Question

A composite resistance of 50 ohm which can carry a current of 4A is to be made from resistances each of resistance 100 ohm which can carry a current of 1 A. The minimum number of resistance to be used is :
(A) 4 (B)8 (C)12 (D)16
Please send your answers with explanation...

Answer 1

Given that,

Composite resistance = 50 ohm

Current = 4 A

Each resistance = 100 ohm

Each resistance carry a current = 1 A

Suppose,

In one pair, two resistance connected in series.

So, the value of one pair of resistance is 200 ohm because each resistance is 100 ohm.

Now, four pairs of resistance connected in parallel

We need to calculate the equivalent resistance

Using formula of parallel

$\dfrac{1}{R_{eq}}=\dfrac{1}{200}+\dfrac{1}{200}+\dfrac{1}{200}+\dfrac{1}{200}$

$\dfrac{1}{R_{eq}}=\dfrac{1}{50}$

$R_{eq}=50\ \Omega$

These pairs are in parallel.

So, The 1 A current would pass through each resistance.

Hence, The minimum number of resistance to be used is 8.

(B) is correct option.