Question

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 

 

Answer 1

(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.