Two resistors with resistance of 2 Ω and 8 Ω are connected in parallel. What is the equivalent resistance (Req) of the combination?
1.6 Ω
For resistors connected in parallel, the reciprocal of the equivalent resistance equals the sum of the reciprocals of the individual resistances: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\).
Substituting \(R_1 = 2\,\Omega\) and \(R_2 = 8\,\Omega\): \(\frac{1}{R_{eq}} = \frac{1}{2} + \frac{1}{8} = \frac{4}{8} + \frac{1}{8} = \frac{5}{8}\).
So \(R_{eq} = \frac{8}{5} = 1.6\,\Omega\). As expected for a parallel combination, this is smaller than the smallest individual resistance (2 Ω).
Check: for two resistors in parallel, \(R_{eq} = \frac{R_1 R_2}{R_1 + R_2} = \frac{2 \times 8}{2 + 8} = \frac{16}{10} = 1.6\,\Omega\), which matches the reciprocal method.
Hence, 1.6 Ω.
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