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We have asked to find equivalent resistance between 1 and 2
★ Equivalent resistance of series connection is given by
R = R₁ + R₂ + ... + Rₙ
★ Equivalent resistance of parallel connection is given by
1/R = 1/R₁ + 1/R₂ + ... + 1/Rₙ
Now see the left part of circuit. Three resistors of resistances 1Ω, 1Ω and 2Ω are connected in series.
➝ R₁ = 1 + 1 + 2
➝ R₁ = 4Ω
After solving this, 4Ω and 4Ω come into parallel.
➝ R₂ = (4 × 4) / (4 + 4)
➝ R₂ = 16/8
➝ R₂ = 2Ω
Finally three resistors of resistances 1Ω, 2Ω and 2Ω come into series.
➝ Req = 1 + 2 + 2
➝ Req = 5Ω
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Explanation:
find the least number that must be added to 1400, to make it a perfect square
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