Making sense of test circuits with Kirchhoff’s laws: part 3

Kirchhoff’s voltage law gives us three equations with three unknowns to solve for loop currents in an unbalanced Wheatstone bridge.

We concluded part 2 of this series by starting to write node and loop equations for a five-resistor Wheatstone bridge circuit, to calculate the currents through each resistor. Figure 1 repeats Figure 3 from part 2, but I’ve assigned component values so we can calculate a numerical result. Note that I’ve retained the V_V and V_X labels, which denote the nodes connecting to the variable and unknown resistors in our version of the bridge with a galvanometer.

Kirchoff current voltage resistor — Figure 1. Red arrows represent branch currents, and blue arrows represent current loops around which voltage drops must sum to zero.

Q: How do we know which directions of the currents, for example, through R₅?
A: We don’t. The current direction through R₅ will depend on the resistor values. If our arrow direction is “wrong,” we will get a negative value of I₅.

Q: So, how do we solve this circuit?
A: We’ll need six equations to compute the six unknown branch currents (red arrows) or three equations to compute the three unknown loop currents. The latter approach looks easier, so we’ll start there, beginning with the loop on the left. Starting at the top and moving clockwise, we note that the current through R₁ is I_A–I_B, so the drop across R1 is (I_A–I_B)R₁. Next, the current through R₃ is I_A–I_C, so R₃ contributes a drop of (I_A–I_C)R₃. Finally, continuing clockwise, we start with our source, which, from our perspective, adds a negative voltage drop. So, we can write our first loop equation summing the voltage drops:

We continue this process for the top right and bottom right loops to get the additional loop equations:

We will probably want to use a program like MATLAB or an online tool to solve these equations, so we’ll put them in the following form, where each variable is preceded by a constant and the right-hand term is a constant:

With some algebra, we come up with these versions of the three equations:

Next, we insert the numerical values:

I used an online solver^[1] to get the results shown in Table 1. Since I entered values in kilohms, the results are in milliamps.

Table 1. Kirchhoff Loop equations and solutions:

Equation		Solved values
Eq 1	8X + -4.7Y + -3.3Z = 1.5	I_A = 0.39253 mA
Eq 2	-4.7X + 7.9Y + -1Z = 0	I_B = 0.25117 mA
Eq 3	-3.3X + -1Y + 11.1Z = 0	I_C = 0.13933 mA

Q: How do we use these results to get our branch currents and node voltages?
A: We quickly notice that I_IN equals I_A, or 0.393 mA, and we can calculate the equivalent impedance of our resistor network as 1.5 V divided by 0.393 mA, or about 3.82 kΩ. We also see that I₄ equals I_C, which equals 0.139 mA, so V_X equals 0.139 mA times 6.8 kΩ, or 947 mV. Similarly, I₂ equals I_B, which equals 0.251 mA, and the drop across R₂ equals 0.251 mA times 2.2 kΩ, or 553 mV.

For the other resistor currents, we’ll need to subtract loop currents. The R₅ current is I_C minus I_B, or -0.112 mA, so the actual current flow is in the opposite direction from the I₅ arrow, and the voltage across R₅ is -112 mV. We can continue with R₁ (for which I₁=I_A–I_B) and R₃ (I₃=I_A–I_C), with the complete results shown in Figure 2.

resistor voltage current — Figure 2. Solving our loop equations enables us to calculate node voltage and branch currents.

Q: Now we’ll have to build this circuit to see if the calculations are correct.
A: My thought was to simulate it in LTspice^[2] as shown in Figure 3. The traces on the left for V_V and V_X are a bit difficult to see, so I have added some notation. The simulation confirms our calculated result.

Figure 3. A simulation confirms our calculated results for *V_V* and *V_X*.

And then I breadboarded the circuit with 1% resistors and confirmed the equivalent resistance, as shown in Figure 4, and the node voltages.

digital multiumeter breadboard Kirchoff resistor — Figure 4. The equivalent resistance of our breadboarded bridge circuit is 3.84 kΩ, within 1% of our calculated value.

Q: Why so many resistors?
A: I didn’t seem to have a 6.8-kΩ version in my collection, so, alluding back to the rat’s nest in part 1, I had to build one out of parts I did have.

Q: What do we look at next?
A: We previously commented on the downside of using a balanced bridge and galvanometer to determine the value of an unknown resistor (in the R₄ position). In part 4, we’ll look at using an unbalanced bridge and a high-impedance voltmeter to make that measurement, a technique useful in strain-gauge and other sensor applications.