Three resistors \(2\Omega ,4\Omega \) and \(5\Omega \) are combined in parallel. What is the total resistance of the combination?
Answer
Total resistance,\(R=\frac{20}{19}\;\Omega\)
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Solution
Given: \({{R}}_{1}=2\;\Omega\)
\(\mathit{R}_2=4\;\Omega\)
\(\mathit{R}_3=5\;\Omega\)
When the resistors are in parallel,
\(\frac{1}{R}=\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}+\frac{1}{{R}_{3}}\)
\(\frac{1}{R}=\frac12+\frac 1 4+\frac 1 5\)
\(\frac{1}{R}=\frac{10+5+4}{20}\)
\(\frac{1}{R}=\frac{19}{20}\)
Total resistance,\(R=\frac{20}{19}\;\Omega\)
\(\mathit{R}_2=4\;\Omega\)
\(\mathit{R}_3=5\;\Omega\)
When the resistors are in parallel,
\(\frac{1}{R}=\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}+\frac{1}{{R}_{3}}\)
\(\frac{1}{R}=\frac12+\frac 1 4+\frac 1 5\)
\(\frac{1}{R}=\frac{10+5+4}{20}\)
\(\frac{1}{R}=\frac{19}{20}\)
Total resistance,\(R=\frac{20}{19}\;\Omega\)
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