Making sense of test circuits with Kirchhoff’s laws: part 2

Figure 1. This circuit contains no series or parallel combinations of resistors.

In part 1 of this series, we looked at ways to simplify resistor networks by identifying series and parallel combinations of resistors. We closed with a look at a version of the Wheatstone bridge, such as the one in Figure 1. Although it has only five resistors, not one of them is in series or parallel with another, so the simple equations we used in part 1 won’t work here. There is a workaround, called a delta-to-wye conversion.^[1] But that just gives you three more equations to memorize, and unless you do the conversion often (if you work with three-phase power, for example), those equations aren’t likely to become second nature, the way the series and parallel resistor equations are for most electrical engineers.

Kirchoff's law Wheatstone Bridge — Figure 2. For a balanced bridge, the galvanometer reads 0.

Q: Wait, what about the version of the Wheatstone bridge with a galvanometer taking the place of R₅?
A: That version, shown in Figure 2, is easier to analyze. Here, R₃ becomes a variable resistor R_V, and R₄ becomes an unknown resistor R_X, whose value we are trying to find.

In the classic Wheatstone bridge application, we manually adjust R_V until the galvanometer or multimeter voltage V_G is 0, at which point we have what we call a balanced bridge. It consists of two simple voltage dividers, and R5 from Figure 1 is essentially infinite. R₁ and R_V make up the first divider, and the voltage across R_V is:

R₂ and R_X make up the second divider, and the voltage across R_X is:

Since the galvanometer reads zero, we can determine R_X as follows:

Q: What’s the drawback to this balanced bridge?
A: If we are trying to measure a parameter such as strain in real time, it’s not practical to manually adjust R_v until V_G is zero, then record the value of R_V, and calculate R_X. We could probably kludge together something based on a voltage-controlled or digitally controlled resistor to replace the manual adjustment, but Kirchhoff’s and Ohm’s laws offer a simpler way for us to derive R_X from the voltages and currents in a fixed-resistor bridge.

voltage current loops — Figure 3. The red arrows represent the current flowing in each circuit branch, while the blue arrows represent current loops around which voltage drops must sum to zero.

Q: How does that work?
A: Let’s go back to our original Figure 1 bridge with five fixed resistors. In Figure 3, I’ve rearranged Figure 1 to make more room for notations showing branch (red arrows) and loop (blue arrows) currents. Kirchhoff gives us two laws that we can apply to solving such a problem. The current law states that the net sum of currents into any node is 0. In Figure 3, for example, that law implies the following:

The second law, Kirchhoff’s voltage law, states that the voltage drops around any closed loop must equal zero. For the loop on the left of Figure 3, this law results in the following equation:

For a given circuit, each law will give us multiple equations on multiple unknowns. Which law or combination of the two laws is easiest for us to use will depend on exactly what we are trying to determine. In our case, we have two goals. First, following up on our work from part 1, what is the equivalent impedance of the bridge if all resistors are fixed and we know their individual values? In other words, what is V_IN/I_IN? And second, of particular interest from a test-and measurement-perspective, is this question: If R₁, R₂, R₃, and R₅ are fixed and known and R₄ is unknown (it might be a strain-gauge element or temperature sensor, for instance), can we calculate R₄ based on the voltage V_V – V_X across R₅? We’ll take a closer look next time.