Three resistors \(1 Ω, 2 Ω\), and \(3 Ω\) are combined in series. What is the total resistance of the combination?
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Three resistors \(1 Ω, 2 Ω\), and \(3 Ω\) are combined in series. What is the total resistance of the combination?
Answer
\(6\;\Omega\)
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Solution
Here we are given three Resistors \(({R}_{1},{R}_{2}\) and \({R}_{3})\) as \(1, 2\) and \(3\; \text{ohms}\)respectively.
\({\text{R}}_{1}=1\;\mathit{\varOmega }\)
\({\text{R}}_{2}=2\varOmega\)
\({\text{R}}_{3}=3\;\varOmega\)
When the resistors are in series, the resistances are added.
\(R={R}_{1}+{R}_{2}+{R}_{3}\)
\(R=1+2+3\)
\(R=6\;\mathit{\varOmega }\)
Hence in series, the resistance will be \(6\mathit{\varOmega }\).
\({\text{R}}_{1}=1\;\mathit{\varOmega }\)
\({\text{R}}_{2}=2\varOmega\)
\({\text{R}}_{3}=3\;\varOmega\)
When the resistors are in series, the resistances are added.
\(R={R}_{1}+{R}_{2}+{R}_{3}\)
\(R=1+2+3\)
\(R=6\;\mathit{\varOmega }\)
Hence in series, the resistance will be \(6\mathit{\varOmega }\).
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