Making sense of test circuits with Kirchhoff’s laws: part 4

We can use a Wheatstone bridge voltage measurement to determine an unknown resistance value.

In part 3 of this series, we used Kirchhoff’s voltage law to derive the branch currents and node voltages for an unbalanced Wheatstone bridge with five known, fixed resistors (Figure 1). Now, we propose to replace R5 with a digital multimeter (DMM) to directly measure V_V – V_X and use that measurement to determine the value of an unknown resistor in the position of R₄.

Circuit shows Kirchoff's laws Whatstone bridge — Figure 1. We can adapt this circuit to use the *V_V* – *V_X* voltage to determine the resistance of an unknown resistor in the position of R₄.

Q: Could we backtrack first? Last time, we solved three loop equations. I was trying to use Kirchhoff’s current law to write six-node equations and got lost. Could you elaborate?

How far did you get?
I_IN = I₁ + I₂, I₁ = I₃ + I₅, I2 = I₄ + I₅, I₃ = I₁ – I₅…

Stop there. Your first three are good, but the last one is just a restatement of the second one. We need an independent equation for I₃.

We know I₃ = V_V/R₃, but we don’t know V_V.
Right, but we can express V_V in terms of one of our original six unknowns, namely I₁. V_V is V_IN minus the voltage drop across R₁, which is I₁ times R₁. Substituting our component values, we can write:

Similarly, we can derive V_X as a function of I₂ and solve for I₄:

And finally, I₅ is V_V minus V_X divided by R₅:

Table 1 shows all six node equations in the format required for the online solver^[1] I’m using, and the solutions are shown in red. With this approach, the solver generates the branch currents directly; we don’t have to derive them from the loop currents. I’ve also included our results from part 3 based on the loop method, with the loop values in blue, and the results agree within a third of a percent.

Table 1. Node equations and solutions.

Node equations		Solved values, node method (mA)	Solved values, loop method (mA)	Difference (%)
Eq.1:	1I_IN – 1I₁ – 1I₂ + 0I₃ + 0I₄ + 0I₅ = 0	I_IN = 0.39331	I_IN = 0.39253	-0.19871
Eq.2:	0I_IN + 1I₁ + 0I₂ – 1I₃ + 0I₄ – 1I₅ = 0	I₁ = 0.14166	I₁ = 0.14136	-0.21222
Eq.3:	0I_IN + 0I₁ + 1I₂ + 0I₃ – 1I₄ + 1I₅ = 0	I₂ = 0.25165	I₂ = 0.25117	-0.19111
Eq.4:	0I_IN + 1.42I₁ + 0I₂ + 1I₃ + 0I₄ + 0I₅ = 0.455	I₃ = 0.25384	I₃ = 0.25320	-0.25276
Eq.5:	0I_IN + 0I₁ + 0.324I₂ + 0I₃ +1 I₄ + 0I₅ = 0.221	I₄ = 0.13947	I₄ = 0.13933	-0.10048
Eq.6:	0I_IN + 4.7 em>I₁ – 2.2I₂ + 0I₃ + 0I₄ + 1I₅=0	I₅ = -0.11218	I₅ = -0.11184	-0.30401

Now, let’s look at the situation where we have an unknown resistor but a known voltage.
Right. Keep in mind that in the 19th century, galvanometers were good at indicating zero current, but not good at quantifying nonzero levels. Consequently, accurate measurements were made under zero-current conditions. We no longer face that limitation, and we can use a measurement setup in Figure 2, which shows a way to make strain-gauge measurements. Here, a DMM, data logger, or data-acquisition instrument replaces the galvanometer from part 2, and R₅ from Figure 1 becomes essentially infinite. R_X, which takes the place of R₄, is a strain-gauge element that has a resistance of 120 Ω when not subjected to strain. To optimize measurement resolution, we assign resistors R₁, R₂, and R₃ the same 120-Ω value.

Bridge circuit with DMM — Figure 2. Given *V_DMM*, we can determine *R_X*.

So, we basically have two voltage dividers.
Right:

and

We can now write an equation for what our DMM would read for a resistance value R_X:

Now we can solve for R_X:

Figure 3 plots this transfer function. Note that for strain-gauge measurements, resistance changes are often less than 1%, and sometimes much less.

Voltage across bridge circuit Kirchoff — Figure 3. Given *V_DMM*, we can determine *R_X*.

Figure 4 shows a simulation result.

Simulation of bridge circuit — Figure 4. A 5-mV DMM reading corresponds to an *R_X* value of 118.4106 Ω.

Q: Why use the bridge—why not just measure the resistance of R_X directly?
A: Temperature compensation is one reason. In Figure 2, R₂ is often a strain-gauge element identical to R_X maintained at the ambient temperature of R_X, but that never undergoes strain. If both R₂ and R_X increase by 0.1% because of a rise in ambient temperature, but R_X continues to be unstrained, the DMM will continue to read zero.

Q: What is strain anyway, and what are its units?
A: Good question. That’s beyond the scope of this series on Kirchhoff’s laws, but we can consider strain measurement in depth in a future series.