Wheatstone bridge & strain gauges: practice problems with step-by-step solutions

Free Wheatstone bridge & strain gauges practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: balance and node voltages, unbalanced bridges, strain gauges and load cells.

Free practice problemEasy
+−9 VR1​ = 200 ΩR2​ = 300 ΩR3​ = 500 ΩR4​ = 200 Ωab+−Vab​ = ?

Find the open-circuit output voltage Vab = Va − Vb of the bridge shown, with Vs = 9 V.

Vs = 9 VR1 = 200 ΩR2 = 300 ΩR3 = 500 ΩR4 = 200 Ω
V
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What you practise

  1. Easybalance and node voltages
  2. Mediumunbalanced bridges
  3. Hardstrain gauges and load cells

How to solve Wheatstone bridge & strain gauges problems

  1. Treat the bridge as two voltage dividers across the same supply.
  2. For balance, set R₁/R₂ = R₃/R₄ and solve for the unknown resistor.
  3. For an unbalanced bridge, find Va and Vb from the dividers and subtract; use the Thévenin equivalent when a meter loads the output.
  4. For strain gauges, convert strain to ΔR/R = GF·ε and use Vs·GF·ε/4, /2 or ×1 for a quarter, half or full bridge.

Common mistakes

  • Pairing the wrong resistors in the balance condition.
  • Using the quarter-bridge formula for a half or full bridge.
  • Forgetting that strain is often given in µε (10⁻⁶).
  • Applying the small-change approximation Vs·ΔR/(4R) to a large resistance change.

Key formulas

Divider voltages
Va = VsR2/(R1 + R2), Vb = VsR4/(R3 + R4)
Output
Vab = Va − Vb
Balance
R1/R2 = R3/R4
Slide-wire bridge
R/Rx = l/(100 − l)
Thévenin resistance
Rth = R1∥R2 + R3∥R4
Gauge factor
ΔR/R = GF·ε
Bridge output
quarter Vs·GF·ε/4, half Vs·GF·ε/2, full Vs·GF·ε
Small change, one arm
Vab ≈ Vs·ΔR/(4R)
Stress
σ = E·ε

Learn Wheatstone bridge & strain gauges from scratch

Before this: Series & parallel resistors, Thévenin & Norton

1What a Wheatstone bridge is

A Wheatstone bridge is two voltage dividers connected to the same supply Vs. The output is the voltage between their midpoints a and b.

+−Vs​R1​R2​R3​R4​ab+−

When the two dividers have the same ratio, a and b sit at the same voltage and the output is zero: the bridge is balanced. A tiny change in one arm then gives a small output that is easy to measure, without the large common voltage on top.

That is why bridges are used to measure unknown resistances and the small resistance changes of sensors such as strain gauges and thermistors.

Self-check All four arms of a bridge are 100 Ω and Vs = 10 V. What is the output Vab?
  1. 0 V
  2. 10 V
  3. 2.5 V
  4. 5 V

Answer: 0 V. Both dividers halve the supply, so Va = Vb = 5 V and Vab = 0. 5 V is the voltage of each midpoint, not the difference.

2Node voltages and the balance condition

Each midpoint is a voltage divider. With R₁ on top of R₂ and R₃ on top of R₄:

Va = Vs·R2/(R1 + R2) Vb = Vs·R4/(R3 + R4) Vab = Va − Vb

The output is zero when the ratios match:

R1/R2 = R3/R4 or R1·R4 = R2·R3

The supply voltage does not appear in the balance condition. So an unknown resistor can be found by adjusting a known one until a galvanometer reads zero.

Self-check A bridge balances with R₁ = 100 Ω, R₂ = 200 Ω and R₃ = 150 Ω. What is R₄?
  1. 133.3 Ω
  2. 300 Ω
  3. 450 Ω
  4. 75 Ω

Answer: 300 Ω. R₄ = R₂·R₃/R₁ = 200·150/100 = 300 Ω. Swapping R₁ and R₂ gives 75 Ω, and pairing the wrong resistors gives 133.3 Ω.

3The method step by step

  1. Label the arms: R₁ and R₂ in the left divider, R₃ and R₄ in the right one, with node a between R₁ and R₂ and node b between R₃ and R₄.
  2. For a balanced bridge, solve R₁·R₄ = R₂·R₃ for the unknown.
  3. For an unbalanced bridge, find Va and Vb from the dividers and subtract.
  4. If a meter with resistance Rg loads the output, replace the bridge by its Thévenin equivalent: Vth = Vab and Rth = R₁∥R₂ + R₃∥R₄. Then Ig = Vth/(Rth + Rg).

4A worked example

Follow every step. Try to predict the next one before you open it.

Worked example

A bridge has R₁ = 200 Ω, R₂ = 300 Ω, R₃ = 400 Ω and R₄ = 400 Ω, with Vs = 12 V. Find the open-circuit output Vab, and the current in an 80 Ω galvanometer connected from a to b.

1

Look at the bridge

+−12 V200 Ω300 Ω400 Ω400 Ωab+−Vab​

The right divider has equal arms, so Vb is half the supply.

2

Node voltages

Va = 12·300/500 = 7.2 V Vb = 12·400/800 = 6 V
3

Output voltage

Vab = 7.2 − 6 = 1.2 V
4

Thévenin resistance

With the supply shorted, R₁ ∥ R₂ runs from a to the rails and R₃ ∥ R₄ from the rails to b.

Rth = 200∥300 + 400∥400 = 120 + 200 = 320 Ω
5

Galvanometer current

Ig = 1.2/(320 + 80) = 0.003 A = 3 mA

It flows from a to b, because Va is higher.

Answer

Vab = 1.2 V and Ig = 3 mA.

5Strain gauges

A strain gauge is a thin metal foil glued to a part. When the part stretches, the foil gets longer and thinner and its resistance rises:

ΔR/R = GF·ε

ε is the strain (often in microstrain, µε = 10⁻⁶) and GF the gauge factor, about 2 for metal foil. The change is tiny: 500 µε on a 350 Ω gauge with GF = 2 gives ΔR = 0.35 Ω.

In a bridge with all arms equal to R, the output is close to:

  • Quarter bridge (one active gauge): Vo ≈ Vs·GF·ε/4.
  • Half bridge (two gauges with opposite strain in one divider): Vo = Vs·GF·ε/2.
  • Full bridge (four active gauges): Vo = Vs·GF·ε.

With Vs = 5 V, GF = 2 and 500 µε, a quarter bridge gives 5·2·0.0005/4 = 1.25 mV. Stress follows from Hooke’s law, σ = E·ε.

Self-check The same gauges (GF = 2, 500 µε, Vs = 5 V) are wired as a half bridge. What is the output?
  1. 0.625 mV
  2. 1.25 mV
  3. 2.5 mV
  4. 5 mV

Answer: 2.5 mV. A half bridge gives Vs·GF·ε/2 = 5·2·0.0005/2 = 2.5 mV, twice the quarter-bridge 1.25 mV. 5 mV would need four active gauges.

6Common mistakes

  • Pairing the wrong arms in the balance condition. Compare the two upper arms with the two lower arms.
  • Using the quarter-bridge factor 4 for a half or full bridge.
  • Forgetting that µε means 10⁻⁶.
  • Using the small-change estimate Vs·ΔR/(4R) for a large resistance change; then work out both dividers exactly.
Self-check A gauge with GF = 2 shows ΔR/R = 0.002. What is the strain?
  1. 0.001 µε
  2. 1000 µε
  3. 2 µε
  4. 4000 µε

Answer: 1000 µε. ε = (ΔR/R)/GF = 0.002/2 = 0.001 = 1000 µε. Writing 0.001 with the unit µε is a factor of a million too small, and multiplying by GF gives 4000 µε.

7How exam questions look

  • Easy: balance and node voltages. You find the unknown or adjustable arm of a balanced bridge, a node voltage, the open-circuit output, the supply current, or the unknown resistor of a slide-wire bridge.
  • Medium: unbalanced bridges. You find the exact or linearised output for a resistance change, the Thévenin resistance, the galvanometer current, the resistance change from a measured output, or the power in a gauge.
  • Hard: strain gauges and load cells. You find strain from ΔR, the output of a quarter, half or full bridge, strain or stress from a reading, the amplifier gain for a converter input, or the output of a load cell rated in mV/V.

For a slide-wire bridge the two wire segments play the role of the ratio arms: R/Rx = l/(100 − l).

Learn it step by step in the appLearn it step by step in the app

Worked examples

Example 1Easy
+−6 VR1​ = 250 ΩR2​ = 300 ΩR3​ = ?R4​ = 150 ΩabG0

The Wheatstone bridge in the diagram is balanced by adjusting R₃. R₁ = 250 Ω, R₂ = 300 Ω and R₄ = 150 Ω. Find the setting R3.

R1 = 250 ΩR2 = 300 ΩR4 = 150 Ω

Find R3 (Ω)

Show hints
  1. At balance both dividers give the same voltage: Va = Vb.
  2. R₁/R₂ = R₃/R₄.
  3. R₃ = R₁·R₄/R₂.
Show the answer and the step-by-step solution

Wheatstone bridge: two voltage dividers share one supply Vs. The left divider R1–R2 gives Va = Vs·R2/(R1 + R2) and the right divider R3–R4 gives Vb = Vs·R4/(R3 + R4); the output is Vab = Va − Vb. The bridge is balanced (Vab = 0, no current in the meter) when R1/R2 = R3/R4, that is R1·R4 = R2·R3. Seen from the output, the bridge is a Thévenin source with Rth = R1∥R2 + R3∥R4.

Balance condition

R₁·R₄ = R₂·R₃

Unknown resistor

R₃ = 250·150/300 = 125 Ω

Answer

R3 = 125 Ω

The supply voltage does not matter at balance.

Example 2Medium
+−10 VR1​ = 150 ΩR2​ = 1000 ΩR3​ = 500 ΩR4​ = 600 ΩabGRg​ = 20 Ω

A galvanometer with a resistance of 20 Ω is connected between a and b of the bridge shown (Vs = 10 V). Find the galvanometer current Ig from a to b, in µA.

Vs = 10 VR1 = 150 ΩR2 = 1000 ΩR3 = 500 ΩR4 = 600 ΩRg = 20 Ω

Find Ig (µA)

Show hints
  1. Replace the bridge by its Thévenin equivalent seen from a–b.
  2. Vth = Va − Vb = 3.241 V, Rth = 403.2 Ω.
  3. Ig = Vth/(Rth + Rg).
Show the answer and the step-by-step solution

Wheatstone bridge: two voltage dividers share one supply Vs. The left divider R1–R2 gives Va = Vs·R2/(R1 + R2) and the right divider R3–R4 gives Vb = Vs·R4/(R3 + R4); the output is Vab = Va − Vb. The bridge is balanced (Vab = 0, no current in the meter) when R1/R2 = R3/R4, that is R1·R4 = R2·R3. Seen from the output, the bridge is a Thévenin source with Rth = R1∥R2 + R3∥R4.

Thévenin voltage

Vth = 10·(1000/1150 − 600/1100) = 3.241 V

Thévenin resistance

Rth = R₁∥R₂ + R₃∥R₄ = 403.2 Ω

Galvanometer current

Ig = 3.241/(403.2 + 20) = 7659 µA

Answer

Ig = 7659 µA
Example 3Hard
+−V350 Ω350 Ω + ΔR350 Ω350 Ωab+−

A 350 Ω strain gauge with gauge factor 2.12 is bonded to a steel bar. Under load its resistance rises by 0.371 Ω. Find the strain ε in microstrain (µε).

R = 350 ΩGF = 2.12ΔR = 0.371 Ω

Find ε (µε)

Show hints
  1. ΔR/R = GF·ε.
  2. ΔR/R = 0.371/350 = 0.00106.
  3. 1 µε = 10⁻⁶.
Show the answer and the step-by-step solution

Strain gauges: a gauge changes its resistance with strain: ΔR/R = GF·ε, where the gauge factor GF is about 2 for metal foil. In a bridge with all arms equal to R, one active gauge (quarter bridge) gives Vo ≈ Vs·GF·ε/4. Two active gauges with opposite strains in adjacent arms (half bridge) give Vs·GF·ε/2, and four (full bridge) give Vs·GF·ε. Stress follows from Hooke’s law, σ = E·ε. Load cells state their sensitivity in mV/V: the output at full load per volt of excitation.

Relative resistance change

ΔR/R = 0.371/350 = 0.00106

Strain

ε = 0.00106/2.12 = 0.0005 = 500 µε

Answer

ε = 500 µε

Questions and answers

What is Wheatstone bridge & strain gauges?

A Wheatstone bridge is two voltage dividers connected to the same supply Vs. The output is the voltage between their midpoints a and b.

When the two dividers have the same ratio, a and b sit at the same voltage and the output is zero: the bridge is balanced. A tiny change in one arm then gives a small output that is easy to measure, without the large common voltage on top.

How do I solve Wheatstone bridge & strain gauges problems step by step?
  1. Treat the bridge as two voltage dividers across the same supply.
  2. For balance, set R₁/R₂ = R₃/R₄ and solve for the unknown resistor.
  3. For an unbalanced bridge, find Va and Vb from the dividers and subtract; use the Thévenin equivalent when a meter loads the output.
  4. For strain gauges, convert strain to ΔR/R = GF·ε and use Vs·GF·ε/4, /2 or ×1 for a quarter, half or full bridge.
What are common mistakes in Wheatstone bridge & strain gauges?
  • Pairing the wrong resistors in the balance condition.
  • Using the quarter-bridge formula for a half or full bridge.
  • Forgetting that strain is often given in µε (10⁻⁶).
  • Applying the small-change approximation Vs·ΔR/(4R) to a large resistance change.
What do the Wheatstone bridge & strain gauges exercises cover?
  • Easy: balance and node voltages
  • Medium: unbalanced bridges
  • Hard: strain gauges and load cells

Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.

How practising works

  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Your progress, mistakes and saved problems are kept, so you can review before the exam.

Topics before and after this one

Learn first

Series & parallel resistorsThévenin & Norton

Courses with this topic

Electronics 1Measurement and InstrumentationMechatronic Systems

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Printable worksheet

Wheatstone bridge & strain gauges worksheet with answers
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