12 min readPathway sheetSection A objective parts (what each rule conserves); Section B long answer (5 marks), often with a choice: Wheatstone bridge balance condition and a metre bridge numerical; also a NEET favourite

Kirchhoff's rules, the Wheatstone bridge and the metre bridge

Junction rule: currents in = currents out. Loop rule: ΣIR = ΣE. A balanced Wheatstone bridge gives P/Q = R/S; the metre bridge uses it.

Junction rule (first rule)The sum of the currents into a junction equals the sum out; it follows from conservation of charge

Loop rule (second rule)Round any closed loop, ΣIR = ΣE; it follows from conservation of energy

Wheatstone bridge balanceP/Q = R/S when the galvanometer shows no current, so S = QR/P

Metre bridgeS = R(100 − l)/l, where l is the balance length in cm from the end next to R; ρ = πd²S/(4L)

Kirchhoff's two rules

Ohm's law alone cannot solve a network with several cells and branches. Kirchhoff gave two rules that solve any such network. First rule (junction rule): the sum of the currents flowing into a junction equals the sum of the currents flowing out. Take currents in as positive and currents out as negative: then the algebraic sum of the currents at a junction is zero. Charge cannot pile up at a junction, so this rule is conservation of charge. Second rule (loop rule): round any closed loop, the algebraic sum of the IR products equals the algebraic sum of the EMFs: ΣIR = ΣE. Put another way, the potential differences round a closed loop add up to zero. This rule is conservation of energy.

Sign rules for the loop rule

You go through	Going this way	Write
A resistor R carrying current I	Along the current	+IR on the IR side
A resistor R carrying current I	Against the current	−IR on the IR side
A cell of EMF E	From its − terminal to its + terminal	+E on the EMF side
A cell of EMF E	From its + terminal to its − terminal	−E on the EMF side

Worked example: a network with two cells

Problem

Two junctions, P at the top and Q at the bottom, are joined by three branches. Left branch: a 10 V cell in series with 2.0 Ω. Middle branch: a 2.0 Ω resistor. Right branch: a 4.0 V cell in series with 1.0 Ω. Both cells have their + terminal towards P, and their internal resistances are included in the resistors. Find the current in each branch.

Working

Mark I₁ up the left branch and I₂ up the right branch. By the junction rule at P, I₁ + I₂ flows down the middle branch.
Left loop (up the left branch, down the middle): +10 − 2.0I₁ − 2.0(I₁ + I₂) = 0, so 2I₁ + I₂ = 5 ... (i)
Right loop (up the right branch, down the middle): +4.0 − 1.0I₂ − 2.0(I₁ + I₂) = 0, so 2I₁ + 3I₂ = 4 ... (ii)
(ii) − (i): 2I₂ = −1, so I₂ = −0.50 A.
From (i): 2I₁ = 5 + 0.50 = 5.5, so I₁ = 2.75 A.
Middle branch: I₁ + I₂ = 2.75 − 0.50 = 2.25 A, flowing down.
The minus sign means 0.50 A flows down through the 4.0 V cell, from its + to its − terminal. The 10 V cell is charging it.
Check: PD across the middle resistor = 2.25 × 2.0 = 4.5 V. Left branch: 10 − 2.75 × 2.0 = 4.5 V. They match.

Answer

2.75 A up the 10 V branch; 2.25 A down the 2.0 Ω middle branch; 0.50 A down through the 4.0 V cell, opposite to the arrow first drawn

Derivation: balance condition of a Wheatstone bridge

Four resistors form a loop ABCD: P in AB, Q in BC, R in AD and S in DC. A cell is joined across A and C, and a galvanometer G across B and D. S is the unknown resistance.

1
Adjust the resistances until G shows no deflection. The bridge is then balanced, and B and D are at the same potential.
2
With no current in G, the junction rule says the current I₁ in P flows on through Q, and the current I₃ in R flows on through S.
3
B and D are at the same potential, so the drop across P equals the drop across R: I₁P = I₃R ... (i)
4
In the same way, the drop across Q equals the drop across S: I₁Q = I₃S ... (ii)
5★
Divide (i) by (ii): P/Q = R/S. So the unknown resistance is S = QR/P.
6
The balance does not depend on the EMF of the cell, and G only has to show zero current, so it need not be calibrated. The bridge is most sensitive when all four resistances are nearly equal.

★ key step

How to draw the Wheatstone bridge

Draw a diamond with A on the left, B at the top, C on the right and D at the bottom.
Put P on AB, Q on BC, R on AD and S on DC. Write each label beside its resistor.
Draw the galvanometer G, a circle with G inside, on a vertical wire from B to D, with a key K₂.
Draw the cell E and a key K₁ on a wire from A to C, outside the diamond.
Mark the currents with arrows: I₁ from A to B, I₃ from A to D, and the galvanometer current Ig from B to D.

Activity: specific resistance of a wire with a metre bridge

A **metre bridge** is a practical Wheatstone bridge. A uniform wire 1 m long, stretched along a metre scale, forms two arms whose resistances are in the ratio of their lengths, l and (100 − l).

1
Put a resistance box R in the left gap and the given wire S in the right gap. Join a cell and a key across the ends of the bridge wire.
2
Join a galvanometer, with a 1000 Ω safety resistor in series, from the middle terminal to a jockey. Draw this circuit with every part labelled.
3
Touch the jockey at the left end, then at the right end. If the galvanometer deflects opposite ways, the circuit is right.
4
Slide the jockey gently along the wire to find the point of no deflection. Then short out the safety resistor and find the point again. Record the balance length l.
5
Calculate S = R(100 − l)/l. Repeat with other values of R, keeping l between 30 cm and 70 cm, and take the mean.
6
Measure the length L of the wire S, and its diameter d with a screw gauge at several places. Calculate ρ = πd²S/(4L).
7★
Sample reading: with R = 4 Ω the balance is at l = 40 cm, so S = 4 × 60 ÷ 40 = 6 Ω. Precautions: close the key only while taking a reading, and never drag the jockey along the wire.

★ key step

In real life

Electricians and laboratory staff measure resistance with bridge circuits, because a null reading does not depend on the battery or on how well the meter is calibrated. Many electronic weighing scales and temperature sensors work as Wheatstone bridges.

Exam angle

1 marks: “On which conservation law is Kirchhoff's loop rule based?” Conservation of energy.

2 marks: “In a metre bridge, a 6 Ω resistor in the left gap gives a balance point at 40 cm. Find the resistance in the right gap.” S = R(100 − l)/l = 6 × 60 ÷ 40 = 9 Ω.

5 marks: “Draw a Wheatstone bridge and derive its balance condition. Why is it called a null method?” Diamond ABCD with P, Q, R and S; G across BD and the cell across AC. At balance Ig = 0, so I₁ flows through P and Q and I₃ through R and S. B and D are at one potential: I₁P = I₃R and I₁Q = I₃S, so P/Q = R/S. It is a null method because the reading is taken when G shows zero current.

Remember it

Junction counts charge, loop counts energy. For the bridge, go round the diamond: P over Q equals R over S.