Question

the resistance of a series combination of three resistance is 8 ohms. find the maximum possible value of equivalent resistance if they are joined in parallel​

Answer 1

Answer:

the answer is the maximum possible value will be 8/9 ohms

Step-by-step explanation:

plzzz...try 2 mark me as brainliest

Answer 2

Answer:

Step-by-step explanation:

en that S = nP …………(1)

S = R(1) + R(2) and P = [R(1)*R(2)]/[R(1) + R(2)]

substitute S and P into equation(1)

R(1) + R(2) = n*[R(1)*R(2)]/[R(1) + R(2)]

[R(1) + R(2)]*[R(1) + R(2)] = n*[R(1)*R(2)]

expanding the LHS:

R(1)^2 + R(2)^2+ 2R(1)R(2) = n*[R(1)*R(2)]

devide through by [R(1)*R(2)]. Gives

[R(1)^2 + R(2)^2]/[R(1)*R(2)] + 2 = n

Since the fraction [R(1)^2 + R(2)^2]/[R(1)*R(2)] > 0 Always, it implies that the minimum value of n is 2.

Hence n > or = 2. implies that the minimum value of n is 2

hope it helps

:)