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Earlier discussion
5 older comments kept for context.
Oluwaseun Adeyemi
Combined resistance of resistors in parallel:
\(\frac{1}{\mathrm{R_T}} = \frac{1}{\mathrm{R_1}} + \frac{1}{\mathrm{R_2}} + \frac{1}{\mathrm{R_3}} + ...\)
David Gomba
This isn't correct and that's not how to solve for combined resistance in parallel.
\(\frac{1}{\mathrm{R_p}} = \frac{1}{\mathrm{R_1}} + \frac{1}{\mathrm{R_2}}\)
\(= \frac{1}{1} + \frac{1}{2} = \frac{3}{2}\)
\(\frac{1}{\mathrm{R_p}} = \frac{3}{2}\)
\(\mathrm{R_p} = \frac{2}{3} = 0.6\Omega\) which is less than R1
Oluwaseun Adeyemi
Thank you for your correction. Adjustments have been made.
Oluwaseun Adeyemi
Two resistors R\(_1\), and R\(_2\) are connected in parallel
R\(_2\) = 2, R\(_1\) = 1 (R\(_2\) being greater than R\(_1\))
Combined resistance \(= 1 + \cfrac{1}{2} = \cfrac{3}{2} = 1.5\)
(greater than R\(_1\), less than R\(_2\))
Oluwaseun Adeyemi
Please, this is not the correct way to solve the question. See David Gomba's explanation.