}{R}+\frac{1}{R}+\frac{1}{R}+\frac{1}{R}+\frac{1}{R}+ntimes$

$R_{eq}^{\prime }=\frac{R}{n}$

Given,

Equivalent resistance in series $-$ Equivalent resistance in parallel $=\frac{x}{y}\Omega$

So, $nR-\frac{R}{n}=\frac{x}{y}$

$implies\frac{R(n^2-1)}{n}=\frac{x}{y}$

$R=\frac{xn}{y(n^2-1)}\Omega$