In a Wheatstone bridge at balance, what does the center meter (between the bridge midpoints) read, assuming an ideal meter and source?

Introduction / Context:
The Wheatstone bridge is a precision circuit for resistance measurement and sensing (e.g., strain gauges). Balance occurs when the ratio of resistances in one leg equals the ratio in the other leg. At this condition, the detector connected between the two midpoints shows no potential difference.

Given Data / Assumptions:

• Ideal DC source energizing the bridge.
• Ideal, high-impedance voltmeter or galvanometer connected between the midpoints.
• Balance condition: R1/R2 = R3/R4 (or equivalently R1 * R4 = R2 * R3).

Concept / Approach:
At balance, node potentials at the two midpoints are equal, so the meter sees zero differential voltage. With no voltage difference, no current flows through the meter. This is the defining criterion used to adjust an unknown resistor until the bridge is balanced.

Step-by-Step Solution:
Let V+ be the supply top node and ground the bottom node.Midpoint left: VL = V+ * (R2 / (R1 + R2)).Midpoint right: VR = V+ * (R4 / (R3 + R4)).Balance condition ensures VL = VR, therefore Vmeter = VL − VR = 0 V.

Verification / Alternative check:
If the bridge is slightly unbalanced, the meter reads a small voltage whose polarity indicates which side is higher—this is how sensitive null measurements are made in practice.

Why Other Options Are Wrong:

• Twice/source/half the source voltage: These imply a nonzero difference, contradicting the equal-potential condition at balance.

Common Pitfalls:

• Confusing equal resistance values with equal ratios; balance requires matched ratios, not necessarily equal resistors.
• Using a low-impedance meter that perturbs the bridge; an ideal or high-impedance detector is assumed.

Final Answer:
zero volts