Three resistances of \( 1\,\Omega \), \( 2\,\Omega \) and \( 3\,\Omega \) are given. In order to obtain an equivalent resistance of \( \displaystyle\frac{11}{3}\,\Omega \) they are to be connected as
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 3\,\Omega \) in series with parallel combination of \( 1\,\Omega \) and \( 2\,\Omega \)
Find the equivalent resistance of each arrangement: resistances in series add, and for two in parallel \( R = R_aR_b/(R_a + R_b) \). The target is \( 11/3 = 3 + \tfrac{2}{3}\,\Omega \), which suggests a \( 3\,\Omega \) in series with something of \( \tfrac{2}{3}\,\Omega \).
\( \displaystyle 1\,\Omega \parallel 2\,\Omega = \frac{1 \times 2}{1 + 2} = \frac{2}{3}\,\Omega \)
\( \displaystyle R_{\mathrm{eq}} = 3 + \frac{2}{3} = \frac{11}{3}\,\Omega \)
This is the arrangement in Option (1).