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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 - Science

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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 

 

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Solution

(a) To get the highest resistance, all the resistors must be connected in series.
      Resistance in a series arrangement is given by R = R1R2 + R3 + R

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.

 


     
                                                                                            

Concept: System of Resistors - Resistors in Series
  Is there an error in this question or solution?

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