Wheatstone Bridge Strain-Gauge Sensitivity
A student builds a Wheatstone bridge to monitor a strain gauge bonded to a metal beam. Three arms are fixed 100 Ω resistors; the fourth arm is the strain gauge, which reads 100.0 Ω when unstrained. The bridge is supplied by a 6.0 V battery of negligible internal resistance. A galvanometer of resistance 50 Ω and threshold sensitivity 10 μA is connected between the midpoints. When the beam is loaded, the gauge resistance rises to 102.0 Ω. The student predicts that the bridge stays balanced because the change is only 2%, and therefore the galvanometer will not move. The lab supervisor asks for the Thevenin voltage across the galvanometer, the total bridge-plus-meter resistance seen by that source, the actual galvanometer current, and a judgment about whether the prediction is correct. The battery and resistors are at room temperature; lead resistances are negligible. The galvanometer coil can safely carry 1 mA.
1. With the gauge at 102.0 Ω, calculate the Thevenin voltage between the bridge midpoints and the Thevenin resistance seen by the galvanometer. Hence find the galvanometer current.
Read the worked answer
Label the top supply node A and the bottom node B, with 6.0 V across them. Left divider: R1 = 100 Ω (top), R2 = 100 Ω (bottom). Right divider: R3 = 100 Ω (top), R4 = 102.0 Ω (bottom).
Taking B as 0 V, V_left = 6.0 × 100/(100 + 100) = 3.000 V. V_right = 6.0 × 102/(100 + 102) = 6.0 × 102/202 = 3.0297 V. The open-circuit (Thevenin) voltage between the midpoints is therefore |3.0297 − 3.000| = 0.0297 V = 29.7 mV.
Short the battery to find the resistance looking into the midpoints: R_th = (100||100) + (100||102) = 50 + (100 × 102)/(202) = 50 + 50.50 = 100.5 Ω. With the galvanometer, R_total = 100.5 + 50 = 150.5 Ω.
The current is I = V_th/R_total = 0.0297/150.5 = 1.97×10^−4 A = 197 μA.
2. Compare the current with the 10 μA threshold and judge the student's prediction. Does the 2% resistance change matter?
Read the worked answer
The galvanometer threshold is 10 μA, and 197 μA is about 20 times larger, so the meter will deflect clearly. The student's prediction that a 2% resistance change cannot move the bridge is wrong.
A Wheatstone bridge is balanced only when the ratios are exactly equal; near balance the output is proportional to the fractional change, not to whether the change is 'small'. The 2 Ω change in 100 Ω produces a 29.7 mV differential because the two dividers no longer sit at the same fraction of the supply. The bridge is a sensitive comparator, not a device that ignores small changes. The 1 mA safe current is also not exceeded (197 μA < 1 mA).
3. Near balance, derive the approximation ΔV ≈ V ΔR/(4R) for this bridge, and use it to estimate the minimum detectable ΔR at 6.0 V.
Read the worked answer
Near balance, with three arms R and the gauge R + ΔR, the midpoint difference is ΔV = V[(R + ΔR)/(2R + ΔR) − 1/2]. For ΔR << R this becomes ΔV ≈ V ΔR/(4R).
At V = 6.0 V and R = 100 Ω, ΔV ≈ 6.0 ΔR/400 = 0.015 ΔR volts per ohm, i.e. 15 mV per ohm.
To reach 10 μA through 150.5 Ω, the required Thevenin voltage is V_th = IR ≈ 10×10^−6 × 150.5 = 1.51×10^−3 V. Thus ΔR ≈ 1.51×10^−3/0.015 = 0.10 Ω.
A change of about 0.1 Ω, or 0.1%, should be detectable. This confirms that the bridge is far more sensitive than the 2% figure suggests, and that increasing the supply voltage or reducing the bridge resistance would improve the resolution further, subject to power and meter limits.