What is the total resistance in the following circuit element ?
- (a)R/2
- (b)3R
- (c)3R/2
- (d)2R/3
Answer
Why
Correct — C, 3R/2. The printed circuit element shows the wire dividing into two branches that each carry a resistor of value R and then rejoin — drawn as a rectangle with one R along its top edge and one R along its bottom edge. Beyond the point where they rejoin, a third resistor R sits in the single wire leading out.
The two branches share both end points, so they are in parallel. For two equal resistances the combination is simply half of one of them:
1/Rp = 1/R + 1/R = 2/R, so Rp = R/2.
The third resistor carries the whole current after the branches have merged, so it is in series and adds directly:
R_total = R/2 + R = 3R/2.
Why the others are wrong
- (a)R/2 — This is only the parallel pair. It stops before the third resistor, which sits in the single wire beyond the junction and must be added on.
- (b)3R — This adds all three resistors as though they formed one series chain. Two of them are branches of a split, so they combine to R/2, not 2R.
- (d)2R/3 — This is the value of two resistors in parallel when one of them is 2R, or the result of inverting the sum in the wrong place. With two equal R in parallel the pair is R/2, and adding the series R gives 3R/2 — a value larger than R, not smaller.
Concept
Two resistors joined at both ends are in parallel and combine by reciprocals; for two equal resistances this reduces to half of one. Resistors carrying the same current one after another are in series and add directly. Any mixed network is solved by collapsing the parallel groups first and then summing what remains along the single path.
A quick sanity check settles the option list before any arithmetic. Adding a series resistor can only raise the total, and the parallel pair is R/2, so the answer must be larger than R/2 but far short of 3R. That leaves 3R/2 as the only sensible candidate. The most common slip is answering R/2 — solving the parallel pair correctly and then forgetting that the circuit continues.
Key facts
- Two equal resistors in parallel give half of one: R/2.
- A series resistor adds directly to whatever precedes it.
- Here R/2 + R = 3R/2.
- A parallel combination is always smaller than its smallest branch.
- Adding a series element always increases the total resistance.
Collapse the parallel pair to R/2, then add the series R — 3R/2, option (c).
Study next
Common traps
- Answering with the parallel pair alone and ignoring the series resistor.
- Adding every resistor as if the circuit were a single chain.
- Forgetting that a parallel combination must fall below its smallest branch.
NDA prints a small element with a parallel pair feeding one series resistor and asks for the total — collapse the pair, then add what follows.
Related PYQs
Consider the following part of an electric circuit: The total electrical resistance in the given part of the electric circuit is
- (a) 15/8 ohm
- (b) 15/7 ohm
- (c) 15 ohm
- (d) 17/3 ohm
Answer(b) 15/7 ohm
The same shape of network one step harder — three unequal branches in parallel feeding a series resistor, solved by exactly this collapse-then-add method.
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 stated in words, for three equal resistors — the same reciprocal law that turns the pair here into R/2.
Practice
- practice — not a real PYQ
Two resistors of R each are in parallel, and two more R are in series after them. The total resistance is
- (a)4R
- (b)5R/2
- (c)R/2
- (d)2R
Answer(b) 5R/2 — the pair gives R/2, and the two series resistors add 2R, making R/2 + 2R = 5R/2. - practice — not a real PYQ
Two equal resistors are connected in parallel. The equivalent resistance is
- (a)twice one of them
- (b)half of one of them
- (c)equal to one of them
- (d)one-fourth of one of them
Answer(b) half of one of them — 1/Rp = 2/R gives Rp = R/2.