Physics · Ch 3 — Current Electricity

Wheatstone Bridge

3.13

Wheatstone Bridge

What is a Wheatstone Bridge?

The Wheatstone bridge is a special arrangement of four resistors used to measure an unknown resistance very precisely. It is a direct application of Kirchhoff’s rules.

The circuit has four resistors: $R_1$ , $R_2$ , $R_3$ , and $R_4$ , connected in a diamond shape.

Battery arm: A source (cell) is connected across one pair of opposite corners (A and C).
Galvanometer arm: A sensitive current-detecting device called a galvanometer (G) is connected across the other pair of opposite corners (B and D).

We assume the cell has no internal resistance for simplicity.

The Key Idea: The Balanced Bridge

The most important case is when the bridge is balanced. This means the resistors are chosen so that no current flows through the galvanometer ( $I_g = 0$ ).

When $I_g = 0$ , the galvanometer shows a null deflection. This is the condition we aim for.

Deriving the Balance Condition

Let’s label the currents in the arms when the bridge is balanced ( $I_g = 0$ ):

Current through $R_1$ = $I_1$
Current through $R_2$ = $I_2$
Current through $R_3$ = $I_3$
Current through $R_4$ = $I_4$

Step 1: Apply Kirchhoff’s Junction Rule

At junction D: $I_1$ enters, $I_3$ leaves. Since $I_g = 0$ , we get $I_1 = I_3$ .
At junction B: $I_2$ enters, $I_4$ leaves. Since $I_g = 0$ , we get $I_2 = I_4$ .

Step 2: Apply Kirchhoff’s Loop Rule

Consider two closed loops that do not include the galvanometer (since $I_g = 0$ , its branch has no potential drop).

Loop ADBA: Going from A → D → B → A. The potential changes are: $-I_1 R_1 + 0 + I_2 R_2 = 0$ This gives:

I_1 R_1 = I_2 R_2 \quad \text{(Equation 1)}

Loop CBDC: Going from C → B → D → C. Using $I_3 = I_1$ and $I_4 = I_2$ , the potential changes are: $I_2 R_4 + 0 - I_1 R_3 = 0$ This gives:

I_1 R_3 = I_2 R_4 \quad \text{(Equation 2)}

Step 3: Combine the Equations

From Equation 1: $\frac{I_1}{I_2} = \frac{R_2}{R_1}$

From Equation 2: $\frac{I_1}{I_2} = \frac{R_4}{R_3}$

Equating the two ratios, we get the balance condition:

\frac{R_2}{R_1} = \frac{R_4}{R_3}

Or, more commonly written as:

R_1 R_4 = R_2 R_3 \quad \text{[Equation 3.64(a)]}

This is the Wheatstone bridge balance condition. When this relation holds, the galvanometer shows zero current.

Practical Use: Measuring an Unknown Resistance …

Figure 3.18The Wheatstone bridge circuit.

Fig. 3.18 — The Wheatstone bridge circuit.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

The figure shows a Wheatstone bridge drawn as a diamond-shaped circuit. The four vertices are labelled A (left), B (top), C (right), and D (bottom). The four arms of the diamond each contain a resistor:

Arm A→B (upper-left): resistor $R_2$
Arm B→C (upper-right): resistor $R_4$ , with the word Unknown written slanted along it
Arm A→D (lower-left): resistor $R_1$
Arm D→C (lower-right): resistor $R_3$ , with the words Standard arm written slanted along it

A galvanometer $G$ (drawn as a circle) connects the vertical diagonal B–D — this is the galvanometer arm. Across the horizontal diagonal A–C, external leads drop down to a battery symbol $\varepsilon$ at the bottom — this is the battery arm. Branch currents are labelled: $I_1$ in arm A–D, $I_2$ in arm A–B, $I_3$ in arm D–C, and $I_4$ in arm B–C.

Physical idea

The Wheatstone bridge is a circuit used to measure an unknown resistance by comparing it with known resistances. The key is to adjust the resistances until the galvanometer shows zero current ( $I_g = 0$ ). This condition is called a balanced bridge. At balance, no current flows through the galvanometer, and the potential difference between B and D is zero.

Key formula derived from the figure

Using Kirchhoff’s rules for the balanced bridge ( $I_g = 0$ ):

Junction rule at D and B gives $I_1 = I_3$ and $I_2 = I_4$ .
Loop rule for loop ADBA:

-I_1 R_1 + 0 + I_2 R_2 = 0 \quad \Rightarrow \quad I_1 R_1 = I_2 R_2

Loop rule for loop CBDC (using $I_3 = I_1$ , $I_4 = I_2$ ):

I_2 R_4 + 0 - I_1 R_3 = 0 \quad \Rightarrow \quad I_1 R_3 = I_2 R_4

Dividing the two equations gives the balance condition:

$\frac{R_2}{R_1} = \frac{R_4}{R_3} \quad \text{or} \quad \boxed{R_2 R_3 = R_1 R_4}$ …

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