ARCHETYPE: Multi-Loop & Two-Battery Circuits

Jump-starting a car is this article's second board example performed with 200-ampere clamps: a healthy 12.6 V battery in parallel with a sagging one, simultaneously running the starter motor and driving current backward through its weak partner — the same minus sign you computed, now smelling faintly of sulphuric acid. Every power bank charging a phone while the phone runs is the canonical three-branch circuit in miniature: source branch, storage branch, load branch, one junction statement deciding who feeds whom. At industrial scale the drill was simply automated. SPICE and its descendants assemble exactly these junction-and-loop equations for billions of elements and solve them millions of times per design; a laptop's power-management firmware juggling wall power, battery charging and load is checked against precisely the two-battery bookkeeping you now do by hand.

Here is the circuit this article masters. Two junctions, $a$ on top and $b$ below, three branches between them: on the left a 12 V battery with $1\ \Omega$ behind it; in the middle a plain $2\ \Omega$; on the right a 6 V battery with $2\ \Omega$. Both batteries point their positive terminals toward $a$, each trying to push current up its own branch.

Test your old tools on it honestly. Is the $1\ \Omega$ in series with the middle $2\ \Omega$? No — junction $a$ sits between them, bleeding current into the right branch. Are the two $2\ \Omega$ resistors in parallel? No — one has a battery riding in its branch. The reduction rules of Resistors in Series & Parallel need bare resistors; a battery inside a branch poisons every shortcut. This is branch-current territory, and three habits define the drill.

Label smart. Give every branch its own current symbol and arrow. Guess directions freely — the algebra is indifferent — but count before you write: 3 branches need 3 equations, and junctions supply only one independent statement (the second junction repeats the first, since what flows into $a$ must flow out of $b$). The rest come from loops: here, two. Smarter still, spend the junction rule at birth — call the currents $I_1$, $I_2$ and write the middle one as $I_1 + I_2$ immediately, working with two unknowns instead of three.

Walk loops with one contract. Choose a direction around each loop and never renegotiate mid-walk. Crossing a resistor with its current arrow costs $-IR$; against it, $+IR$. Crossing a battery from its negative plate to its positive plate gains $+\varepsilon$ regardless of which way the current flows; the reverse crossing costs $-\varepsilon$. (Why these signs work is Kirchhoff's Rules territory; here we spend them, fast.)

Trust the minus sign. If a current solves negative, your guess was backwards; the magnitude is already correct. Do not redraw arrows — that way lies a fifteen-minute spiral. Negative currents through a battery are physics: that battery is being charged, absorbing $\varepsilon I$ into chemistry, exactly as EMF, Terminal Voltage & Internal Resistance describes from the single-battery side.

That node-voltage cross-check deserves promotion to a habit — it is how you answer the JEE favourite "find the potential difference between $a$ and $b$." Potential bookkeeping along a branch: start at $b$ and add potential changes as you walk to $a$. Up the right branch: $V_a = V_b + 6 - I_2 \cdot 2$. With $I_2 = -0.75\ \mathrm{A}$ (0.75 A flowing downward, against the walk):

where: the $+1.5\ \mathrm{V}$ is the resistor's contribution — walking against the true current, potential rises across a resistor. Read it physically: the branch's terminals sit at 7.5 V though its EMF is only 6 V, because a charging battery's terminal voltage exceeds its EMF by $Ir$ — the charging twin of the $V = \varepsilon - Ir$ discharge law. Walking the left branch instead gives $12 - 4.5 \times 1 = 7.5\ \mathrm{V}$: same answer, as single-valued potential demands. Two branches, one number — a third independent check, after the junction audit and the power ledger.

Board level — one loop, two opposing batteries (CBSE/ISC pattern, 3 marks)

A battery of EMF $9\ \mathrm{V}$, internal resistance $0.5\ \Omega$, sits in a single loop with a battery of EMF $3\ \mathrm{V}$, internal resistance $0.5\ \Omega$, and an external $R = 5\ \Omega$, EMFs opposing. Find the current, each terminal voltage, and each battery's role.

Step 1. One loop, no junctions: walk once in the direction the 9 V pushes. $+9 - 3 - I(0.5 + 0.5 + 5) = 0$.

Step 2. Solve: $6 = 6I$, so $I = \boxed{1.0\ \mathrm{A}}$, flowing the way the stronger EMF pushes.

Step 3. Terminal voltages, by the twin laws. Discharging: $V_1 = 9 - 0.5 = \boxed{8.5\ \mathrm{V}}$. The 3 V battery has current driven into its positive terminal — charging — so $V_2 = 3 + 0.5 = \boxed{3.5\ \mathrm{V}}$.

Step 4. Ledger: the 9 V source delivers $9\ \mathrm{W}$; resistors burn $I^2 \times 6 = 6\ \mathrm{W}$; the 3 V battery banks $3\ \mathrm{W}$ as chemistry. $9 = 6 + 3$ ✓. This loop is every battery charger you own.

Board level — two batteries in parallel charging one another (ISC/A-Level pattern)

Two cells, $\varepsilon_1 = 8\ \mathrm{V}$ ($r_1 = 1\ \Omega$) and $\varepsilon_2 = 4\ \mathrm{V}$ ($r_2 = 1\ \Omega$), are joined in parallel (positive to positive) across a $2\ \Omega$ load. Find all three currents and each cell's role.

Step 1. Label the shared top node $a$, bottom node $b$; set $V_b = 0$, call $V_a = V$. Branch currents toward $a$: from cell 1, $(8 - V)/1$; from cell 2, $(4 - V)/1$; down the load, $V/2$.

Step 2. Junction statement at $a$: $\dfrac{8 - V}{1} + \dfrac{4 - V}{1} = \dfrac{V}{2}$. Multiply by 2: $16 - 2V + 8 - 2V = V$, so $24 = 5V$ and $V = \boxed{4.8\ \mathrm{V}}$.

Step 3. Currents: $I_1 = 8 - 4.8 = \boxed{3.2\ \mathrm{A}}$ out of cell 1; $I_2 = 4 - 4.8 = \boxed{-0.8\ \mathrm{A}}$ — cell 2 receives 0.8 A, because the node sits above its EMF; load $I_3 = 4.8/2 = \boxed{2.4\ \mathrm{A}}$. Junction audit: $3.2 - 0.8 = 2.4$ ✓.

Step 4. Ledger: cell 1 delivers $25.6\ \mathrm{W}$; heat is $10.24 + 0.64 + 11.52 = 22.4\ \mathrm{W}$; cell 2 stores $3.2\ \mathrm{W}$. $25.6 = 22.4 + 3.2$ ✓. Moral: parallel cells of unequal EMF quietly waste power charging the weaker one — why Combining Cells: Series & Parallel wants matched EMFs in parallel banks.

Competitive level — one current, ninety seconds (JEE-Main pattern, ~2.5 minutes)

In the canonical circuit (12 V behind $1\ \Omega$; 6 V behind $2\ \Omega$; middle branch $2\ \Omega$), find the current in the middle branch.

Step 1. Superposition route. Kill the 6 V (wire in its place, $2\ \Omega$ stays): the 12 V sees $1 + (2 \| 2) = 2\ \Omega$, drives 6 A, and the node voltage $6\ \mathrm{V}$ splits it equally: $3\ \mathrm{A}$ down the middle.

Step 2. Kill the 12 V ($1\ \Omega$ stays): the 6 V sees $2 + (2 \| 1) = 8/3\ \Omega$, drawing $2.25\ \mathrm{A}$; node voltage $1.5\ \mathrm{V}$ sends $0.75\ \mathrm{A}$ down the middle.

Step 3. Both contributions run the same way: $I_3 = 3 + 0.75 = \boxed{3.75\ \mathrm{A}}$. Timing note: budget 2.5 minutes. Superposition wins because both sub-circuits are plain reductions — no simultaneous equations. Had the question asked for all three currents or any power, the branch-current drill is faster overall; and never superpose powers — currents first, square after.

Competitive level — the dead branch and the overfilled battery (JEE-Advanced stretch, ~4 minutes)

In the canonical topology the right-branch EMF is unknown: 12 V behind $1\ \Omega$ left, $\varepsilon$ behind $2\ \Omega$ right, $2\ \Omega$ middle. (a) For what $\varepsilon$ does the right branch carry no current? (b) For $\varepsilon = 6\ \mathrm{V}$, find the right battery's terminal voltage and state.

Step 1. Part (a): a dead branch draws nothing and drops nothing across its own resistance, so its terminal pair sits at a potential difference of exactly $\varepsilon$. Meanwhile the left-plus-middle loop runs alone: $I = \dfrac{12}{1 + 2} = 4\ \mathrm{A}$, putting the node at $V_a = 4 \times 2 = 8\ \mathrm{V}$ above $b$.

Step 2. Dead-branch condition: $\varepsilon = V_a - V_b = \boxed{8\ \mathrm{V}}$. (Both ways: $\varepsilon > 8$ pushes current up the right branch; $\varepsilon < 8$ lets the node push current down into it.)

Step 3. Part (b): $\varepsilon = 6 < 8$, so current flows down the right branch — the battery charges. From the master solution that current is $0.75\ \mathrm{A}$ and $V_a = 7.5\ \mathrm{V}$.

Step 4. Terminal voltage of the charging battery: $V = \varepsilon + I r = 6 + 0.75 \times 2 = \boxed{7.5\ \mathrm{V}}$ — matching $V_a - V_b$, as it must. A charging battery's terminals read above its EMF; on discharge, below. Examiners flip this sign on one option every year. Timing note: ~4 minutes. Part (a) is pure node thinking — no loop equations once you see that a dead branch pins its terminal difference to its EMF. That idea ("$I = 0$ through a source ⇒ terminals show the bare EMF") also powers the potentiometer's null method, coming later in this chapter.

Renegotiating the walk. Choosing clockwise, then flipping direction halfway because a battery "felt backwards" — the classic self-inflicted wound. Decide the tour, write every sign from the crossing, let wrong-feeling terms stand.

EMF signs from current direction. The sign of $\varepsilon$ depends only on which plate you cross first: minus-to-plus is $+\varepsilon$, even if the current inside runs plus-to-minus (a charging battery). Tying it to the current is the most common single error in board scripts.

Using both junction equations. With two junctions, the second statement repeats the first; adding it produces a degenerate system. Junctions give $(\text{number of junctions}) - 1$ independent statements — here, one.

Killing a source but not keeping its $r$. In superposition an EMF dies to a plain wire but its internal resistance stays. Deleting the whole branch changes the topology and every answer downstream.

Superposing power. Heat goes as $I^2$; currents add, their squares do not. Total current first, then square. A JEE-Main option equal to "$P' + P''$" is bait.

Terminal voltage of a charging battery. $V = \varepsilon + Ir$ on charge, $\varepsilon - Ir$ on discharge. If your two-branch walks of $V_a - V_b$ disagree, this sign is almost always the culprit.

Fraction discipline. These circuits are engineered for clean fractions ($15/4\ \mathrm{A}$, $24/5\ \mathrm{V}$). Keep them until the last line; a rounded $0.7$ for $0.75$ makes the power ledger miss by watts and costs the audit mark.

Texts. NCERT Physics Class 12, Ch. 3 · HC Verma, Concepts of Physics Vol. 2, Ch. 32 · Halliday, Resnick & Walker, Ch. 27 (multiloop circuits).

Provenance. All results recomputed from scratch. A verifier should confirm: the canonical solution $4.5$, $-0.75$, $3.75\ \mathrm{A}$ ($V_a = 7.5\ \mathrm{V}$; ledger $54 = 49.5 + 4.5\ \mathrm{W}$); the superposition split ($6\ \mathrm{A}$/$2.25\ \mathrm{A}$ sub-circuits, sums matching); board example 1 ($1.0\ \mathrm{A}$; $8.5$/$3.5\ \mathrm{V}$; $9 = 6 + 3\ \mathrm{W}$); board example 2 ($4.8\ \mathrm{V}$; $3.2$, $-0.8$, $2.4\ \mathrm{A}$; $25.6 = 22.4 + 3.2\ \mathrm{W}$); the dead-branch condition $\varepsilon = 8\ \mathrm{V}$; the charging terminal voltage $7.5\ \mathrm{V}$.

Prerequisites. Kirchhoff's Rules · EMF, Terminal Voltage & Internal Resistance. Leads to. ARCHETYPE: Battery Problems — Terminal V, Max Power & Efficiency · Combining Cells: Series & Parallel · Redraw the Circuit (technique, coming up) · Kirchhoff's Rules (UG deep dive) — the university treatment, including where the loop rule can legitimately fail.