Question

Three resistors of 6Ω,3Ω and 2Ω are connected together so that the total resistance is greater than 6Ω but less than 8Ω. Draw a diagram to show this arrangement and calculate its total resistance in each case.

Answer 1

Given : Three resistors of 6Ω,3Ω and 2Ω are connected together so that the total resistance is greater than 6Ω but less than 8Ω

To find : Arrangement of resistors & total resistance

Solution:

Three resistors of 6Ω,3Ω and 2Ω

if all connected in series the total resistance = 11 Ω

adding all in parallel will result in less than 2Ω

6Ω can not be in parallel as resistance required more than 6Ω  

only way to get  resistance between 6Ω  &  8Ω

is that add 6Ω  in series with parallel combination of 3Ω and 2Ω

as 3Ω and 2Ω  

               _____3 Ω   ______

_______|                                | ____6Ω_______

              |   _____2 Ω   ____|

Net resistance = 6  +   1/(1/2 + 1/3)

= 6 +  6/(3 + 2)

= 6 + 6/5

= 6 + 1.2

= 7.2 Ω

7.2 Ω   is  total resistance

all other possible arrangement  

6 + 3 + 2 = 11 Ω

1/(1/6 + 1/3 + 1/2) = 1 Ω

3 + 1/(1/6 + 1/2) = 4.5 Ω

2 + 1/(1/6 + 1/3) = 4 Ω

1/(1/(2 + 3) + 1/6)  = 30/11 Ω

1/(1/(2 + 6) + 1/3)  = 24/11 Ω

1/(1/(6 + 3) + 1/2)  = 18/11 Ω

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Answer 2

Answer:

We connect the resistance of 3Ω and 2Ω in parallel to each other and then connect this (parallel combination) to the 6Ω resistance in series. The required arrangement is as shown in figure. The equivalent resistance of 3Ω and 2Ω in parallel.

1/R = 1/3 + 1/2 = 2 + 3/6 = 5/6

Or R = 6/5Ω = 1.2Ω.

Thus, the total resistance of the set up is (1.2 + 6)Ω = 7.2Ω

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