Question

3. Three resistances of 2Ω, 3Ω and 5Ω are connected in electric circuit. Find (a) maximum effective resistance. (b) minimum effective resistence

Answer 1

Answer:

Maximum will be in series circuit i.e 2 + 3 + 5 = 10 ohm

Minimum will be in parallel circuit i.e 1/2 + 1/3 + 1/5 = 1/R = >where R = 0.967 ohm

Hope this helps

Answer 2

We have to find the highest and lowest total resistance that can be secured by the combinations of 2 ohm, 3 ohm and 5 ohm.

Take R1 = 2ohm, R2 = 3ohm, R3 = 5ohm.

There are two ways to find the resistance. First by using series combination and second by using parallel combination.

To find lowest resistance we have to add the resistors in the parallel combination and to find the highest resistance we have to add the resistors in the series combination.

For highest resistance:

Rs = R1 + R2 + R3

Rs = 2 + 3 + 5

Rs = 10 ohm

For lowest resistance:

1/Rp = 1/R1 + 1/R2 + 1/R3

1/Rp = 1/2 + 1/3 + 1/5

1/Rp = (15 + 10 + 6)/30

1/Rp = 31/30

Rp = 30/31

Rp = 0.97 ohm

Therefore, total highest resistance is 10 ohm and total lowest resistance is 0.97 ohm.