Wheatstone bridge

Curricula: A-level (extension), Sek II

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Aim

Balance the bridge and find an unknown resistance from the balance condition.

The bench

A 5 V source. The upper arm has \(R_1\) and \(R_2\), 1 kΩ each. The lower arm has \(R_3\), 1 kΩ, and an unknown resistor \(R_x\). \(R_3\) acts as a resistance box: its value is changed in the settings panel. The voltmeter is across the diagonal of the bridge.

Procedure

1. Note the voltmeter reading.
2. Select \(R_3\) and change its resistance in the settings panel until the voltmeter shows zero.
3. Note \(R_3\) at balance.
4. Calculate \(R_x = R_3 \cdot R_2 / R_1\) and compare it with the nominal value of \(R_x\) in its settings panel.

Expected result

With \(R_3\) = 1 kΩ the voltmeter shows about 0.50 V: the midpoint of the lower arm is at 3.0 V, that of the upper arm at 2.5 V. Zero is obtained at \(R_3\) = 1.5 kΩ: \(R_x\) = 1500 · 1000 / 1000 = 1500 Ω, as in the settings. The remainder is almost zero, due to the resistance of the cables.

Questions

1. What must the ratio of the arm resistances be for the voltmeter to show zero?
2. Why is a measurement with a bridge more accurate than a measurement using Ohm's law?

Answers

1. Both arms must divide the source voltage in the same ratio, \(R_1 / R_2 = R_3 / R_x\): then the two midpoints are at the same potential, which gives \(R_x = R_3 \cdot R_2 / R_1\). Here \(R_1 = R_2\), so the bridge balances when \(R_3 = R_x\) = 1.5 kΩ.
2. At balance the voltmeter only has to detect zero, and no current flows through it, so its own resistance and scale error do not enter the result; neither does the exact source voltage. \(R_x\) follows from the ratio of known resistances alone. With Ohm's law, \(R = U / I\) carries the errors of two meters, plus the voltage drop across the ammeter or the current drawn by the voltmeter.