Question

you are given several identical resistors each having resistance of 10 ohm and a maximum current of 1a is allowed to pass through each resistor. it is required to make a suitable combination of these resistances of 5 ohm which can carry a current of 4a. find the number of such resistances that will be required.

Answer 1

To convey a current of 4 amperes, we require four ways, each conveying a current of one ampere. Give r a chance to be the obstruction of every way. These are associated in parallel way. Subsequently, their equal resistance will be r/4. As per the given issue  

r/4=5  

r=20  

For this propose two protections ought to be associated. There are four such blends. Henceforth, the aggregate number of resistance = 4 x 2 =8

Another method of this is given in the following image.

Answer 2

8 resiastors will be required

as current through resistor is 1A so minimum 4 parallel circuit required to make it 4 A

1/r + 1/r + 1/r + 1/r = 1/5

so 4/r = 1/5

r = 20

20 = 10+10

so two resistors in each parallel circuit and total 4 circuits

2*4= 8 resistors required