Question

When two resistors of resistances `R_1 and R_2` are connected in parallel, the net resistance is `3 Omega`. When connected in series, its value is `16 Omega`. Calculate the values of `R_1 and R_2`.

Answer 1

When `R_1` and `R_2` are connected in parallel, net resistance `(R_p)` is given by
`(1)/(R_p) = (1)/(R_1) + (1)/(R_2) = (R_1 + R_2)/(R_1 R_2)`
or `R_p = (R_1 R_2)/(R_1 + R_2) = 3` …(i)
When `R_1` and `R_2` are connected in series, net resistance `(R_s)` given by
`R_s = R_1 + R_2 = 16` ...(ii)
From eqns. (i) and (ii), `(R_1 R_2)/(16) = 3` or `R_1 R_2 = 48`
or `R_1(16 - R_1) = 48` ("as" `R_1 + R_2 = 16, R_2 = 16 - R_1`)
or `16 R_1 - R_1^2 = 48`
or `R_1^2 - 16 R_1 + 48 = 0`
or `R_1^2 - 4 R_1 - 12 R_1 + 48 = 0`
or `R_1(R_1 - 4) - 12 (R_1 - 4) = 0` or `(R_1 - 12)(R_1 - 4) = 0`
From eqn. (i), `R_2 = 4 Omega` or `12 Omega`
Therefore, the ressistance of two resistors are `4 Omega` and `12 Omega`.