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?
Solution
There are three resistors of resistances 2 Ω, 3 Ω, and 6 Ω respectively.
(a) The following circuit diagram shows the connection of the three resistors.
H
ere, 6 Ω and 3 Ω resistors are connected in parallel.
Therefore, their equivalent resistance will be given by
`1/(1/6+1/3)=(6xx3)/(6+3)=2 Ω`
This equivalent resistor of resistance 2 Ω is connected to a 2 Ω resistor in series.
Therefore, the equivalent resistance of the circuit = 2 Ω + 2 Ω = 4 Ω
Hence the total resistance of the circuit is 4 Ω.
(b) The following circuit diagram shows the connection of the three resistors.
All the resistors are connected in series. Therefore, their equivalent resistance will be given as
`1/(1/2+1/3+1/6)=1/((3+2+1)/6)=6/6=1Omega`