Bridge Reading That Disagrees with Calculation
A final-year lab group is characterizing an unbalanced Wheatstone bridge. The left divider across a regulated 10.00 V supply is R1 = 1.00 kΩ from the positive rail to node A and R2 = 2.00 kΩ from node A to ground. The right divider is R3 = 3.00 kΩ from the positive rail to node B and R4 = 4.00 kΩ from node B to ground. All four resistors are 1% metal-film types. The group computes the open-circuit bridge voltage V_AB and expects about 0.95 V. They measure with a data-acquisition card whose differential input resistance is 10 kΩ, and the card reports 0.77 V. They have already swapped resistors and repeated the measurement with the same result. The supervisor says the bridge is probably fine and the instrument is the problem. The students want to know whether a 1% resistor could explain a 0.18 V discrepancy, and what input resistance would be needed for a 1% loading error.
1. Compute the bridge Thévenin voltage and Thévenin resistance seen by the DAQ card, predict the card reading, and decide whether a faulty 1% resistor is a plausible explanation.
Read the worked answer
With the left divider, V_A = 10 × 2/(1+2) = 6.667 V. With the right divider, V_B = 10 × 4/(3+4) = 5.714 V. The open-circuit bridge voltage is V_AB = 6.667 − 5.714 = 0.952 V, so the students’ 0.95 V expectation is correct.
The resistance seen looking into terminals A–B is R_th = (1 kΩ || 2 kΩ) + (3 kΩ || 4 kΩ) = 0.667 kΩ + 1.714 kΩ = 2.381 kΩ. With a 10 kΩ meter, the reading is V_meas = 0.952 × 10/(10 + 2.381) = 0.769 V. That matches the observed 0.77 V almost exactly.
A 1% resistor tolerance can shift the open-circuit voltage by roughly 0.1 V in the worst case, but it would not produce a repeatable value that agrees with the finite-meter-loading prediction to within 0.001 V. The evidence points to meter loading, not a faulty resistor. The bridge is probably fine.
2. Find the minimum DAQ input resistance for which the loading error is below 1% of the open-circuit voltage, and comment on using a 10 MΩ DMM instead.
Read the worked answer
The fractional loading error is R_th/(R_in + R_th). Requiring this to be below 0.01 gives R_in > 99 R_th = 99 × 2.381 kΩ ≈ 235.7 kΩ. A 10 kΩ card is therefore far too low for this bridge.
A 10 MΩ DMM gives an error of about 2.381 kΩ/10 MΩ ≈ 0.024%, which is negligible compared with the 1% resistor tolerance. The students should either use a high-impedance meter or correct the reading using the known R_th. If they must use the 10 kΩ card, the corrected open-circuit voltage is 0.77 × (10 + 2.381)/10 ≈ 0.95 V.