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Wheatstone bridge

Curricula: A-level (extension), Sek II

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Aim

Balance a Wheatstone bridge and find an unknown resistance from three known ones by a null method.

Theory

Wheatstone bridge: R1 and R2 in the upper row, R3 and Rx in the lower row, the voltmeter V between the junctions of the two rows, the source U below, its "+" on the left

A Wheatstone bridge is two potential dividers connected to the same source. In the upper divider the current flows through \(R_1\) and then \(R_2\); in the lower one, through \(R_3\) and then \(R_x\). Each of the four resistors is an arm of the bridge. Relative to the "−" terminal of the source, the junction of the upper divider is at a potential of \(U \cdot R_2 / (R_1 + R_2)\), and the junction of the lower one at \(U \cdot R_x / (R_3 + R_x)\) (see the lab Potential divider with a potentiometer). A voltmeter connected between the two junctions forms the diagonal of the bridge and reads the potential difference between them.

The bridge is balanced when the two junctions are at the same potential: the voltmeter shows zero, and no current flows along the diagonal. Equal potentials mean \(R_2 / (R_1 + R_2) = R_x / (R_3 + R_x)\). Turning both fractions upside down and subtracting 1 from each side gives the balance condition \(R_1 / R_2 = R_3 / R_x\), so the unknown resistance is \(R_x = R_3 \cdot R_2 / R_1\). Written as \(R_1 \cdot R_x = R_2 \cdot R_3\), the condition says that the products of opposite arms are equal: \(R_1\) and \(R_x\) on one side, \(R_2\) and \(R_3\) on the other.

\(R_1\) and \(R_2\) are the ratio arms, and \(R_3\) is the variable arm. In a school laboratory \(R_3\) is a resistance box: precision resistors selected with one dial each for units, tens, hundreds and thousands of ohms. The ratio arms set the range and the resolution of the bridge. With \(R_1 / R_2\) = 10, \(R_3\) balances at ten times \(R_x\), so a box that steps in whole ohms finds \(R_x\) to a tenth of an ohm. With \(R_1 / R_2\) = 1/10 the same box measures resistances ten times beyond its own range. In a metre bridge the ratio arms are the two parts of one uniform wire, divided by a sliding contact. The resistance of a wire is proportional to its length, so \(R_1 / R_2 = l_1 / l_2\) and \(R_x = R_3 \cdot l_2 / l_1\).

The sign of the reading shows which way to change \(R_3\). With the "+" socket of the voltmeter on the lower junction, the reading is positive while \(R_3\) is too small, because the lower junction is then at a higher potential than the upper one. When \(R_3\) is too large, the reading is negative. The balance point lies between two values of \(R_3\) that give readings of opposite sign. Each halving keeps the half that still contains the balance point, so the interval shrinks by half at every step.

Measuring at balance is a null method. The scale error of the meter does not matter, because it only has to show zero, and its resistance does not matter, because no current flows through it. The source voltage cancels out of the balance condition. The accuracy depends on how well \(R_1\), \(R_2\) and \(R_3\) are known and on how small an imbalance the meter can detect, and that imbalance gives a larger reading at a higher source voltage. A measurement by Ohm's law, \(R = U / I\), carries the scale errors of two meters, and one of them always reads more than belongs to \(R_x\): either the voltmeter includes the voltage across the ammeter, or the ammeter includes the voltmeter's current (see the lab Connecting an ammeter and a voltmeter). The traditional detector of a bridge is a centre-zero galvanometer, a sensitive meter for small currents. On our bench a voltmeter that reads down to microvolts does the same job.

Samuel Hunter Christie described the circuit in 1833, and Charles Wheatstone made it widely known in 1843. Strain-gauge and resistance-thermometer circuits still use the bridge in deflection mode: the bridge is balanced at a reference state, the sensor forms one arm, and the voltage across the diagonal follows small changes in its resistance.

The bench

The diagram in Theory keeps the rows and sides of the bench: the same two rows, "+" on the left and "−" on the right. A 5 V source with an internal resistance of 0.5 Ω, its "+" on the buses on the left and its "−" on the buses on the right. The upper divider runs along the top row, with \(R_1\) = 1 kΩ next to "+" and \(R_2\) = 1 kΩ next to "−". The lower divider runs along the bottom row, with \(R_3\) = 1 kΩ next to "+" and the unknown resistor \(R_x\) next to "−". \(R_x\) has no colour bands, and selecting it shows its value, so leave it until step 4. \(R_3\) serves as the resistance box and is set in ohms: its parameter card opens in kΩ, so choose Ω in the unit list each time \(R_3\) is selected. A reading near −2.5 V means the value went in as kΩ. The voltmeter is in the diagonal, its "+" socket on the bus between \(R_3\) and \(R_x\) and its "−" socket on the bus between \(R_1\) and \(R_2\). A positive reading has no sign on the screen.

Procedure

  1. Write down the voltmeter reading and its sign.
  2. Set \(R_3\) to 2000 Ω and write down the reading and its sign.
  3. If the readings at 1000 Ω and 2000 Ω have opposite signs, the balance point lies between them. Set \(R_3\) to the middle of the interval, and keep the half whose ends give readings of opposite sign. Repeat, rounding to whole ohms, until the ends are 1 Ω apart, and record each value of \(R_3\) with its reading in a table. Write down the value of \(R_3\) whose reading is closer to zero.
  4. Calculate \(R_x = R_3 \cdot R_2 / R_1\). Select \(R_x\) to read its nominal value and compare it with the result.
  5. Leaving \(R_3\) at balance, raise the source voltage to 10 V and write down the reading. Then set \(R_3\) 1 Ω above the balance value and compare the reading with the one at 5 V from step 3. Set \(R_3\) back to the balance value and the source back to 5 V.
  6. Set \(R_1\) to 2 kΩ. From \(R_x\) found in step 4, predict the value of \(R_3\) that balances the bridge now. Set \(R_3\) to the prediction and check that the readings 1 Ω either side of it have opposite signs.
Expected result

With \(R_3\) = 1000 Ω the voltmeter shows about 227 mV, a positive reading. At 2000 Ω it shows about −625 mV: \(R_3\) has passed the balance point. Halving the interval gives −278 mV at 1500 Ω, −51 mV at 1250 Ω and 81 mV at 1125 Ω, and after about ten settings ends between 1200 Ω and 1201 Ω. The reading is about 18 µV at 1200 Ω and −1.02 mV at 1201 Ω, so \(R_3\) = 1200 Ω and \(R_x\) = 1200 × 1000 / 1000 = 1200 Ω, the nominal value of \(R_x\). The reading does not reach exactly zero, because the cables, about 0.02 Ω each, add to the arms unequally.

At 10 V the bridge stays balanced, and the reading is about 36 µV, although both junctions have risen to 5.0 V. At 1201 Ω the reading is about −2.05 mV, twice the −1.02 mV at 5 V: the balance point does not move, but a higher source voltage makes the null sharper.

With \(R_1\) = 2 kΩ the bridge balances at \(R_3\) = \(R_x \cdot R_1 / R_2\) = 2400 Ω, as predicted: \(R_x\) = 2400 × 1000 / 2000 = 1200 Ω. The readings are about 0.46 mV at 2399 Ω and −0.46 mV at 2401 Ω. A change of 1 Ω in \(R_3\) is now 0.5 Ω of \(R_x\), so the same box measures \(R_x\) twice as finely.

Questions

  1. Calculate the voltmeter readings at \(R_3\) = 1000 Ω and 2000 Ω from the divider formula and compare them with the readings from steps 1 and 2.
  2. From the readings at the two ends of your last 1 Ω interval in step 3, find by linear interpolation, to a tenth of an ohm, the value of \(R_3\) at which the reading would be zero. If \(R_1\), \(R_2\) and \(R_3\) of a real bridge each have a tolerance of 0.1%, what is the uncertainty in \(R_x\)?
  3. \(R_x\) is a strain gauge, and the bridge is balanced as at the end of step 3. The gauge is stretched, and its resistance rises by 1 Ω. What does the voltmeter show, and why does a strain-gauge circuit use the bridge rather than a voltmeter across the gauge alone?
Answers
  1. The upper junction is at 5 V × 1000 / 2000 = 2.500 V for any \(R_3\). At \(R_3\) = 1000 Ω the lower one is at 5 V × 1200 / 2200 ≈ 2.727 V, and the voltmeter shows 2.727 − 2.500 ≈ 0.227 V. At 2000 Ω the lower junction is at 5 V × 1200 / 3200 = 1.875 V, and the reading is 1.875 − 2.500 = −0.625 V. The calculated values differ from the readings by less than 1 mV. The difference comes mainly from the 0.5 Ω internal resistance of the source, and to a lesser extent from the 10 MΩ of the voltmeter and the cables.
  2. The reading is about 18 µV at 1200 Ω and −1.02 mV at 1201 Ω, so it falls by about 1.04 mV per ohm. Zero lies 0.018 / 1.04 ≈ 0.02 of the way from 1200 Ω to 1201 Ω, at \(R_3\) ≈ 1200.0 Ω. The meter locates the null well inside one step of the box, so the 1 Ω steps limit the result: 1 Ω in 1200 Ω is less than 0.1%. In a real bridge the percentage uncertainties of the three resistors add in \(R_3 \cdot R_2 / R_1\), and the uncertainty in \(R_x\) can reach 0.3%.
  3. Near balance a change \(\Delta R_x\) changes the reading by \(U \cdot R_3 / (R_3 + R_x)^2 \cdot \Delta R_x\): 5 V × 1200 / 2400² × 1 Ω ≈ 1.04 mV, a positive reading. Across the gauge alone the same 1 Ω would change a voltage of about 2.5 V by about 1 mV, and reading it would take a meter that resolves 0.04% of its reading, with a source that does not drift by as much. The bridge subtracts the constant 2.5 V: the change starts from zero, a sensitive range shows it, and a drift of the source voltage does not move the zero.