The Wheatstone Bridge: Deriving the Balance Condition

Current Electricity | Class 12 CBSE Physics | Concept Notes

Why the Wheatstone bridge exists

Ever wondered how to precisely measure an unknown electrical resistance? While multimeters are handy, sometimes they lack the precision needed for certain applications. That’s where the Wheatstone bridge comes in! This ingenious circuit, dating back to the 19th century, provides a highly accurate way to determine an unknown resistance by comparing it to known resistances. In essence, it uses a balance principle – just like a balance scale – to find the value of an unknown resistance.The Wheatstone bridge sidesteps this problem entirely with a clever trick: instead of measuring a voltage or current directly, it finds the point of zero current in a sensitive branch. At that null point, the measuring instrument (a galvanometer) draws no current at all — so it can’t disturb the circuit, no matter how imperfect it is. This is the same null-method philosophy you’ll meet again in the potentiometer.

Setting up the bridge

The image below shows a typical Wheatstone bridge circuit. It consists of four resistors – let’s call them P, Q, R, and S – arranged in a diamond shape.

Resistors: There are two known resistors, let’s call them R (known variable resistor) and S (known fixed resistor). Then there are two resistors with values we want to find. Let’s call them P (unknown resistor) and Q (the resistor whose value we are trying to determine).

Galvanometer: A sensitive current-detecting device called a galvanometer (G) is connected between two opposite junctions (B and D).

Power Source: A battery or DC power source is connected to the other two junctions (A and C).

Objective: The goal is to adjust the known variable resistor (R) until the current through the galvanometer (G) is zero. This condition is called the “balanced condition.”

Deriving the balance condition

Now, let’s dive into the fascinating derivation of the balance condition. It’s surprisingly simple yet powerful.

1. Balanced State: When the bridge is balanced, no current flows through the galvanometer. This means the potential at point B is equal to the potential at point D.
2. Voltage Drops: Since points B and D are at the same potential, the voltage drop across resistors P and R must be equal. Likewise, the voltage drop across resistors Q and S must be equal.
3. Currents:
   • Let the current through P be I1. Since no current goes through G, all of I1 also flows through Q. So, current through P = current through Q = I1.
   • Similarly, let the current through R be I2. All of I2 also flows through S. So, current through R = current through S = I2.
4. Using Ohm’s Law (V=IR):
   • Voltage drop across P: VP = I1 * P
   • Voltage drop across R: VR = I2 * R
   • Since VP = VR in a balanced state, I1 * P = I2 * R.
5. Applying the Same to the Other Side:
   • Voltage drop across Q: VQ = I1 * Q
   • Voltage drop across S: VS = I2 * S
   • Since VQ = VS in a balanced state, I1 * Q = I2 * S.
6. Dividing the Equations:
   • Now, let’s divide the equation in step 4 by the equation in step 5:(I1 * P) / (I1 * Q) = (I2 * R) / (I2 * S)
7. Simplifying:
   • The current terms (I1 and I2) cancel out, leaving us with:P / Q = R / S

This simple yet elegant ratio is the famous balance condition of the Wheatstone bridge! It means that when the bridge is balanced, the ratio of resistances in adjacent arms is equal. This allows us to find an unknown resistance if we know the values of the other three.

Worked Example 1

Let’s look at a few examples to see how this derivation works in practice.

• Example 1: Suppose P = 10 Ω, R = 20 Ω, and S = 40 Ω. Find Q.
  • Using the balance condition: P / Q = R / S
  • 10 / Q = 20 / 40
  • 10 / Q = 1/2
  • Q = 20 Ω

Worked Example 2 — the trap question

A Wheatstone bridge has resistors P = 5 Ω, Q = 15 Ω, and S = 30 Ω. The bridge is balanced. What is the value of R?

• Using the balance condition: P / Q = R / S
• 5 / 15 = R / 30
• 1/3 = R / 30
• R = 30 / 3 = 10 Ω

Worked Example  3

In a Wheatstone bridge, R = 100 Ω, S = 200 Ω, and Q = 300 Ω. What is P when the bridge is balanced?

• Using the balance condition: P / Q = R / S
• P / 300 = 100 / 200
• P / 300 = 1/2
• P = 300 / 2 = 150 Ω

From bridge to meter bridge

The Wheatstone bridge principle is so versatile that it’s also the foundation of another useful instrument called the meter bridge.

• Concept: A meter bridge uses a long uniform resistance wire instead of discrete resistors for two arms (R and S) of the bridge.
• Bridge Construction: A meter bridge features a 1-meter long wire mounted on a wooden board. A movable sliding contact called a “jockey” can be moved along this wire.
• Analogy: The jockey splits the wire into two sections:
  • Length l: One section represents resistor R, with its resistance proportional to length l.
  • Length (100 - l): The other section represents resistor S, with its resistance proportional to (100 - l).
• Balance Condition in Meter Bridge: By moving the jockey and finding the point of zero deflection in the galvanometer, we establish the balance condition. At this point:P / Q = l / (100 - l)
  • This simple equation allows for the direct determination of an unknown resistance by just measuring lengths l and (100 - l).

Common errors to watch for

Error	What’s actually true
“Zero galvanometer current means zero current everywhere”	Current still flows through all four arms; only the galvanometer branch carries zero current
Writing P/Q = R/S with arms mismatched to the actual circuit diagram	Always re-derive from your specific labeled diagram — the positions of P, Q, R, S relative to the galvanometer determine which ratio equals which
Assuming the balance condition depends on the cell’s EMF	It doesn’t — EMF and the branch currents cancel out of the derivation entirely (see Step 4)
Ignoring end-resistance/contact resistance in a meter-bridge realization	These act like small additional resistances at the ends of the wire and are a genuine source of experimental error, especially when the balance point falls near either end

Quick Practice

Q1. In a Wheatstone bridge, P = 4 Ω, Q = 6 Ω, S = 9 Ω. Find R at balance.

 Answer: P/Q = R/S ⟹ R = S × P/Q = 9 × 4/6 = 6 Ω

Q2. Why is the Wheatstone bridge method considered more accurate for measuring resistance than a simple ammeter-voltmeter method? 

Answer: Because it’s a null method — at balance, the galvanometer carries zero current, so its own resistance (and any imperfection in it) doesn’t affect the measurement. An ammeter-voltmeter method, by contrast, always carries some current through instruments that have non-zero (or non-infinite) resistance, introducing a systematic error.

Q3. If you swapped the positions of the galvanometer and the cell in a Wheatstone bridge (i.e., drove current through B–D and used A–C as the “detector” branch), would the same balance condition P/Q = R/S still apply? 

Answer: This is a genuine transfer question — students should recognize that swapping the roles changes which junctions the KCL/KVL argument is applied to, and re-deriving (rather than assuming the same formula still applies) is the physically correct approach. In fact, for the standard bridge topology, the balance condition is reciprocity-symmetric in this particular swap, but the reasoning* — not just the memorized formula — is what should transfer.*

Summary

Step	What it establishes
KCL at B (I_g=0)	Current through P = current through R
KCL at D (I_g=0)	Current through Q = current through S
V_B = V_D (zero galvanometer deflection)	Potential drops A→B and A→D are equal; potential drops B→C and D→C are equal
Combine and cancel currents	P/Q = R/S

The Wheatstone bridge balance condition is not a separate law — it’s KCL and KVL, applied at the specific moment the galvanometer reads zero, with the currents on both sides cancelling algebraically. That’s what makes it worth deriving rather than memorizing: the same two laws you already know are doing all the work.

https://edu-physics.com/current-electricity-master-hub/