Question

A man has five resistors each of value $\frac{1}{5}\Omega$. What is the minimum resistance that one can obtain by connecting all the five resistances?

• A. $\frac{1}{10}\Omega$
• B. $\frac{1}{5}\Omega$
• C. $\frac{1}{50}\Omega$
• D. $\frac{1}{25}\Omega$

Answer 1

The correct option is D

$\frac{1}{25}\Omega$

Step 1: Given data,

R=$\frac{1}{5}\Omega$

Number of resistors (n) = $5$

Step 2: Finding equivalent resistance of parallel combination,

We can arrange these resistors in two ways, either in series or in parallel connection

The parallel connection gives equivalent resistance of

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+....+\frac{1}{R_n}$

If all resistors are equal i.e.$R_1=R_2=R_3=R_n=R$

⇒$\frac{1}{R_{eq}}=\frac{n}{R}$

⇒$R_{eq}=\frac{R}{n}$

Putting the known value,

⇒$R_{eq}=\frac{1}{5\times 5}\Omega$

⇒$R_{eq}=\frac{1}{25}\Omega$

Step 3: Finding equivalent resistance of series combination,

$R_{eq}=R_1+R_2+....+R_n$

If all resistors are equal i.e.$R_1=R_2=R_3=R_n=R$

⇒$R_{eq}=nR$

⇒$R_{eq}=5\times \frac{1}{5}\Omega$

⇒$R_{eq}=1\Omega$

Since the resistance in parallel is lesser than the series resistance.

The minimum equivalent resistance is achieved by connecting all the resistors in a parallel combination.

So, the minimum resistance he can obtain by connecting them is $\frac{1}{25}\Omega$.

Hence, option (D) is correct.