Which of the following arrangement of resistors offers minimum effective resistance between points X and Y ?

Correct — A. Work out all four and compare:

(a) three 3 Ω in parallel: 1/R = 1/3 + 1/3 + 1/3 = 1, so R = 1 Ω.
(b) two 3 Ω in parallel: R = 3/2 = 1·5 Ω.
(c) two 1 Ω in series: R = 1 + 1 = 2 Ω.
(d) three 1 Ω in series: R = 1 + 1 + 1 = 3 Ω.

The smallest is 1 Ω, arrangement (a).

The result is worth reading as a principle rather than four sums. Putting resistors in parallel gives the current extra routes and always lowers the resistance below the smallest branch; putting them in series forces the current through every one in turn and always raises it. So the minimum must come from the parallel arrangement with the most branches — and here that is (a), even though its individual resistors are the larger 3 Ω ones.

Why the others are wrong

• (b)
  Two resistors of 3 Ω each, connected in parallel between X and Y — Parallel, so on the right track, but with only two branches it comes to 1·5 Ω. Adding the third 3 Ω branch in (a) brings it down to 1 Ω.
• (c)
  Two resistors of 1 Ω each, connected in series between X and Y — Series adds: 1 + 1 = 2 Ω. The small individual values are misleading — a series connection can never fall below its largest resistor.
• (d)
  Three resistors of 1 Ω each, connected in series between X and Y — The largest of the four at 3 Ω. Each extra resistor in series adds to the total, so this is the maximum here, not the minimum.

Series resistors add, so the total always exceeds the largest member. Parallel resistors combine by reciprocals, so the total always falls below the smallest member, and every additional branch lowers it further. For n equal resistors R in parallel the result is simply R/n.

Key facts

• n equal resistors R in parallel give R/n.
• Series always exceeds the largest resistor; parallel always falls below the smallest.
• (a) 3 Ω x3 parallel = 1 Ω; (b) 3 Ω x2 parallel = 1·5 Ω.
• (c) 1 Ω x2 series = 2 Ω; (d) 1 Ω x3 series = 3 Ω.
• Each extra parallel branch lowers the total; each extra series element raises it.

Common traps

• Picking the option with the smallest individual resistors regardless of arrangement.
• Forgetting that every extra parallel branch lowers the total further.
• Treating a parallel result as if it could exceed one branch.

• practice — not a real PYQ

  Four resistors of 8 Ω each are connected in parallel. The effective resistance is

  • (a)32 Ω
  • (b)8 Ω
  • (c)4 Ω
  • (d)2 Ω
  Answer(d) 2 Ω — for n equal resistors in parallel, R/n = 8/4 = 2 Ω.
• practice — not a real PYQ

  Which arrangement of three equal resistors R gives the greatest resistance?

  • (a)All three in parallel
  • (b)All three in series
  • (c)Two in parallel with the third in series
  • (d)Two in series with the third in parallel
  Answer(b) All three in series — series adds, giving 3R, the largest possible.