Question

You are given several identical resistance each of value $R=10$ and each capable of carrying a maximum current of one ampere. It is required to make a suitable combination of these resistance of $5$ which can carry a current of $4$ ampere. The minimum number of resistance's of the type $R$ that will be required for this job is

• A. $4$
  
• B. $10$
  
• C. $8$
  
• D. $20$
  

Answer 1

The correct option is C $8$

Resistance of each resistor $R=10$

Equivalent resistance of each segment $R^{\prime }=R+R=2R$

$\therefore$ $R^{\prime }=2\times 10=20$

Such $4$ segments are connected in parallel to each other.

Thus equivalent resistance of the circuit $R_p=\frac{R^{\prime }}{4}$

$⟹$ $R_p=\frac{20}{4}=5$

Maximum current flowing through each resistor is one ampere i.e. $i=1A$

Thus total current flowing through the circuit $I=i+i+i+i$

$\therefore$ $I=4i=4A$

Thus minimum $8$ resistors are required to satisfy the above conditions.