Question

a) how will you join three resisitor of resistance 4 ohm , 6 ohm , and 12 ohm to get an equivalent resistance of 8 ohm.
b) what would be the highest and lowest equivalent resistances possible by joining these resistors.


NEED OF URGENT ANSWER ....​

Answer 1

a) in the image

b) connect all in parallel to get least = 2 ohm

connect all in series to get most = 22 ohm

Ask any other doubt freely buddy

Answer 2

Given,

Three resistors with resistance 4 Ω, 6 Ω, and 12 Ω.

To find,

1. How to join three resistors to get an equivalent resistance of 8 ohms.
2. The highest and lowest equivalent resistances possible by joining these resistors.

Solution,

R1 = 4Ω, R2 = 6Ω , R3 = 12Ω

If we connect the resistor R3 = 12 Ω and R2 = 6 Ω in parallel we would get,

⇒   1/R(parallel)  = 1/R2 + 1/R3.

⇒   1/R(parallel)  = 1/6 + 1/12

⇒   1/R(parallel) = (1(2)+1(1)) /12

⇒   1/R(parallel) = 3/12

⇒   1/R(parallel) = 1/4

⇒   R(parallel)   = 4 Ω.

Now, the Resistor R1 = 4Ω is connected in series with the equivalent resistance of 12 Ω and 6 Ω in parallel.

⇒   R(equivalent) = 4 Ω + 4 Ω

⇒   R(equivalent) = 8 Ω

Hence to get the equivalent resistance of 8 Ω we can connect 12 Ω and 6 Ω in parallel and then connect 4 Ω in series.

To get the highest equivalent resistance we should always connect the given resistors in series.

⇒   R(series) = 4 Ω + 6 Ω + 12 Ω

⇒   R(series) = 22 Ω.

To get the lowest equivalent resistance we should always connect the given resisters in parallel.

⇒   1/R(parallel)  = 1/R1 + 1/R2 + 1/R3

⇒   1/R(parallel)  = 1/4 + 1/6 + 1/12

⇒   1/R(parallel) = (3 + 2 + 1)/12

⇒   1/R(parallel) = 6/12

⇒   1/R(parallel) = 1/2

R(parallel)   = 2 Ω.

Hence R(max) = 22 Ω and R(min) = 2 Ω.