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Solution for You Are Given Three Resistances of 1, 2 and 3 Ohms. Shows by Diagrams, How with the Help of These Resistances You Can Get: (I) 6 ω (Ii) `6/11` ω (Iii) 1.5 ω - CBSE Class 10 - Science

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Question

You are given three resistances of 1, 2 and 3 ohms. Shows by diagrams, how with the help of these resistances you can get:

(i) 6 Ω
(ii) `6/1` Ω
(iii) 1.5 Ω

Solution

To get a value of 6 Ω, all the resistors are to be connected in series as shown below:

  

Net resistance, R = R1 + R2 + R3
                        R = 1 Ώ + 2 Ώ + 3 Ώ = 6 Ώ  

To get a value of 6 / 11 Ω, all the resistors are to be connected in parallel as shown below: 

 

When tree resistance is conencted in parallel their resultant is given by 

`1/R=1/R_1+1/R_2+1/R_3` 

`R1=3 Ω

`R_2=2`Ω

`R_3=1` Ω

`1/R=1/3+1/2+1/1` 

`1/R=(2+3+6)/6`

`1/R=11/6`

`R=6/11`Ω          

  To get a value of 1.5 Ω, the 1 Ω and 2 Ω resistors should be connected in series and this arrangement should be connected in parallel with the 3 Ω resistor as shown below:       

     

The resistors of resistance 1 Ώ and 2 Ώ are in series. Therefore, their net resistance is: 

`R=R_1+R_2` 

`R=1I+2I` 

`R=3I`

This 3 Ώ is connected in parallel with another 3 Ώ resistance.
Therefore, net resistance will be: 

`1/R=1/R_1+1/R_1` 

`1/R=1/3+1/3` 

`1/R=2/3` 

`R=3/2=1.5`Ω  

`1R=1R_1+1R_2  1R=13+131R=23R=32=1.5Ω`

  Is there an error in this question or solution?

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Solution You Are Given Three Resistances of 1, 2 and 3 Ohms. Shows by Diagrams, How with the Help of These Resistances You Can Get: (I) 6 ω (Ii) `6/11` ω (Iii) 1.5 ω Concept: Resistance of a System of Resistors - Resistors in Parallel.
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