First slide

combination of resistance

Question

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

Moderate

A

4
B

10
C

8
D

20

Solution

To carry a current of 4 amperes, we need four paths, each carrying a current of one ampere, Let r be the resistance of each path. These are connected in parallel. Hence their equivalent resistance will be r / 4. According to the given problem
$\frac{r}{4} = 5 or r = 20 Ωa$
For this purpose two resistances should be connected. There are four such combinations. Hence, the total number of resistances = 4x 2= 8

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App