Consider the following electric circuit : The current in the above electric circuit is :

Q: What is the answer to NDA (I) 2024 GAT Q126?

A. Correct — A, 1 A. The printed circuit runs from a 10 V battery through a 5 Ω resistor, then through a block of three resistors drawn one above the other between the same two nodes, and then through a second 5 Ω resistor back to the battery. The three parallel resistors are labelled 10 Ω, **0 Ω** and 10 Ω. That middle branch is the whole question. A 0 Ω branch is a plain wire joining the two nodes — a **short circuit**. Current takes it in preference to anything else, and the resistance of a group containing a short is zero: 1/Rp = 1/10 + 1/0 + 1/10 → the 1/0 term dominates, so Rp = **0 Ω**. The two 10 Ω resistors are still physically present but carry no current, because both their ends sit at the same potential. What remains is the two series resistors: R_total = 5 + 0 + 5 = **10 Ω**, so I = V/R = 10/10 = **1 A**. (Provisional — pending the official UPSC key.)

(a)1 A
(b)(10/15) A
(c)2 A
(d)1·5 A

Answer

Why

Correct — A, 1 A. The printed circuit runs from a 10 V battery through a 5 Ω resistor, then through a block of three resistors drawn one above the other between the same two nodes, and then through a second 5 Ω resistor back to the battery.

The three parallel resistors are labelled 10 Ω, 0 Ω and 10 Ω. That middle branch is the whole question. A 0 Ω branch is a plain wire joining the two nodes — a short circuit. Current takes it in preference to anything else, and the resistance of a group containing a short is zero:
1/Rp = 1/10 + 1/0 + 1/10 → the 1/0 term dominates, so Rp = 0 Ω.

The two 10 Ω resistors are still physically present but carry no current, because both their ends sit at the same potential.

What remains is the two series resistors: R_total = 5 + 0 + 5 = 10 Ω, so
I = V/R = 10/10 = 1 A.

Why the others are wrong

(b)
(10/15) A — This comes from taking the parallel group as 5 Ω — which is what 10 ∥ 10 would give if the 0 Ω branch were ignored — making the total 15 Ω. The printed 0 Ω branch cannot be ignored; it shorts the group out.
(c)
2 A — This needs a total of 5 Ω, as if only one of the two series resistors were present. Both 5 Ω resistors carry the full current, so both count.
(d)
1·5 A — This does not follow from any consistent reading of the network. With the short in place the total is fixed at 10 Ω and the current at exactly 1 A.

Concept

A branch of zero resistance placed across two nodes is a short circuit: it forces the two nodes to the same potential, so any element connected between them carries no current and the whole group behaves as 0 Ω. In the reciprocal formula for parallel resistance a zero-resistance branch contributes an infinite conductance, which swamps every other term.

Printing '0 Ω' among ordinary values is a deliberate test of whether a candidate computes mechanically or reads the circuit. Anyone who applies 10 ∥ 10 = 5 without noticing the middle branch reaches 15 Ω and option (b), which is exactly the trap. A quick way to see it: ask whether any current would bother flowing through a 10 Ω resistor when a bare wire connects the same two points.

Key facts

A 0 ohm branch between two nodes is a short circuit.
Any resistance in parallel with a short is bypassed and carries no current.
A parallel group containing a 0 ohm branch has zero resistance.
Here the total is 5 + 0 + 5 = 10 ohm.
I = V/R = 10/10 = 1 A.

The 0 Ω branch decides it

Parallel group: 10 Ω, 0 Ω, 10 Ω→The 0 Ω branch shorts the other two

Group resistance→0 Ω — the 10 Ω resistors carry no current

Series total→5 + 0 + 5 = 10 Ω

Current→I = 10 V / 10 Ω = 1 A

Spot the short before computing — the two 10 Ω resistors drop out entirely, option (a).

Study next

Short circuits and zero-resistance branchesSeries and parallel reductionWhy a shorted element carries no currentOhm's law applied to a reduced network

Common traps

Computing 10 ∥ 10 and ignoring the printed 0 Ω branch.
Assuming every drawn resistor must carry current.
Dropping one of the series resistors from the total.

NDA prints a network containing an unusual value such as 0 Ω — read the labels before reducing, because a short removes whole branches from the calculation.

Related PYQs

NDA_GAT_2023_II_Q1092023Open & attempt →

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

(a) [circuit diagram: three 3Ω resistors connected in parallel between X and Y]
(b) [circuit diagram: two 3Ω resistors connected in parallel between X and Y]
(c) [circuit diagram: two 1Ω resistors connected in series between X and Y]
(d) [circuit diagram: three 1Ω resistors connected in series between X and Y]

Answer(a) [circuit diagram: three 3Ω resistors connected in parallel between X and Y]

The same series-versus-parallel reasoning used to rank four arrangements rather than to reduce one — and the same lesson that the arrangement, not the printed values, decides the total.

Practice

practice — not a real PYQ
A 6 Ω resistor is connected in parallel with a piece of wire of negligible resistance. The combination has resistance
- (a)6 Ω
- (b)3 Ω
- (c)0 Ω
- (d)12 Ω
Answer(c) 0 Ω — the wire shorts the resistor, so the pair behaves as a plain connection.
practice — not a real PYQ
In a circuit, a resistor is short-circuited by a copper wire. The current through that resistor is
- (a)maximum
- (b)unchanged
- (c)zero
- (d)doubled
Answer(c) zero — both its ends are at the same potential, so no current passes through it.