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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
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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 Î + 2 Î + 3 Î = 6 Î
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 Î and 2 Î are in series. Therefore, their net resistance is:
`R=R_1+R_2`
`R=1I+2I`
`R=3I`
This 3 Î is connected in parallel with another 3 Î 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`
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