How can three resistors of resistances 2 Ω , 3 Ω , and 6 Ω be connected to give a total resistance of (a) 4 Ω , (b) 1 Ω?
R1 = 2 ohm R2 = 3 ohm R3 = 6 ohm (a) When R2 and R3 are connected in parallel with R1 in series we get 1/R = 1/R2 + 1/R3 g= 1/3 + 1/6 = 1/2 Thus, R = 2 ohm Resistance in series = R + R1 = 2 + 2 = 4 ohm (b) When R1,R2, R3 are connected in parallel we get 1/R = 1/R1 + 1/R2 + 1/R3 = 1/2 + 1/3 + 1/6 = 1 ohm.
