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Question

A composite resistance of 50ohm 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

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B

8

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C

12

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D

16

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Solution

The correct option is B

8


Step 1: Given data

Composite resistance = 50Ω
Current through 50Ω=4A
Value of each resistance =100Ω
Current through each 100Ω resistor=1A

Step 2: Formula used

1Req=1R1+1R2+1R3++1RN Where, Req is the equivalent resistance and R1,R2,R3,.....RNare the values of 1st,2nd,3rd,...nthresistance respectively.

Step 3: Find the minimum number of resistances to be used

If we consider 4 pairs of resistances connected in parallel. The value of each pair will be 200Ω.

Then the equivalent resistance will be,
1Req=1R1+1R2+1R3+1R4
Substituting the values in above equation we get,
1Req=1200+1200++1200++1200
1Req=4200
Req=2004
Req=50Ω

Now, as the resistances are connected in parallel. The total current entering the system will be equal to the current leaving the system. Thus, the total current flowing through these 8 resistors will be 4A.
Therefore, the minimum number of resistances to be used is 8.

Hence, option (B) is the correct answer.


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