You are given three resistors of \( 1\,\Omega, 2\,\Omega \) and \( 3\,\Omega \). Arrange the following combinations in the increasing order of equivalent resistance.
(A) \( 2\,\Omega \) and \( 3\,\Omega \) are connected in parallel to each other and \( 1\,\Omega \) is connected in series with them.
(B) \( 1\,\Omega \) and \( 2\,\Omega \) are connected in series and \( 3\,\Omega \) is connected in parallel with them.
(C) \( 1\,\Omega \) and \( 3\,\Omega \) are connected in series and \( 2\,\Omega \) is connected in parallel with them.
(D) \( 1\,\Omega \) and \( 3\,\Omega \) are connected in parallel to each other and \( 2\,\Omega \) is connected in series with them.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (C), (B), (A), (D)
Resistors in series add, \( R = R_1 + R_2 \); two resistors in parallel give \( R = R_1R_2/(R_1+R_2) \). Work out each combination:
(A) \( 2\,\Omega \) and \( 3\,\Omega \) in parallel, \( 1\,\Omega \) in series — \( \displaystyle R_A = 1\,\Omega + \frac{2\times3}{2+3}\,\Omega = 1\,\Omega + 1.2\,\Omega = 2.2\,\Omega \)
(B) \( 1\,\Omega \) and \( 2\,\Omega \) in series, \( 3\,\Omega \) in parallel — \( \displaystyle R_B = \frac{(1+2)\times3}{(1+2)+3}\,\Omega = 1.5\,\Omega \)
(C) \( 1\,\Omega \) and \( 3\,\Omega \) in series, \( 2\,\Omega \) in parallel — \( \displaystyle R_C = \frac{(1+3)\times2}{(1+3)+2}\,\Omega = \frac{8}{6}\,\Omega \approx 1.33\,\Omega \)
(D) \( 1\,\Omega \) and \( 3\,\Omega \) in parallel, \( 2\,\Omega \) in series — \( \displaystyle R_D = 2\,\Omega + \frac{1\times3}{1+3}\,\Omega = 2\,\Omega + 0.75\,\Omega = 2.75\,\Omega \)
Increasing order: \( 1.33\,\Omega \lt 1.5\,\Omega \lt 2.2\,\Omega \lt 2.75\,\Omega \), that is (C), (B), (A), (D).