Chapter 3 · Current Electricity

3.4 Kirchhoff's Rules and the Wheatstone Bridge

Two rules — what flows into a junction flows out, and the potential changes round any closed loop add up to zero — solve every d.c. circuit, however tangled. The Wheatstone bridge uses them to compare resistances with a galvanometer that reads zero.

A car has two sources of electrical energy — the battery and the alternator — feeding the same lamps, radio and fan at once. A solar home system connects a panel, a battery and several appliances; a laptop charger tops up the battery while also running the computer. In circuits like these the resistors are neither simply in series nor simply in parallel, and the series–parallel rules of Class 10 are not enough. We need a method that works for any network of cells and resistors.

In 1845 Gustav Kirchhoff, then a student, stated two rules that do exactly that. They are nothing more than conservation of charge and conservation of energy, written for circuits. This section states Kirchhoff's junction rule and Kirchhoff's loop rule, turns them into a step-by-step method, and applies them to the most famous network of all: the Wheatstone bridge, which measures an unknown resistance by adjusting a known one until a galvanometer reads zero.

Kirchhoff's Rules

First some words for the parts of a network.

📘 Definition 3.9 — Junction, branch and loop
In an electric network, a junction (or node) is a point where three or more conductors meet. A branch is a path between two junctions containing no other junction; in a steady state every element of a branch carries the same current. A loop is any closed path through the network that does not pass through the same junction twice.

Charge cannot pile up at a junction in a steady state: if it did, the potential there would keep changing and the currents would not be steady. So every coulomb that arrives must leave.

📐 Law 3.8 — Kirchhoff's junction rule

At any junction of a circuit carrying steady currents, the sum of the currents entering the junction equals the sum of the currents leaving it: $$\sum I_\text{in}=\sum I_\text{out}\qquad\text{or}\qquad\sum I=0$$ (counting currents entering as positive and leaving as negative). The rule expresses conservation of charge.

Now take a charge once round a closed loop and back to its starting point. The potential at a point has a single value, so the charge ends where it began: the rises and falls of potential along the way must cancel. (This is a property of the electrostatic field, developed in Chapter 2: the work it does round a closed path is zero. Equivalently, the energy a charge gains from the cells in the loop equals the energy it gives up in the resistors.)

📐 Law 3.9 — Kirchhoff's loop rule

The algebraic sum of the changes in potential around any closed loop of a circuit is zero: $$\sum\Delta V=0,\qquad\text{i.e.}\qquad\sum\varepsilon=\sum IR\ \text{ round the loop (with signs).}$$ The rule expresses conservation of energy. Sign convention, travelling round the loop in a chosen direction:

• through a resistor in the direction of its assumed current: change $-IR$; against it: $+IR$;

• through a cell from its $-$ terminal to its $+$ terminal: $+\varepsilon$; from $+$ to $-$: $-\varepsilon$ (and its internal resistance is treated like any other resistor).

🧭 Problem-solving strategy — solving a network with Kirchhoff's rules
  1. Draw and label. Mark every junction and give each branch its own current with an arrow, $I_1, I_2,\dots$. Guess the directions — a wrong guess just gives a negative answer.
  2. Junctions. Use the junction rule to express as many currents as possible in terms of others. A network with $j$ junctions gives $j-1$ independent junction equations.
  3. Loops. Choose enough independent loops (each containing at least one branch not used before) to match the number of unknowns. Go round each in a chosen direction, applying the sign convention.
  4. Solve the simultaneous equations. A negative current flows opposite to its arrow; keep the sign while substituting.
  5. Check with a loop you did not use, and with power: the power supplied by the cells equals the power used.
✏️ Example 1 — A two-loop network
Two junctions $A$ (top) and $B$ (bottom) are joined by three branches: the left branch has a $12$ V cell and a $2.0\ \Omega$ resistor, the middle branch a $4.0\ \Omega$ resistor, the right branch a $6.0$ V cell and a $2.0\ \Omega$ resistor. Both cells have their positive terminals towards $A$ and negligible internal resistance. Find the current in each branch and the potential difference $V_A-V_B$.
Label Let $I_1$ flow up the left branch, $I_2$ up the right branch, and $I_3$ down the middle branch. Junction $A$: $I_3=I_1+I_2$.
Left loop Go up the left branch and down the middle: $+12-2.0I_1-4.0I_3=0$, i.e. $6I_1+4I_2=12$.
Right loop Up the right branch, down the middle: $+6.0-2.0I_2-4.0I_3=0$, i.e. $4I_1+6I_2=6$.
Solve Multiply the first by $3$ and the second by $2$ and subtract: $18I_1-8I_1=36-12$, so $I_1=2.4$ A. Then $4I_2=12-6(2.4)=-2.4$, $I_2=-0.60$ A, and $I_3=2.4-0.60=1.8$ A.
Interpret $I_2$ is negative: $0.60$ A flows down through the $6.0$ V cell, from its $+$ to its $-$ terminal — the $12$ V cell is charging it. $V_A-V_B=I_3(4.0)=7.2$ V.
Check Outer loop (up the left branch, down the right): $12-2.0(2.4)+2.0(-0.60)-6.0=12-4.8-1.2-6.0=0$ ✓ (the term $+I_2R_2$ appears because we go against the arrow of $I_2$). Also $V_A-V_B=6.0+(0.60)(2.0)=7.2$ V along the right branch ✓. Power: the $12$ V cell supplies $12(2.4)=28.8$ W; resistors take $2(2.4)^2+4(1.8)^2+2(0.6)^2=11.52+12.96+0.72=25.2$ W and the $6.0$ V cell stores $6(0.6)=3.6$ W; $25.2+3.6=28.8$ W ✓. Interactive 1 opens on this circuit.
✏️ Example 2 — Potential differences round a single loop
A cell of emf $9.0$ V and internal resistance $0.50\ \Omega$ and a cell of emf $3.0$ V and internal resistance $0.50\ \Omega$ are connected in a loop with a $5.0\ \Omega$ resistor, with their positive terminals facing each other (opposing). Find the current and the terminal voltage of each cell.
Loop rule Go round in the direction the $9.0$ V cell drives current: $+9.0-0.50I-3.0-0.50I-5.0I=0$ (we cross the $3.0$ V cell from $+$ to $-$). So $6.0I=6.0$ and $I=1.0$ A.
9 V cell Delivering current: $V_1=9.0-(1.0)(0.50)=8.5$ V.
3 V cell Current is forced into its positive terminal, so it is being charged: $V_2=3.0+(1.0)(0.50)=3.5$ V.
Check Going round from the $9$ V cell's terminals: $8.5=3.5+5.0(1.0)$ ✓.
Interactive 1 — Kirchhoff Solver for a Two-Loop Network

The network of Example 1. The solver applies the junction rule at $A$ and the loop rule to the two loops (dashed arrows), and solves for the three branch currents. Arrows show the actual directions; the moving dots go at speeds proportional to the currents. The readouts check both rules with the computed currents. Set a negative $\varepsilon_2$ to turn the right cell round.

126224
$I_1$ up left branch (A)—
$I_2$ up right branch (A)—
$I_3$ down middle (A)—
junction $A$: $I_1+I_2-I_3$—
loop sums $\sum\Delta V$ (V)—
$V_A-V_B$ (V)—
⚠️ Common mistakes with Kirchhoff's rules

1. Sign errors in the loop rule. The sign of $IR$ depends on whether you cross the resistor with or against the assumed current, not the actual one. The sign of $\varepsilon$ depends only on which terminal you meet first.

2. Changing a current's direction halfway. Once you have drawn an arrow for $I_2$, use it in every equation. If $I_2$ comes out negative, the arrow was wrong — the magnitude is still right.

3. Too many junction equations, too few loops. With $j$ junctions only $j-1$ junction equations are independent; the rest must come from loops.

4. Forgetting internal resistances — they are resistors in the branch of their cell.

The Wheatstone Bridge

Four resistors $P$, $Q$, $R$ and $S$ are joined in a ring $ABCD$: $P$ between $A$ and $B$, $Q$ between $B$ and $C$, $R$ between $A$ and $D$, and $S$ between $D$ and $C$. A cell is connected across $A$ and $C$, and a galvanometer (a sensitive current detector) across $B$ and $D$ — the "bridge". In general current flows through the galvanometer. But for one special combination of resistances $B$ and $D$ are at the same potential, and the galvanometer reads zero: the bridge is balanced.

At balance, $I_g=0$. Then the current $I_1$ in $P$ continues unchanged through $Q$, and the current $I_2$ in $R$ continues through $S$. Apply the loop rule to loop $ABDA$ (through $P$, the galvanometer and $R$) and loop $BCDB$ (through $Q$, $S$ and the galvanometer); the galvanometer terms vanish: $$-I_1P+0+I_2R=0\quad\Rightarrow\quad I_1P=I_2R,$$ $$-I_1Q+I_2S+0=0\quad\Rightarrow\quad I_1Q=I_2S.$$ Dividing the first equation by the second removes the currents.

📐 Result 3.10 — Balance condition of the Wheatstone bridge

A Wheatstone bridge with arms $P$ ($AB$), $Q$ ($BC$), $R$ ($AD$) and $S$ ($DC$), a cell across $AC$ and a galvanometer across $BD$, is balanced (no current through the galvanometer) if and only if $$\frac PQ=\frac RS.$$ The condition does not depend on the emf or internal resistance of the cell, nor on the resistance of the galvanometer.

If $S$ is unknown, adjust the known resistor $R$ until the galvanometer reads zero; then $S=R\,\dfrac QP$. $P$ and $Q$ are the ratio arms: with $Q/P=10$, $1$ or $0.1$, a single variable resistor covers three ranges. This is a null method: we do not need to know how much current a given deflection means, only to see that there is none. That makes it very precise, which is why bridges are still used inside strain gauges, weighing scales, gas sensors and resistance thermometers. Notice also that, at balance, the galvanometer branch carries no current and can be removed: the network then reduces to $(P+Q)$ in parallel with $(R+S)$ — a useful trick for finding equivalent resistances.

✏️ Example 3 — Measuring an unknown resistance
In a Wheatstone bridge the ratio arms are $P=100\ \Omega$ and $Q=200\ \Omega$. The galvanometer reads zero when the variable resistor is set to $R=150\ \Omega$. Find the unknown resistance $S$. What would happen to the balance if the cell's emf were doubled?
Balance $\dfrac PQ=\dfrac RS\Rightarrow S=R\dfrac QP=150\times\dfrac{200}{100}=300\ \Omega$.
Doubling the emf Every current doubles, but $V_B$ and $V_D$ are still equal: the bridge stays balanced. (A larger emf does make an unbalance easier to see, so it improves sensitivity — as long as the resistors do not heat up.)
✏️ Example 4 — An unbalanced bridge
A bridge has $P=Q=R=100\ \Omega$ and $S=150\ \Omega$. A $2.0$ V cell of negligible internal resistance is connected across $AC$ ($A$ positive), and a galvanometer of resistance $50\ \Omega$ across $BD$. Find the current through the galvanometer and its direction.
Setup Take $V_C=0$, so $V_A=2.0$ V. Let the unknown potentials be $V_B$ and $V_D$. (Kirchhoff's junction rule at $B$ and $D$ — current in equals current out — is the quickest route here.)
Junction B $\dfrac{2.0-V_B}{100}=\dfrac{V_B}{100}+\dfrac{V_B-V_D}{50}\ \Rightarrow\ 4V_B-2V_D=2.0$.
Junction D $\dfrac{2.0-V_D}{100}+\dfrac{V_B-V_D}{50}=\dfrac{V_D}{150}\ \Rightarrow\ -6V_B+11V_D=6.0$ (after multiplying by $300$).
Solve From the first, $V_B=0.5+0.5V_D$; substituting, $-3-3V_D+11V_D=6$, so $V_D=1.125$ V and $V_B=1.0625$ V.
Answer $I_g=\dfrac{V_B-V_D}{50}=\dfrac{-0.0625}{50}=-1.25\times10^{-3}$ A: a current of $1.25$ mA flows from $D$ to $B$. Why that way? $S$ is larger than it should be for balance ($S=100\ \Omega$), so a larger share of the cell's voltage falls across $S$ and $D$ sits at a higher potential than $B$. Check: set these values in Interactive 2.
✏️ Example 5 — A balanced bridge inside a network
Find the equivalent resistance between $A$ and $C$ of the network $AB=2.0\ \Omega$, $BC=4.0\ \Omega$, $AD=3.0\ \Omega$, $DC=6.0\ \Omega$, with a $5.0\ \Omega$ resistor joining $B$ and $D$.
Spot the bridge It is a Wheatstone bridge with $P=2.0$, $Q=4.0$, $R=3.0$, $S=6.0\ \Omega$ and the $5.0\ \Omega$ resistor in place of the galvanometer. $\dfrac PQ=\dfrac12=\dfrac RS$: balanced.
Remove BD At balance $V_B=V_D$, so no current flows in the $5.0\ \Omega$ resistor; it can be removed without changing anything.
Combine $(2.0+4.0)=6.0\ \Omega$ in parallel with $(3.0+6.0)=9.0\ \Omega$: $R_\text{eq}=\dfrac{6.0\times9.0}{15}=3.6\ \Omega$. The value of the middle resistor does not matter at all.
✏️ Example 6 — A strain gauge in a bridge
A strain gauge is a thin metal foil glued to a beam; when the beam stretches, the foil gets longer and thinner and its resistance rises. A $120\ \Omega$ gauge forms arm $S$ of a bridge whose other arms are fixed $120\ \Omega$ resistors, with a $5.0$ V supply across $AC$. A high-resistance voltmeter across $BD$ reads zero when the beam is unloaded. Under load the gauge's resistance becomes $120.24\ \Omega$. Find the voltmeter reading.
Concept The voltmeter draws (almost) no current, so each side of the bridge is a simple voltage divider. With $V_C=0$: $V_B=5.0\times\dfrac{Q}{P+Q}$ and $V_D=5.0\times\dfrac{S}{R+S}$.
Potentials $V_B=2.5$ V; $V_D=5.0\times\dfrac{120.24}{240.24}\approx2.50250$ V.
Reading $V_D-V_B\approx2.5\times10^{-3}$ V $=2.5$ mV, for a resistance change of only $0.2\%$. For small changes $\Delta V\approx\dfrac{V_\text{supply}}{4}\dfrac{\Delta S}{S}=\dfrac{5.0}{4}(0.002)=2.5$ mV.
Why a bridge? Measuring $120.24\ \Omega$ directly to the necessary accuracy is hard; measuring a few millivolts away from zero is easy. Digital kitchen scales, crane load cells and bridge-monitoring sensors all work this way.
Interactive 2 — Balancing a Wheatstone Bridge

A $2.0$ V cell drives the bridge; the galvanometer ($50\ \Omega$) sits across $BD$. The full network is solved with Kirchhoff's rules at every setting, and the needle shows the galvanometer current. Choose an unknown $S$ and a ratio $Q/P$, then turn the variable resistor $R$ until the needle stands at zero — or press Play and let the needle hunt for the null. It opens on Example 4.

100
galvanometer current $I_g$ (B → D)—
$V_B$, $V_D$ (V)—
$P/Q$ and $R/S$—
balance at $R=SP/Q$—
🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — the metre bridge and the potentiometer

The metre bridge is listed by JEE Main, NEET and JEE Advanced (in CBSE it is a practical only, not in the theory paper). The potentiometer is not in the 2026 CBSE, JEE Main, NEET or JEE Advanced syllabi; read that part as background.

Metre bridge. A Wheatstone bridge in which the ratio arms $P$ and $Q$ are the two parts of a uniform resistance wire $1$ m long, divided at the point where a sliding contact (the jockey) gives zero galvanometer current. A known resistance $R$ is in one gap and the unknown $S$ in the other. If the balance point is at length $l$ (in cm) from the end next to $R$, the wire's resistances are proportional to their lengths, and $$\frac{R}{S}=\frac{l}{100-l}\qquad\Longrightarrow\qquad S=R\,\frac{100-l}{l}.$$ The result is most accurate when the balance point is near the middle of the wire; interchanging $R$ and $S$ and averaging removes errors from the end connections.

Potentiometer. A long uniform wire carries a steady current from a driver cell, so the potential falls uniformly along it: the potential gradient is $k=V_\text{wire}/L$. A cell of emf $\varepsilon$ connected (through a galvanometer) between the start of the wire and a sliding contact is balanced at the length $l$ where $\varepsilon=kl$ — at balance it draws no current, so the potentiometer measures the true emf. Two cells compare as $\dfrac{\varepsilon_1}{\varepsilon_2}=\dfrac{l_1}{l_2}$. If a resistor $R$ is then connected across a cell, the balance length falls from $l_1$ to $l_2$ and the internal resistance is $$r=R\left(\frac{l_1}{l_2}-1\right).$$ Neither instrument is required for the board examination.

Exercises

I
Kirchhoff's rules. In the "two-loop network" parts, junctions $A$ (top) and $B$ (bottom) are joined by three branches: left, cell $\varepsilon_1$ with $R_1$; middle, $R_3$; right, cell $\varepsilon_2$ with $R_2$; both cells have their positive terminals towards $A$ and negligible internal resistance.
  • 1 Currents of $3$ A and $2$ A flow into a junction and $4$ A flows out along a third wire. Find the current in the fourth wire and its direction.
  • 2 Going round a loop you cross a $5.0\ \Omega$ resistor in the direction of its $2.0$ A current, and then a $6.0$ V cell from its $+$ to its $-$ terminal. Write the two potential changes.
  • 3 A $12$ V cell of internal resistance $1.0\ \Omega$ is in a loop with $2.0\ \Omega$ and $3.0\ \Omega$ resistors. Use the loop rule to find the current.
  • 4 Cells of $10$ V and $4.0$ V, each of internal resistance $1.0\ \Omega$, are in a loop with a $4.0\ \Omega$ resistor, opposing each other. Find the current and the terminal voltage of the $4.0$ V cell.
  • 5 Two-loop network with $\varepsilon_1=8.0$ V, $R_1=4.0\ \Omega$, $\varepsilon_2=4.0$ V, $R_2=4.0\ \Omega$, $R_3=2.0\ \Omega$. Find the three branch currents.
  • 6 For part 5, find $V_A-V_B$.
  • 7 Explain why the junction rule expresses conservation of charge and the loop rule conservation of energy.
  • 8 Solving a network, you find $I_2=-0.30$ A. What does the minus sign tell you?
1 In: $5$ A; out so far: $4$ A. The fourth wire carries $1$ A out of the junction.
2 Resistor, with the current: $-IR=-10$ V. Cell, $+$ to $-$: $-6.0$ V.
3 $12-1.0I-2.0I-3.0I=0\Rightarrow I=2.0$ A.
4 $10-1.0I-4.0-1.0I-4.0I=0\Rightarrow I=1.0$ A. The $4.0$ V cell is being charged: $V=4.0+(1.0)(1.0)=5.0$ V.
5 Left loop: $8=4I_1+2(I_1+I_2)\Rightarrow6I_1+2I_2=8$. Right loop: $4=4I_2+2(I_1+I_2)\Rightarrow2I_1+6I_2=4$. Solving: $I_1=1.25$ A, $I_2=0.25$ A (both upwards), $I_3=1.5$ A downwards in the middle.
6 $V_A-V_B=I_3R_3=(1.5)(2.0)=3.0$ V. (Check through the right branch: $4.0-(0.25)(4.0)=3.0$ V ✓.)
7 In a steady state charge cannot accumulate at a junction, so the charge arriving per second equals the charge leaving. Round a closed loop a charge returns to the same potential, so the energy it gains from the cells equals the energy it loses in resistors (and in cells being charged).
8 The current has magnitude $0.30$ A but flows opposite to the arrow drawn for $I_2$. Keep the negative value when substituting into other equations.
II
The Wheatstone bridge. Arms $P$ ($AB$), $Q$ ($BC$), $R$ ($AD$), $S$ ($DC$); cell across $AC$ with $A$ positive; galvanometer across $BD$.
  • 1 $P=10\ \Omega$, $Q=20\ \Omega$ and the bridge balances at $R=15\ \Omega$. Find $S$.
  • 2 With $P=Q=100\ \Omega$ the bridge balances at $R=47.5\ \Omega$. Find $S$.
  • 3 Show that a balanced bridge stays balanced if the cell and the galvanometer are interchanged.
  • 4 Give two reasons why the Wheatstone bridge measures resistance accurately.
  • 5 Find the equivalent resistance between $A$ and $C$ when $P=3.0$, $Q=6.0$, $R=4.0$, $S=8.0\ \Omega$ and a $10\ \Omega$ resistor joins $B$ and $D$.
  • 6 $P=Q=R=10\ \Omega$, $S=20\ \Omega$, galvanometer resistance $20\ \Omega$, ideal $3.0$ V cell. Find the galvanometer current and its direction.
  • 7 Five equal resistors $R$ form a bridge (four arms and the middle link). Find the resistance between $A$ and $C$.
  • 8 A balanced bridge has $P=Q$, and $S$ is a copper coil. The coil warms up. Which of $B$ and $D$ is now at the higher potential, and which way does current flow through the galvanometer?
1 $S=R\dfrac QP=15\times2=30\ \Omega$.
2 $S=R=47.5\ \Omega$.
3 With the cell across $BD$ and the galvanometer across $AC$, the balance condition is $\dfrac PR=\dfrac QS$. This is the same equation as $\dfrac PQ=\dfrac RS$ (cross-multiply both: $PS=QR$).
4 It is a null method — only the absence of current is detected, so the galvanometer need not be calibrated; and the balance condition does not depend on the emf or internal resistance of the cell (or the galvanometer's resistance).
5 $\dfrac36=\dfrac48$: balanced, so the $10\ \Omega$ link carries no current. $R_\text{eq}=\dfrac{9.0\times12}{21}\approx5.1\ \Omega$.
6 With $V_C=0$, $V_A=3.0$ V. Junction $B$: $\dfrac{3-V_B}{10}=\dfrac{V_B}{10}+\dfrac{V_B-V_D}{20}$; junction $D$: $\dfrac{3-V_D}{10}+\dfrac{V_B-V_D}{20}=\dfrac{V_D}{20}$. Solving: $V_B\approx1.579$ V, $V_D\approx1.895$ V, so $I_g=\dfrac{V_B-V_D}{20}\approx-0.0158$ A: about $16$ mA from $D$ to $B$.
7 Balanced ($R/R=R/R$), so the middle link carries no current: $2R$ in parallel with $2R$ gives $R$.
8 $V_D=\varepsilon\dfrac{S}{R+S}$ rises when $S$ rises, while $V_B$ is unchanged. So $D$ is higher and current flows through the galvanometer from $D$ to $B$.
III
Applications.
  • 1 Modelling. A bridge has three fixed $350\ \Omega$ arms and a $350\ \Omega$ strain gauge, with a $10$ V supply. Under load the gauge becomes $350.7\ \Omega$. Find the reading of a high-resistance voltmeter across $BD$.
  • 2 A platinum thermometer ($100\ \Omega$ at $0\ ^\circ\text{C}$, $\alpha=3.85\times10^{-3}\ ^\circ\text{C}^{-1}$) is arm $S$ of a bridge with $P=Q$. Balance is found at $R=119.25\ \Omega$. Find the temperature.
  • 3 Modelling. A $14$ V charger (internal resistance $0.50\ \Omega$) and a $10$ V battery (internal resistance $0.50\ \Omega$) are joined in parallel, positive to positive, across a $2.0\ \Omega$ lamp. Find the current in the charger, the battery and the lamp. Is the battery being charged?
  • 4 Three branches join junctions $A$ and $B$: $10$ V with $2.0\ \Omega$; $6.0$ V with $2.0\ \Omega$; $2.0$ V with $1.0\ \Omega$ (all positive terminals towards $A$). Find $V_A-V_B$ and the current in each branch.
  • 5 In the network of Example 1 ($12$ V, $2.0\ \Omega$; $4.0\ \Omega$; $\varepsilon_2$, $2.0\ \Omega$), what value of $\varepsilon_2$ makes the current through the right-hand cell zero?
  • 6 A bridge has $P=10$, $Q=20$, $R=30$, $S=40\ \Omega$, galvanometer resistance $10\ \Omega$ and an ideal $2.0$ V cell. Find the galvanometer current and its direction.
1 $V_B=5.0$ V; $V_D=10\times\dfrac{350.7}{700.7}\approx5.0050$ V. Reading $\approx5.0$ mV.
2 $S=R=119.25\ \Omega$, so $T=\dfrac{119.25-100}{(3.85\times10^{-3})(100)}=\dfrac{19.25}{0.385}=50.0\ ^\circ\text{C}$.
3 Let $I_1$ (charger) and $I_2$ (battery) flow out of their positive terminals, $I_3=I_1+I_2$ through the lamp. $14=0.5I_1+2I_3$ and $10=0.5I_2+2I_3$ give $I_1\approx6.67$ A, $I_2\approx-1.33$ A, $I_3\approx5.33$ A. Yes: $1.33$ A flows into the battery's positive terminal, charging it, while the lamp gets $10.7$ V.
4 Junction rule with each branch current $I_k=(\varepsilon_k-V)/R_k$ (out of the cell's $+$ terminal), where $V=V_A-V_B$: $\dfrac{10-V}{2}+\dfrac{6-V}{2}+\dfrac{2-V}{1}=0\Rightarrow V=5.0$ V. Currents: $2.5$ A and $0.5$ A up through the $10$ V and $6.0$ V branches, $3.0$ A down through the $2.0$ V branch (charging that cell).
5 If the right branch carries no current, $12$ V drives $\dfrac{12}{2.0+4.0}=2.0$ A through the left and middle branches, so $V_A-V_B=8.0$ V. With no current, the right branch's potential difference is just its emf: $\varepsilon_2=8.0$ V.
6 Junction equations at $B$ and $D$ (with $V_C=0$, $V_A=2.0$ V) give $V_B\approx1.296$ V and $V_D\approx1.239$ V, so $I_g\approx5.6$ mA from $B$ to $D$. (Without the galvanometer, $V_B=1.33$ V and $V_D=1.14$ V, so $B$ is higher — consistent.)
IV
More practice.
  • 1 Two-loop network with $\varepsilon_1=8.0$ V, $R_1=2.0\ \Omega$, $\varepsilon_2=6.0$ V, $R_2=3.0\ \Omega$, $R_3=2.0\ \Omega$. Find the branch currents.
  • 2 For part 1, find the power supplied by each cell and show that it equals the total power dissipated in the resistors.
  • 3 Find the equivalent resistance between $A$ and $C$ of the network $AB=1.0$, $BC=2.0$, $AD=2.0$, $DC=4.0\ \Omega$, with $3.0\ \Omega$ joining $B$ and $D$.
  • 4 A $6.0$ V cell of internal resistance $1.0\ \Omega$ supplies $4.0\ \Omega$ and $12\ \Omega$ resistors in parallel. Use Kirchhoff's rules to find the current in each resistor.
1 $8=2I_1+2(I_1+I_2)$ and $6=3I_2+2(I_1+I_2)$: $4I_1+2I_2=8$, $2I_1+5I_2=6$. So $I_1=1.75$ A, $I_2=0.50$ A, $I_3=2.25$ A.
2 Supplied: $8(1.75)+6(0.50)=14+3=17$ W. Dissipated: $2(1.75)^2+3(0.50)^2+2(2.25)^2=6.125+0.75+10.125=17$ W ✓.
3 $\dfrac12=\dfrac24$: balanced, the $3.0\ \Omega$ link carries nothing. $R_\text{eq}=\dfrac{3.0\times6.0}{9.0}=2.0\ \Omega$.
4 Let $I$ leave the cell and split into $I_a$ ($4.0\ \Omega$) and $I_b$ ($12\ \Omega$). Junction: $I=I_a+I_b$. Loops: $4I_a=12I_b$ and $6=1.0I+4I_a$. Then $I_a=3I_b$, $I=4I_b$, $6=4I_b+12I_b$: $I_b=0.375$ A, $I_a=1.125$ A (total $1.5$ A).
🎯 Exam corner — Kirchhoff's rules and the Wheatstone bridge BoardJEE MainJEE AdvancedNEET

E1. Board · 3 marks State Kirchhoff's junction and loop rules. Use them to find the currents in a network in which junctions $A$ and $B$ are joined by three branches: a $6.0$ V cell with a $3.0\ \Omega$ resistor; a $2.0\ \Omega$ resistor; a $4.0$ V cell with a $2.0\ \Omega$ resistor (both cells with their positive terminals towards $A$, internal resistances negligible).

Rules Junction: the sum of currents entering a junction equals the sum leaving. Loop: the algebraic sum of potential changes round any closed loop is zero.
Equations $I_1$ up the $6.0$ V branch, $I_2$ up the $4.0$ V branch, $I_3=I_1+I_2$ down the $2.0\ \Omega$. Loops: $6=3I_1+2(I_1+I_2)$, $4=2I_2+2(I_1+I_2)$, i.e. $5I_1+2I_2=6$ and $2I_1+4I_2=4$.
Solve Doubling the first and subtracting the second: $8I_1=8$, so $I_1=1.0$ A, $I_2=0.50$ A, $I_3=1.5$ A ($V_A-V_B=3.0$ V). Exam tip: show the circuit with current arrows and the loop directions — that diagram earns a mark.

E2. Board · 2 marks Using Kirchhoff's rules, obtain the balance condition of a Wheatstone bridge.

Setup Arms $P$ ($AB$), $Q$ ($BC$), $R$ ($AD$), $S$ ($DC$); galvanometer across $BD$. At balance $I_g=0$, so $P$ and $Q$ carry the same current $I_1$, and $R$ and $S$ the same current $I_2$.
Loops $ABDA$: $-I_1P+I_2R=0$. $BCDB$: $-I_1Q+I_2S=0$. Dividing: $\dfrac PQ=\dfrac RS$.
Exam tip State clearly why the same current flows in $P$ and $Q$ — the junction rule at $B$ with $I_g=0$.

E3. JEE Main In a network, $AB=4\ \Omega$, $BC=8\ \Omega$, $AD=6\ \Omega$, $DC=12\ \Omega$, and a $10\ \Omega$ resistor joins $B$ and $D$. The equivalent resistance between $A$ and $C$ is (A) $7.2\ \Omega$ (B) $10\ \Omega$ (C) $5.0\ \Omega$ (D) $8.4\ \Omega$.

Concept $\dfrac48=\dfrac6{12}$: a balanced bridge, so the $10\ \Omega$ link carries no current.
Combine $12\ \Omega$ in parallel with $18\ \Omega$: $\dfrac{12\times18}{30}=7.2\ \Omega$.
Answer (A). Exam tip: always test the ratio before starting a long Kirchhoff calculation.

E4. NEET A Wheatstone bridge has $P=2\ \Omega$, $Q=3\ \Omega$ and $R=4\ \Omega$. For balance, and after the cell and galvanometer are then interchanged: (A) $S=6\ \Omega$; still balanced (B) $S=6\ \Omega$; no longer balanced (C) $S=2.7\ \Omega$; still balanced (D) $S=8\ \Omega$; no longer balanced.

Balance $S=R\dfrac QP=4\times\dfrac32=6\ \Omega$.
Interchange The new condition $\dfrac PR=\dfrac QS$ is the same equation ($PS=QR$): still balanced.
Answer (A). Exam tip: (C) is $R\,P/Q$ — the ratio upside down.

E5. JEE Advanced (One or more correct.) Junctions $A$ and $B$ are joined by three branches: an $8.0$ V cell with $2.0\ \Omega$; a $2.0\ \Omega$ resistor; a $6.0$ V cell with $1.0\ \Omega$ (positive terminals towards $A$, ideal cells). Then (A) the current in the $2.0\ \Omega$ middle resistor is $2.5$ A (B) the $6.0$ V cell is being charged (C) $V_A-V_B=5.0$ V (D) the $8.0$ V cell supplies $12$ W.

Solve Quickest by the junction rule at $A$ with $V=V_A-V_B$: $\dfrac{8-V}{2}+\dfrac{6-V}{1}=\dfrac V2\Rightarrow 4-\dfrac V2+6-V=\dfrac V2\Rightarrow V=5.0$ V.
Currents Middle: $5.0/2.0=2.5$ A. $8.0$ V branch: $(8-5)/2=1.5$ A out of its $+$ terminal. $6.0$ V branch: $(6-5)/1=1.0$ A out of its $+$ terminal — it is discharging.
Answer (A), (C), (D) ($8.0\times1.5=12$ W); (B) is false. Exam tip: when all branches join the same two junctions, one unknown — the potential difference — is enough.

E6. JEE Main In a metre bridge, a $2.0\ \Omega$ resistor is in the left gap and an unknown resistor in the right gap. The balance point is $40$ cm from the left end. The unknown resistance is (A) $1.3\ \Omega$ (B) $3.0\ \Omega$ (C) $2.0\ \Omega$ (D) $0.80\ \Omega$.

Concept The metre bridge is beyond the NCERT syllabus (see the 🚀 box), but it is just a Wheatstone bridge: $\dfrac RS=\dfrac{l}{100-l}$.
Numbers $S=2.0\times\dfrac{60}{40}=3.0\ \Omega$.
Answer (B). Exam tip: (A) comes from pairing the $40$ cm length with the wrong gap; the length measured from the end next to $R$ goes with $R$.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. How small a stretch of a steel beam can a strain-gauge bridge detect — and how much better is a bridge with four active gauges than one with a single gauge?

Why it matters. Strain gauges in Wheatstone bridges weigh lorries at toll plazas, measure thrust in rocket-engine tests, monitor the cables of suspension bridges and the wings of aircraft, and sit inside every digital kitchen scale. Structural-health monitoring of bridges and buildings is an active research field worldwide, and its first design question is always the same: how big is the bridge's output signal compared with the noise of the electronics?

What to do.

  1. A metal-foil gauge has a "gauge factor" of about $2$: its fractional resistance change is $\Delta R/R=2\epsilon$, where the strain $\epsilon$ is the fractional change in length. Find $\Delta R/R$ for a strain of $500\times10^{-6}$ (a steel beam near its working load).
  2. Quarter bridge: one active $350\ \Omega$ gauge and three fixed $350\ \Omega$ arms, supply $5.0$ V. Using voltage dividers (as in Example 6) find the exact output, and compare with the approximation $\dfrac{V}{4}\dfrac{\Delta R}{R}$.
  3. Full bridge: four active gauges — two stretched ($+\Delta R$) on top of the beam and two compressed ($-\Delta R$) underneath, placed so that their effects add. Show that the output is exactly $V\,\dfrac{\Delta R}{R}$.
  4. If the measuring electronics can resolve $1\ \mu\text{V}$, what is the smallest strain each bridge can detect?

Tools: Calculator or spreadsheet; optionally a load cell kit and a microcontroller with an amplifier module.

Hand in: Your calculations, a comparison table of the two bridges, and a paragraph explaining why designers prefer full bridges (consider also temperature changes, which affect all four gauges equally).

Going further. Why does a temperature change of the whole beam produce no output in a full bridge but does in a quarter bridge? Relate your answer to the balance condition $P/Q=R/S$.

Instructor note: $\Delta R/R=1.0\times10^{-3}$. Quarter bridge exact output $V\dfrac{x}{2(2+x)}$ with $x=\Delta R/R$: $1.249$ mV (approximation $1.25$ mV). Full bridge: $5.0$ mV exactly — four times larger and linear. Minimum strains for $1\ \mu$V: quarter bridge $\epsilon_\text{min}=\dfrac{4(10^{-6})}{5.0\times2}=4\times10^{-7}$; full bridge $1\times10^{-7}$. A uniform temperature change multiplies all four arms by the same factor, leaving $P/Q$ and $R/S$ unchanged — self-compensation; in a quarter bridge only one arm changes.

🧪 Self-check quiz

0/5 correct

Q1. Kirchhoff's junction rule is a consequence of the conservation of

Q2. Kirchhoff's loop rule is a consequence of the conservation of

Q3. A Wheatstone bridge with arms $P$ ($AB$), $Q$ ($BC$), $R$ ($AD$), $S$ ($DC$) and the galvanometer across $BD$ is balanced when

Q4. Solving a circuit with Kirchhoff's rules gives one current as $-0.5$ A. This means

Q5. Five resistors, each $R$, form a Wheatstone bridge (four arms and the galvanometer link). The resistance between the two supply points is

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