Show how you would connect three resistors each of resistance 6 ft, so that the combination has a resistance of
(a) 9 Ω, and
(b) 4 Ω.
(a) In order to get a resistance of 9 ft from the given three resistors, two resistors can be connected in parallel and this parallel combination in series with the third one.
Let Rp be the equivalent resistance of parallel combination. Then
\(\frac{1}{R_p}\) = \(\frac{1}{6}\) + \(\frac{1}{6}\)
= \(\frac{2}{6}\) = \(\frac{1}{3}\)
or Rp = 3Ω
∴ Equivalent resistance of series combination,
Rs = 3 + 6 = 9Ω
(b) In order to get a resistance of 4 ft from the given three resistors, two resistors can be connected in series and this series in parallel with the third one.
Let Rs be the equivalent resistance of series combination. Then,
Rs = 6 + 6 = 12Ω
∴ Equivalent resistance of parallel combination,
\(\frac{1}{R_p}\) = \(\frac{1}{12}\) + \(\frac{1}{6}\)
= \(\frac{1 + 2}{12}\) = \(\frac{3}{12}\) = \(\frac{1}{4}\)
or Rp = 4Ω