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
Why
Correct — B, 15/7 ohm. The printed circuit shows current entering at the left and immediately splitting three ways: a 2 ohm branch along the top, a 4 ohm branch through the middle and an 8 ohm branch along the bottom. The three branches rejoin at a single point, and from there a 1 ohm resistor carries the current on to the output. So it is a three-way parallel group followed by one resistor in series.
Combine the parallel group with the reciprocal rule:
1/Rp = 1/2 + 1/4 + 1/8. Putting these over eight gives 4/8 + 2/8 + 1/8 = 7/8, so Rp = 8/7 ohm.
Note the sense-check: 8/7 ohm is about 1·14 ohm, comfortably below 2 ohm, the smallest branch — a parallel combination always comes out under its smallest member.
Now add the series resistor, which simply adds:
R_total = 8/7 + 1 = 8/7 + 7/7 = 15/7 ohm.
Why the others are wrong
- (a)15/8 ohm — This comes from writing 1/Rp = 7/8 and then treating 8/7 as 7/8 — inverting the reciprocal sum the wrong way, then adding 1. The reciprocal sum is 7/8, so Rp is its inverse, 8/7.
- (c)15 ohm — This is what you get by adding all four resistors in series: 2 + 4 + 8 + 1 = 15. But the first three are branches of a parallel split, not a chain, so they cannot simply be added.
- (d)17/3 ohm — This does not follow from either rule applied to these values. The parallel group is fixed at 8/7 ohm and the series step adds exactly 1, so the only consistent total is 15/7.
Concept
Resistors sharing both end points are in parallel and combine by reciprocals, 1/Rp = 1/R1 + 1/R2 + 1/R3; resistors carrying the same current one after another are in series and simply add. A mixed network is solved from the inside out — reduce each parallel cluster to a single value first, then treat that value as an ordinary resistor in the series chain that remains.
The arithmetic is easiest if you choose a common denominator that all three branches divide into. Here 2, 4 and 8 all divide 8, so the sum 4/8 + 2/8 + 1/8 = 7/8 falls out with no fractions to wrestle. The single most common error is stopping at 7/8 and offering it, or its near neighbour 15/8, as the answer — 7/8 is the sum of the reciprocals, so it still has to be inverted. Checking that the parallel result sits below the smallest branch catches that slip immediately.
Key facts
- Parallel: 1/Rp = 1/R1 + 1/R2 + 1/R3; series: R = R1 + R2.
- Here 1/Rp = 1/2 + 1/4 + 1/8 = 7/8, so Rp = 8/7 ohm.
- Adding the series 1 ohm gives 8/7 + 1 = 15/7 ohm.
- A parallel combination is always smaller than its smallest branch.
- The reciprocal sum must be inverted — 7/8 is not the resistance.
Collapse the parallel trio to 8/7 ohm, then add the series 1 ohm — 15/7 ohm, option (b).
Study next
Common traps
- Offering the reciprocal sum (7/8) without inverting it.
- Adding every resistor as though the whole circuit were in series.
- Applying the product-over-sum shortcut to three resistors at once — it only holds for two.
NDA prints a small parallel cluster feeding a series resistor and asks for the total — invert the reciprocal sum first, then add what follows in series.
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 stated in words — three equal resistors in parallel give one-third of a single one, the same reciprocal law applied here to unequal branches.
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
A 50 ohm wire cut into five pieces and re-paralleled — the same reciprocal arithmetic, reached through how resistance scales with length.
Practice
- practice — not a real PYQ
Resistors of 3 ohm and 6 ohm are joined in parallel, and a 2 ohm resistor is joined in series with the pair. The total resistance is
- (a)11 ohm
- (b)4 ohm
- (c)2 ohm
- (d)9 ohm
Answer(b) 4 ohm — the parallel pair is (3x6)/(3+6) = 2 ohm, and the series 2 ohm brings the total to 4 ohm. - practice — not a real PYQ
Three resistors of 1 ohm, 2 ohm and 2 ohm are connected in parallel. The equivalent resistance is
- (a)5 ohm
- (b)2 ohm
- (c)0·5 ohm
- (d)1 ohm
Answer(c) 0·5 ohm — 1/R = 1 + 1/2 + 1/2 = 2, so R = 0·5 ohm, below the smallest branch.