Question

A composite resistance of $50ohm$ which can carry a current of $4A$ is to be made from resistances each of resistance $100ohm$ which can carry a current of $1A$. the minimum number of resistances to be used is:

• A. $4$
• B. $8$
• C. $12$
• D. $16$

Answer 1

The correct option is B

$8$

Step 1: Given data

Composite resistance = $50\mathrm{\Omega }$
Current through $50\mathrm{\Omega }=4\mathrm{A}$
Value of each resistance $=100\mathrm{\Omega }$
Current through each $100\mathrm{\Omega }$ resistor$=1A$

Step 2: Formula used

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\dots +\frac{1}{R_N}$ Where, $R_{eq}$ is the equivalent resistance and $R_1,R_2,R_3,.....R_N$are the values of $1^{st},2^{nd},3^{rd},...n^{th}$resistance respectively.

Step 3: Find the minimum number of resistances to be used

If we consider $4$ pairs of resistances connected in parallel. The value of each pair will be $200\mathrm{\Omega }$.

Then the equivalent resistance will be,
${\displaystyle \frac{1}{R_{eq}}}={\displaystyle \frac{1}{R_1}}+{\displaystyle \frac{1}{R_2}}+{\displaystyle \frac{1}{R_3}}+{\displaystyle \frac{1}{R_4}}$
Substituting the values in above equation we get,
${\displaystyle \frac{1}{R_{eq}}}={\displaystyle \frac{1}{200}}+{\displaystyle \frac{1}{200}}++{\displaystyle \frac{1}{200}}++{\displaystyle \frac{1}{200}}$
$\Rightarrow {\displaystyle \frac{1}{R_{eq}}}={\displaystyle \frac{4}{200}}$
$\Rightarrow R_{eq}={\displaystyle \frac{200}{4}}$
$\Rightarrow R_{eq}=50\mathrm{\Omega }$

Now, as the resistances are connected in parallel. The total current entering the system will be equal to the current leaving the system. Thus, the total current flowing through these $8$ resistors will be $4A$.
Therefore, the minimum number of resistances to be used is $8$.

Hence, option (B) is the correct answer.