# How do you actually use a Maxwell's bridge to determine an unknown impedance?

Below is depicted a Maxwell bridge:

The resistor $R_2$ and capacitor $C_2$ are variable, while $R_1$ and $R_4$ are fixed and known. The resistance $R_3$ and inductance $L_3$ are unknown.

The circuit has the property that the voltmeter in the middle of the circuit will read $0$ volts iff we have both $R_3=R_1R_4/R_2$ and $L_3=R_1R_4C_2$.

Supposedly we are supposed to adjust the variable capacitance $C_2$ and variable resistance $R_2$ until the voltmeter reads $0$ volts, and then we can use the above formulas to calculate the unknowns $R_3$ and $L_3$.

My question is how is this actually possible in practice, as you are varying two numbers simultaneously. If for example you fix $R_2$ and vary $C_2$, well the voltmeter won't read $0$ volts unless you were lucky enough to have chosen the right value of $R_2$.

First, a remark: that bridge and similar others are no longer used to measure impedances. Modern measurements are routinely performed with vector impedance meters (LCR meters), or, for more accuracy, with specialized bridges like transformer ratio bridges, current comparator bridges and digital bridges. Accurate vector impedance meters, which cost around 10 k\$, can reach accuracies of the order of$10^{-4}\div 10^{-3}$in a wide frequency range (from a few hertz to tens of megahertz). Specialized bridges can reach accuracies of the order of$10^{-8}\div 10^{-5}$in the audio frequency range. As you have rightly guessed, it's not easy to balance AC bridges. This for two reasons: 1. Contrary to DC bridges, in AC bridges you have to adjust two elements instead of one. This, of course, should be expected: for an AC bridge to be balanced, the voltages across two adjacent branches (e.g., branch 1 and 3 in your schematic) should be equal both in magnitude and phase. 2. The actions of the variable elements$R_2$and$C_2$are not orthogonal, and this makes it hard to guess how to change the elements, if the balance is performed manually. In general, to balance AC bridges there are two possible strategies that depend on the type of detector (voltmeter) available. 1. If the detector is a scalar voltmeter (i.e., a voltmeter capable of measuring only the magnitude of an AC voltage) and the adjustment is performed manually, one can adjust the two variable components iteratively, searching for the minimum magnitude of the equilibrium voltage at each iteration. For example, in your case, you can start adjusting$C_2$and you adjust it until you reach a minimum of the detector reading; then you adjust$R_2$until you reach a lower minimum; then you adjust again$C_2$and you continue in this way until you reach zero. Of course, this procedure can take a lot of time. 2. If, instead, the voltmeter is a synchronous detector (i.e., a voltmeter capable of measuring both the in- and out-of-phase components of a voltage with respect to a reference), it is possible to implement a root-finding algorithm in the complex plane. This strategy can be automated, and it's the strategy that is commonly employed in modern specialized bridges. For more on modern impedance measurements, you can have a look at: [1] L. Callegaro, Electrical Impedance: Principles, Measurement, and Applications, CRC press, 2013. Disclaimer: The author and I work together. • Interesting. That's exactly what I likened it to in my head: finding a root of a two-variable function, which is very hard compared to the same task for a one-variable function. I guess in the case of the scalar voltmeter, to demonstrate that the "algorithm" actually works, we should verify that the voltmeter-reading$v(C_2,R_2)$only has one local minimum. Otherwise we could get stuck at a local minimum which is not with$v=0$. Commented Jan 11, 2017 at 21:22 • @JoshuaBenabou In old bridges used for primary measurements (accuracy of the order of$10^{-7}\$), the first balance, done manually, could have taken even... one month! Nowadays, with synchronous detectors and digital sources adjustable in magnitude and phase, a bridge balance can be attained in less than 1 minute, with the root-finding algorithm. Commented Jan 11, 2017 at 21:27