Current Electricity · Tier 1 · Topic 1 of 12
The load-bearing skill of the whole chapter. Everything after this — Kirchhoff, bridges, bulbs, cells — is this topic wearing a different hat.
Imagine a school corridor full of children walking from one end to the other.
Current is how many children walk past you every second. Voltage is how hard they are being pushed from behind. Resistance is how hard the corridor makes it for them — a long narrow corridor with lots of furniture in the way is high resistance; a short wide empty one is low resistance.
Now the whole of this topic is just two questions asked over and over: what happens if you put two corridors one after the other? and what happens if you open a second corridor beside the first?
If a child must walk through corridor A and then corridor B, they meet both lots of furniture. The total difficulty is the sum. And every single child who leaves A must enter B — nobody vanishes in between — so the current is the same everywhere in a series chain.
The push, though, gets used up in stages. Some of the push is spent getting through A, the rest getting through B. So voltages add up and split themselves between the resistors: the bigger resistor eats the bigger share.
Now open a second door. The children can choose. More doors means it is easier to get through, not harder — so the combined resistance is smaller than either corridor alone. This is the one result that feels wrong the first time and must be memorised as a sanity check: a parallel answer bigger than the smallest resistor is always an arithmetic mistake.
Both corridors start at the same place and end at the same place, so the voltage across each is identical. The children split up, with more of them choosing the easier corridor. Currents add up.
Animation 1 · Series vs parallel, watching the charge
Orange dots are the charge carriers. Watch what stays the same and what splits.
A plain wire in a circuit has no resistance. So two points joined only by a wire are not two places at all — they are one place. Physicists call such a place a node.
This is why NEET circuits look terrifying: they draw the same simple circuit in a twisted shape and hope you won't recognise it. Before you write a single formula, redraw the circuit with every node collapsed to a single dot. Nine times out of ten a monster turns into two resistors in parallel.
Animation 2 · Collapsing nodes — the redraw that solves the circuit
Step through. The circuit never changes electrically; only the drawing gets honest.
When a network looks the same from two branches — a cube, a hexagon, a square — the current entering must split equally between branches that are mirror images of each other. Points that sit at the same potential can then be joined together with a wire (they were going to have zero current between them anyway) or the connection between them can be cut. Either move is free, and both turn a mess into series-and-parallel.
Twelve equal resistors R form a cube; the battery is across a body diagonal. Current 3I enters corner A and splits into three equal parts I (the three edges from A are indistinguishable). Each of those I splits into two halves I/2. At the far end the halves recombine into I, then into 3I. Walking one path A → B → C → C′:
Same argument gives 3R/4 across a face diagonal and 7R/12 across a single edge. These three numbers are worth memorising outright — NEET has never needed a fourth.
Four resistors in a diamond with a fifth resistor bridging the middle. If the two side-paths divide the voltage in exactly the same ratio, the two middle points sit at the same potential, so no current crosses the bridge. The bridge resistor is then doing nothing and may be deleted (or shorted — both give the same answer).
Animation 3 · Hunting the balance point
Drag S. Watch the bridge current die exactly when P/Q = R/S.
Take a wire of total resistance R and bend it into a ring. Touch two points on the ring. A child at one point can walk left or right — two routes, so the two arcs are in parallel. Each arc's resistance is just its share of the ring: an arc covering a fraction f of the circle has resistance fR.
The maximum you can ever get is when the two arcs are equal (diametrically opposite): R/2 in parallel with R/2, giving R/4. Touch two points right next to each other and you get almost zero. So the answer always sits between 0 and R/4 — a free sanity check.
Animation 4 · The ring of wire
Move the second contact around the ring and watch the two arcs trade resistance.
Animation 5 · Why the max-to-min ratio is n²
n identical resistors. All in series is the biggest; all in parallel is the smallest.
Series. NCERT §3.4 argues it with geometry: place two identical slabs of length l end to end. The combination has length 2l, the same current flows through both, and the potential differences add, so RC = 2V/I = 2R. Since R = ρl/A and length has doubled, resistance doubles. Generalising, resistances in a chain add.
Parallel. The same section splits one slab lengthwise into two of area A/2 each. Each half carries I/2 at the full voltage V, so each has resistance 2R — halving the area doubles the resistance. Run it backwards: putting two R's side by side doubles the available area and halves the resistance.
Notice both derivations are just Ohm's law plus one conservation statement — charge is conserved at a junction (parallel) and energy is conserved around a loop (series). That is Kirchhoff's two rules in disguise, which is why Topic 6 will feel familiar.
Two results worth having on instant recall, because they save a full circuit solve:
The swap in the parallel formula is deliberate and is the single most common slip: the current through R₁ carries R₂ on top of the fraction, because the easier branch gets more traffic.
A wire of resistance R cut into n equal pieces gives n pieces of R/n each. Put those n pieces in parallel and you divide again by n:
And for n identical resistors, series gives nR while parallel gives R/n, so the largest obtainable resistance is n² times the smallest — a result NEET asks directly.