How can three resistors of resistances 2 Ω, 3 Ω and 6 Ω be connected to give a total resistance of (a) 4 Ω and (b) 1Ω?
(a) In order to get a resistance of 4 Ω, we can connect resistors of 3 Ω and 6 Ω in parallel and this parallel combination in series with 2 Ω resistor.
Let Rp be the equivalent resistance of parallel combination. Then,
\(\frac{V}{R_p}\) = \(\frac{V}{R_1}\) + \(\frac{V}{R_2}\)
= \(\frac{1}{3}\) + \(\frac{1}{6}\) = \(\frac{2+1}{6}\)
= \(\frac{3}{6}\) = \(\frac{1}{2}\)
or R = 2Ω
∴ Equivalent resistance of series combination,
Rs = 2 + 2 = 4Ω
(b) In order to get 1 Ω resistance, we can connect all the three resistors in parallel.
Let R' be the equivalent resistance of parallel combination. Then
\(\frac{1}{R'}\) = \(\frac{1}{2}\) + \(\frac{1}{3}\) + \(\frac{1}{6}\)
= \(\frac{3+2+1}{6}\) = \(\frac{6}{6}\) = 1
or R' = 1Ω