Subtopic: Measurement of force, stress, torque, speed, displacement and acceleration on machine elements – shafts, couplings, keys, bearings and knuckle joints
Question Statement:
A load cell employing Wheatstone bridge circuit has two fixed resistors and two strain gauges all of which have a value of 120 ohm. The gauge factor is 2.1. The strain in each of the two strain gauges, one in tension and the other in compression, is 1.65 × 10⁻⁴. If the battery current in the initial balanced condition of the bridge is 50 mA, the sensitivity of the load cell in v/strain will be
Options:
(A) 0.001
(B) 0.002
(C) 6.3
(D) 12.6
Official Answer Key:
D
Step-by-Step Solution & Explanation
Excitation voltage: with 4 equal 120 Ω arms and a balanced-bridge current of 50 mA flowing through each series branch (2 resistors in series per branch), Vex = I × 2R = 0.05 × 240 = 12 V. With two active gauges in adjacent arms (one +Δε, one −Δε — a half-bridge, push–pull configuration), the bridge output is doubled relative to a single active arm: Vo = (Vex/2)·GF·ε = (12/2) × 2.1 × 1.65×10⁻⁴ = 6 × 2.1 × 1.65×10⁻⁴ = 2.079×10⁻³ V. Sensitivity = Vo/ε = 2.079×10⁻³ / 1.65×10⁻⁴ = 12.6 V/strain, matching option (D).
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