Consider the following circuit : Which one of the following is the value of the resistance between points A and B in the circuit given above?
- (a)2/5 R
- (b)3/5 R
- (c)3/2 R
- (d)4R
Answer
Why
Correct — B, 3/5 R. The printed circuit holds four equal resistors, each of value R, arranged in two routes from A to B. On the upper route the wire leaving A splits into two branches that each carry one R and then rejoin, and a third R continues from that junction on to B. On the lower route a single fourth R runs from A straight across to B, bypassing the other three.
Take the upper route first. The two branches that split and rejoin are in parallel, so together they count as R/2. The third R sits in series after them, so the whole upper route is R/2 + R = 3R/2.
That upper route and the lower single R both join A to B, so the two routes are themselves in parallel. For two resistances in parallel the combined value is their product over their sum: (3R/2 x R) / (3R/2 + R) = (3R^2/2) / (5R/2) = 3R/5.
Why the others are wrong
- (a)2/5 R — This is what the product-over-sum rule gives if the upper route is taken as R rather than 3R/2 and paired with a lower 2R/3 — a mis-grouping. The upper route is R/2 followed by a series R, which is 3R/2, and 3R/2 in parallel with R is 3R/5, not 2R/5.
- (c)3/2 R — This is the upper route on its own — the parallel pair R/2 plus the series R. It forgets that the fourth resistor offers a second, independent route from A to B. Adding a parallel path can only lower the resistance, so the answer must come out below 3R/2.
- (d)4R — This treats all four resistors as one long series chain. They are not: two of them split and rejoin, and a fourth spans A to B on its own, so nothing here is a simple sum.
Concept
Resistors in series add directly, R_total = R1 + R2, because the same current passes through each in turn. Resistors in parallel combine as reciprocals, 1/R_total = 1/R1 + 1/R2, because the current divides between them. A network is solved by collapsing it inwards — reduce each simple group to a single equivalent resistance, then treat that result as one resistor in the next stage.
The step that decides this item is spotting that the circuit has two separate routes from A to B, not one. Candidates who follow only the visible upper chain reach 3R/2 and stop, which is exactly option (c). A useful check: adding any parallel path always lowers the total resistance, so once the upper route is 3R/2 the final answer has to be smaller than that. Only options (a) and (b) are, and the arithmetic picks (b).
Key facts
- Series resistors add: R + R = 2R.
- Two equal resistors in parallel give half of one of them: R/2.
- For exactly two resistances in parallel, R_total = product / sum.
- Adding a parallel branch always reduces the total resistance below the smaller branch.
- Here: upper route = R/2 + R = 3R/2; that in parallel with R gives 3R/5.
Collapse the parallel pair, add the series resistor, then put that route in parallel with the lone R — 3R/5, option (b).
Study next
Common traps
- Solving only the obvious chain and missing a second route between the same two points.
- Adding all resistors as if the whole network were in series.
- Applying product-over-sum to more than two resistors at once.
- Forgetting that a parallel branch must lower the total.
NDA prints a small network of equal resistors and asks for the resistance between two marked points — collapse parallel groups first, then add what is in series, and check that each added parallel path lowers the total.
Related PYQs
Three equal resistors are connected in parallel configuration in a closed electrical circuit. Then the total resistance in the circuit becomes
- (a) one-third of the individual resistance.
- (b) two-third of the individual resistance.
- (c) equal to the individual resistance.
- (d) three times of the individual resistance.
Answer(a) one-third of the individual resistance.
The parallel rule on its own — three equal resistors in parallel give one-third of a single resistance, the same reciprocal law used here to collapse the split pair to R/2.
An electric wire of resistance 50 ohm is cut into five equal wires. These wires are then connected in parallel. What is the equivalent resistance of this combination?
- (a) 2 ohm
- (b) 10 ohm
- (c) 0·5 ohm
- (d) 5 ohm
Answer(a) 2 ohm
Cutting a 50 ohm wire into five equal pieces and paralleling them — the same parallel arithmetic, reached through the fact that resistance falls with length.
Practice
- practice — not a real PYQ
Two resistors of R each are connected in parallel, and a third R is joined in series with that pair. What is the total resistance?
- (a)3R
- (b)3R/2
- (c)2R/3
- (d)R/2
Answer(b) 3R/2 — the parallel pair is R/2, and the series R adds directly to give 3R/2. - practice — not a real PYQ
A resistance of 3R/2 is connected in parallel with a resistance of R. What is the equivalent resistance?
- (a)5R/2
- (b)2R/5
- (c)3R/5
- (d)R/3
Answer(c) 3R/5 — product over sum: (3R/2 x R) / (5R/2) = 3R/5.