Wheatstone Bridge

The circuit, described for the diagram mark

• Four resistors form the four arms of a quadrilateral ABCD: $P$ in AB, $Q$ in BC, $R$ in AD and $S$ in DC. In CBSE's alternative labelling these are $R_1, R_2, R_3, R_4$ respectively.
• Across one pair of diagonally opposite corners, A and C, a cell (with a key) is connected — this is the battery arm.
• Across the other pair, B and D, a galvanometer G is connected — this is the galvanometer arm.
• Mark the currents: $I$ from the cell divides at A into $I_1$ through $P$ and $(I - I_1)$ through $R$; a current $I_g$ flows through G from B to D.

Deriving the balance condition

• Of special interest is the balanced bridge, in which the resistances are such that $I_g = 0$.
• With $I_g = 0$, the junction rule at B and at D gives immediately that the current through $P$ also flows through $Q$, and the current through $R$ also flows through $S$: call them $I_1$ and $I_2$.
• Apply the loop rule to the closed loop ABDA. With no current in G there is no potential drop across it, so

$$-I_1 P + I_2 R = 0 \qquad \Rightarrow \qquad I_1 P = I_2 R$$

• Apply the loop rule to the closed loop BCDB:

$$-I_1 Q + I_2 S = 0 \qquad \Rightarrow \qquad I_1 Q = I_2 S$$

• Dividing the first by the second,

$$\dfrac{P}{Q} = \dfrac{R}{S} \qquad \text{i.e.} \qquad \dfrac{R_1}{R_2} = \dfrac{R_3}{R_4}, \qquad R_1R_4 = R_2R_3$$

• This is the balance condition — the condition for the galvanometer to show zero, or null, deflection.
• The same result follows from the general loop equations without assuming $I_g = 0$ in advance: writing loop ABDA as $I_1R_1 + I_gG - (I-I_1)R_3 = 0$ and loop BCDB as $(I_1-I_g)R_2 - (I-I_1+I_g)R_4 - I_gG = 0$, then putting $I_g = 0$ and dividing, gives $R_1R_4 = R_2R_3$.

Why the bridge is used

• Keeping $R_1$ and $R_2$ known in the first two arms and putting the unknown in the fourth arm, vary $R_3$ until the galvanometer reads zero. Then

$$R_4 = R_3\,\dfrac{R_2}{R_1}$$

• The balance condition involves only the four resistances. It does not contain the emf of the cell, its internal resistance, or the resistance or sensitivity of the galvanometer — this is why the null method is accurate and why the bridge is preferred to measuring $V$ and $I$ separately.
• At balance, B and D are at the same potential, so the galvanometer arm carries no current and may be removed altogether, or replaced by a wire, without changing any current in the rest of the circuit. That is what makes balanced-bridge networks easy to reduce.
• Interchanging the cell and the galvanometer does not disturb the balance.

Metre bridge — the practical form

• A metre bridge is a Wheatstone bridge in which two of the arms are the two parts of a uniform wire exactly one metre long, stretched along a metre scale between two thick copper strips.
• The unknown resistance $R$ is in the left gap and a known resistance box $S$ in the right gap; a cell with a key is across the whole wire, and a galvanometer runs from the midpoint of the gaps to a sliding jockey that can touch the wire anywhere along its length.
• Since the wire is uniform, the resistances of its two parts are proportional to their lengths. At the null (balance) point, at a distance $l$ cm from the left end,

$$\dfrac{R}{S} = \dfrac{l}{100 - l} \qquad \Rightarrow \qquad R = S\,\dfrac{l}{100 - l}$$

• Measure the length $L$ and the diameter $d$ of the unknown wire; then its specific resistance (resistivity) is

$$\rho = \dfrac{RA}{L} = \dfrac{R\,\pi d^{2}}{4L}$$

• Precautions worth a line: keep the jockey pressed only momentarily, take the balance point near the middle of the wire so that the fractional error is least, and use a plug key so that current flows only while a reading is taken.

Worked cases

• Recognising a balanced bridge. Arms $AB = 1\ \Omega$, $BC = 2\ \Omega$, $AD = 2\ \Omega$, $DC = 4\ \Omega$, with a $5\ \Omega$ resistor across BD. Since $AB/BC = 1/2$ and $AD/DC = 2/4 = 1/2$, the bridge is balanced, so no current flows through the $5\ \Omega$ arm and it may be dropped. Then $ABC = 1+2 = 3\ \Omega$ and $ADC = 2+4 = 6\ \Omega$ in parallel:

$$R_{eq} = \dfrac{3\times6}{3+6} = 2\ \Omega, \qquad I = \dfrac{4}{2} = 2\ \text{A}$$

• An unbalanced bridge (NCERT Example 3.7). $AB = 100\ \Omega$, $BC = 10\ \Omega$, $CD = 5\ \Omega$, $DA = 60\ \Omega$, galvanometer $15\ \Omega$ across BD, 10 V across AC. Mesh BADB: $100I_1 + 15I_g - 60I_2 = 0$. Mesh BCDB: $10(I_1-I_g) - 15I_g - 5(I_2+I_g) = 0$. Mesh ADCEA: $60I_2 + 5(I_2+I_g) = 10$. Solving the three gives $I_g = 4.87\ \text{mA}$.

Standard results

• Balance condition: $P/Q = R/S$, i.e. $R_1/R_2 = R_3/R_4$, i.e. $R_1R_4 = R_2R_3$.
• At balance $I_g = 0$ and $V_B = V_D$; the galvanometer arm can be removed or shorted.
• The condition is independent of the cell's emf and of the galvanometer's resistance.
• Metre bridge: $R/S = l/(100-l)$; specific resistance $\rho = RA/L = R\pi d^2/4L$.
• Balanced bridge with a resistor across the galvanometer arm: ignore that resistor and reduce the two branches in parallel.