Question

Two resistances when connected in parallel give resultant value of 2 ohm; when connected in series the value becomes 9 ohm. Calculate the value of each resistance.in​

Answer 1

Answer:

Let two resistance be p and q

when connected in parallel

(1/p) + (1/q) = 1/2

(p+q)/(pq) = 1/2....(1)

when connected in series

p +q = 9...…(2)

put the value of p + q from (2) in (1)

9/(pq) = (1/2)

pq = 18.…...(3)

(p-q)^2 = (p+q)^2 - 4pq = 9^2 - 4×18 (by putting values from 2 and 3) = 81 - 72 = 9

(p - q)^2 = 9

p-q = 3....(4)

Add (2) and (4)

2p = 12

p = 6

Subtract (4) from (2)

2q = 6

q = 3

Two resistances are 6 and 3 ohm