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`