What is (a) highest, and (b)Ω lowest, resistance which can be obtained by combining four resistors having the following resistances?
4 Ω, 8 Ω, 12 Ω, 24 Ω
Solution
(a) To get the highest resistance, all the resistors must be connected in series.
Resistance in a series arrangement is given by R = R1+ R2 + R3 + R4
R=4Ω+8Ω+12Ω+24Ω
R=48Ω
Therefore, the highest resistance is 48 Ω.
(b) To get the lowest resistance, all the resistors must be connected in parallel.
Resistance in a parallel arrangement is given by:
`1/R=1/R_1+1/R_2+1/R_3+1/R_4`
Here`R_1=4`Ω
`R_2=8` Ω
`R_3=12` Ω
`R_4=24`Ω
`1/R=1/4+1/8+1/12+1/24`
`1/R=(6+3+2+1)/24`
`1/R=12/24`
R=2Ω
Therefore, the lowest resistance of the arrangement is 2 Ω.
