Question

Find the ratio of the maximum to the minimum resistance for a combination of 'n' equal resistors.

• A. n
• B. $\frac{1}{n^2}$
• C. $n^2$
• D. $\frac{1}{n}$

Answer 1

The correct option is C $n^2$

Equivalent resistance is maximum when connected in series and minimum when connected in parallel.

Equivalent resistance when the resistors are connected in series will be:
$R_s=R+R+R+...\text{n times}=nR$
Equivalent resistance when the resistors are connected in parallel will be
$\frac{1}{R_p}=\frac{1}{R}+\frac{1}{R}+\frac{1}{R}+...\text{n times}$
$\frac{1}{R_p}=\frac{n}{R}$
$R_p=\frac{R}{n}$

Ratio of maximum to minimum resistance will be:
$\frac{R_{max}}{R_{min}}=\frac{R_s}{R_p}=\frac{nR}{\left(\frac{R}{n}\right)}=n^2$
Therefore the ratio will be $(n^2:1)$