A composite resistance of 50 ohm 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.

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 the formula of parallel

1/Req = 1/200+1/200+1/200+1/200

1/Req = 1/50

Req = 50Ω

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

(B) is the correct option.

Was this answer helpful?

  
   

0 (0)

Upvote (0)

Choose An Option That Best Describes Your Problem

Thank you. Your Feedback will Help us Serve you better.

Leave a Comment

Your Mobile number and Email id will not be published. Required fields are marked *

*

*

BOOK

Free Class

Ask
Question