Wheatstone Bridge & Meter Bridge - Physics Class 12 - NEET PDF Download

Wheatstone Bridge

The Wheatstone bridge is a four-resistor network designed to measure an unknown resistance accurately by comparison. It was designed by Charles Wheatstone. The four resistances P, Q, R and S are connected as the four arms of a quadrilateral ABCD. A cell (source) with key K₁ is connected across one pair of opposite corners A and C, and a sensitive galvanometer with key K₂ is connected across the other pair of opposite corners B and D.

• When K₁ is closed, a current i from the cell reaches junction A and splits into two currents i₁ (through AB) and i₂ (through AD).
• At junction B the current i₁ further splits into (i₁ - i_g) along BC and i_g through the galvanometer BD.
• At junction D the currents i₂ and i_g combine to give (i₂ + i_g) along DC.
• If the galvanometer shows a deflection, current i_g ≠ 0 and the bridge is unbalanced; by adjusting the arms P, Q, R and S one seeks the position where the galvanometer reading becomes zero (i_g = 0).
• When the galvanometer shows zero deflection the bridge is balanced. In that condition the same current flows through AB and BC (i₁), and the same current flows through AD and DC (i₂), so P and Q form one series pair and R and S form the other series pair.
• The arms BD and AC are called conjugate arms because interchanging the galvanometer and the cell does not affect the balance condition.
• The sensitivity of the bridge depends on the values of the four resistances; the sensitivity is maximum when all four resistances are of the same order of magnitude.

Condition for Balance

Apply Kirchhoff's laws to the two meshes containing the galvanometer branch. Let G denote the galvanometer resistance and i_g the galvanometer current. For mesh ABDA:

i₁P + i_gG - i₂R = 0.

For mesh BCDB:

(i₁ - i_g)Q - (i₂ + i_g)S - i_gG = 0.

When the bridge is balanced, i_g = 0. The equations reduce to

i₁P - i₂R = 0 → i₁P = i₂R

i₁Q - i₂S = 0 → i₁Q = i₂S

Dividing these two relations gives the standard balance condition:

P / Q = R / S

Or equivalently,

P × S = Q × R

Thus, if three resistances are known, the fourth (unknown) resistance can be determined. Arms P and Q are commonly called ratio arms; R is the known arm and S is the unknown arm (or vice versa depending on lab notation).

Examples and Worked Problems

Ex.22 Find equivalent resistance of the circuit between the terminals A and B.

Sol.

The given circuit is a Wheatstone bridge in a balanced condition.

Verify balance: 10 × 3 = 30 = 6 × 5, so P×Q = R×S and the central branch carries no current.

Hence the bridge simplifies to series and parallel combinations as shown below.

Ex.23 Find the equivalent resistance between A and B

Sol.

The arrangement is a balanced Wheatstone bridge, therefore the diagonal branch has zero current and may be removed. The circuit reduces to the equivalent simpler network illustrated.

Unbalanced Wheatstone Bridge

Ex.24

Find equivalent resistance ?

Sol.

Let the electric potential at point B be x and at point E be y with respect to reference. Apply Kirchhoff's current law (KCL) at nodes as shown in the diagram.

Applying KCL at B gives:

8x - 5y = v ...(1)

Applying KCL at E gives:

8y - 5x = 2v ...(2)

Solve equations (1) and (2) for x and y to obtain node potentials.

After solving we obtain the values of x and y shown in the figures.

Currents from branches BC and EF add to give the total current i flowing from the source.

i = i₃ + i₄ =

=

Using the obtained total current and applied voltage v, the equivalent resistance is

R_eq =

Ladder (Infinite Network) Problems

Find the effective resistance between A and B for an infinite chain of resistors. Let the effective resistance be R_E. Because the network is infinite, removing the first section leaves the remaining infinite network unchanged; hence the resistance between corresponding points in the remaining network is also R_E.

The original infinite chain is equivalent to a single resistor R in series with the parallel combination of R and R_E. Thus the effective resistance satisfies a quadratic relation.

Derive the equation:

R_ER + R_E² = R² + 2RR_E

Rearrange:

R_E² - R R_E - R² = 0

Ex.25

Find the equivalent resistance between A and B when each next section increases resistance by a factor k.

Sol.

Let R_E be the effective resistance between A and B. Because the network is infinite, removing one section leaves the remainder equivalent to kR_E. The effective resistance satisfies the relation obtained by replacing the remainder with kR_E.

Therefore the effective resistance between A and B is given by the equation shown below; solving yields the final expression for R_E.

Symmetrical Circuits

Circuit symmetry often simplifies analysis: mirror symmetry and folding symmetry can reveal equipotential points and branches with zero current, allowing simplification without lengthy algebra.

Ex.26 Find the equivalent Resistance between A and B

Sol. I Method : mirror symmetry

The branches AC and AD are symmetric, so currents through them are equal. The left and right halves are mirror images about the central plane CD. From symmetry it follows that the current in the resistor between C and E is zero and similarly in ED is zero. The circuit simplifies as shown.

II Method : folding symmetry

Because V_C = V_D, points C and D are at the same potential and can be considered the same node. Folding reduces the circuit to a simpler balanced Wheatstone bridge.

The simplified bridge is balanced and further reduces to the circuit below:

In the folding method it is not necessary to know the currents in CA and DA explicitly.

Ex.27 Find the equivalent Resistance between A and B

Sol.

The circuit has symmetry in the two input branches AC and AD, so currents in them are equal at the input. However the outputs are not mirror images (left has R, right has 2R), so the overall distribution differs. Nevertheless V_C = V_D, and points C and D can be treated as the same potential, reducing the circuit to a balanced Wheatstone bridge.

The bridge then simplifies to the equivalent network:

Which leads to the final equivalent resistance

Ex.28 Find the equivalent Resistance between A and B

Sol.

Here V_A = V_C and V_B = V_D, so the circuit simplifies as shown:

Therefore the equivalent resistance is

Ans.

Ex.29

Twelve equal resistors each of resistance R are connected along the edges of a cube. Find the equivalent resistance of the network for different pairs of terminals.

(a) When current enters at vertex 1 and leaves at the opposite vertex 6 (body diagonal)

Sol.

By symmetry, nodes 2, 4 and 8 are equipotential (they are reached from 1 by an edge and from 6 by a face diagonal), and nodes 3, 5 and 7 are equipotential. Grouping equipotential nodes reduces the network to a simpler equivalent.

(b) When current enters at vertex 1 and leaves at adjacent vertex 2

Sol.

Nodes 3 and 7 are equipotential with respect to 1 and 2 (reachable along faces/edges); similarly 4 and 8 are equipotential. Use these equalities to collapse nodes and compute the equivalent resistance.

(c) When current enters at vertex 1 and leaves at vertex 3

Sol.

Cutting the cube along the plane through 2, 4, 5 and 7 and using mirror symmetry yields the reduced configuration shown below. The equivalent resistance is then computed from that simpler network.

Meter Bridge

The meter bridge is an application of the Wheatstone bridge used in the laboratory to measure an unknown resistance by comparison with a standard resistance using a uniform wire of known length (commonly 1 metre). The meter bridge provides a convenient practical method because the resistance of a uniform wire is proportional to its length.

Typical arrangement: a uniform wire AC of length 1 m (100 cm) is stretched and its ends are connected to a known resistance P and an unknown resistance S. A cell is connected across the combination and a galvanometer is connected between a movable jockey and the junction between P and S. By sliding the jockey to find a balance point B where the galvanometer shows zero current, the lengths AB = ℓ and BC = (100 - ℓ) are measured and used to compute the unknown resistance.

If the wire has resistance per unit length r, then

P = r ℓ

Q = r(100 - ℓ)

At the balance condition the Wheatstone bridge balance relation gives:

Thus

P / Q = S / Rwire (or in the usual meter-bridge notation)

Solving for the unknown resistance S gives:

S = P × (BC / AB) = P × (100 - ℓ) / ℓ

Here P is the known resistance, ℓ is the balancing length from A to the jockey in cm, and 100 - ℓ is the remainder of the 1 m wire. This formula is used in laboratory practice to obtain precise values of an unknown resistance.

Summary

The Wheatstone bridge is a precise method to determine unknown resistances using comparison and the balance condition P/Q = R/S. The meter bridge is a practical laboratory implementation based on the same principle using a uniform wire. Circuit symmetry and recognition of equipotential points can greatly simplify equivalent resistance calculations in complex resistor networks. For infinite networks, self-similarity leads to algebraic equations for the effective resistance which are solved to obtain the result.