A student has three resistors of resistances 2Ω, 4Ω and 6Ω. How can he connect all these resistors to get equivalent resistance of 1.09Ω?
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Given :
A student has three resistors of resistances 2Ω, 4Ω and 6Ω.
To Find :
How should we connect them to get equivalent resistance of 1.09Ω.
Solution :
First of all we need to know following formulae.
★ Equivalent of series :
R = R₁ + R₂ + ... + Rₙ
★ Equivalent of parallel :
1/R = 1/R₁ + 1/R₂ + ... + 1/Rₙ
In order to get equivalent resistance of 1.09Ω, we have to connect them in parallel to each other as shown in attachment.
Let's find equivalent resistance;
= 1/R = 1/2+1/4+1/6
= 1/R = 6/12+3/12+2/12
= 1/R = 11/12
= R = 12/11
= R = 1.09Ω
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