**SAN LUIS OBISPO, Calif.** Last month, we used Thevenin
equivalents to analyze a simple resistive T network. One of the more
common applications of Thevenin equivalents is in the analysis of an
unbalanced Wheatstone Bridge.

The * Wheatstone bridge * is named after Sir Charles
Wheatstone (1802-1875), an English physicist and inventor.
The Wheatstone bridge was first described by Samuel
Hunter Christie (1784-1865) in his paper * Experimental Determination
of the Laws of Magneto-electric Induction* (1833). The circuit was
brought into general use by Wheatstone in 1843. Samuel Christie was the
son of James Christie, founder of the well-known auction galleries
(see *Dictionary of Scientific Biography*; Charles Coulston Gillispie,
editor). The
bridge was (and is) used chiefly to measure resistance. Substituting
resistances and reactances for the reference resistance in the bridge,
and driving the bridge with AC, allows it to be used to measure impedance.

Figure 1 shows a simple * balanced * Wheatstone bridge with
a voltmeter as a null detector. Let's determine the voltage indicated by
the voltmeter.

The voltage on the left side of the voltmeter (marked R for the red
or positive lead) with respect to ground can be determined using the
voltage divider formula as

The voltage on the right side of the voltmeter (marked B for the
black, negative, or *reference* lead) with respect to ground can also
be determined using the voltage divider formula as

V_{RB} is the voltage a point R with respect to point B.
Further, V_{RB} = V_{R} - V_{B}. So, in this case,
V_{RB} = 0 volts. We say the bridge is * balanced.*

We can see that the bridge is balanced if V_{R} =
V_{B}. We can generalize this as

We now see that the ratio of the resistances in each "arm" of the bridge determines whether the bridge is balanced. Further, if the bridge is balanced, we can substitute an ammeter (ideally zero resistance) for the voltmeter and read zero current. If we have the same voltage at two points in a circuit and connect them together (through the ammeter), no current will flow.

Note that the voltage is * negative. * If we were to reverse
the voltmeter leads, the voltage would be positive. However, as long as
we specify the voltage as V_{RB}, the voltage is negative, and
polarity * is * important.

Although it is quite possible (and appears standard) to bring the bridge down to a single Thevenin equivalent, I prefer to make two equivalent circuits, one for the left half of the bridge and one for the right half. This approach allows us to determine all the circuit node voltages from the equivalent circuit while the typical "single equivalent" approach only yields the "load current" (the current through R5).

In figure 4, we've split V_{S} into two equivalent sources.
At this point, we * could * determine the circuit parameters (V_{A
} and V_{B} would be most useful) using superposition. Give
it a try! Apply the voltage sources one at a time, substituting a short for
the voltage source that is out of circuit. Determine V_{A} and
V_{B} when each of the sources is present (V_{A} * due to
* the left source and * due to * the right source), then add the
results to yield the total V_{A} and total V_{B}. It
should match what we come up with.

Figure 5 shows a the circuit of figure 4 with Thevenin equivalents for each "half" of the circuit. V12 is the Thevenin voltage formed by the left 10 volt source, R2, and R1. It is the voltage measured at the junction of R1 and R2 (point A) with R5 disconnected. Since this then becomes a simple voltage divider, V12 is (R2/(R1+R2))*10V, or 5 volts.

R12 is the Thevenin resistance of the left half of the circuit. This is determined by shorting out all voltage sources and measuring the resistance between point A and ground, with R5 removed. The shorted voltage source results in R1 being in parallel with R2, so the Thevenin resistance is 1K//1K or 500 ohms.

Similarly, V34 is the Thevenin voltage formed by the right 10 volt source, R4, and R3. It is the voltage measured at the junction of R3 and R4 (point B) with R5 disconnected. Since this also becomes a simple voltage divider, V34 is (R4/(R3+R4))*10V, or 6.667 volts.

In figure 5, these Thevenin equivalents are substituted for the
original voltage sources and R1, R2, R3, and R4. R5 is carried to the
equivalent circuit. The unbalanced loaded bridge has become a simple
series circuit that can be analyzed using "differential Ohm's law". The
current through the resistors is (V_{tail}-V_{tip})/R, or
(6.667V-5V)/4.833K = 344.8uA. This current can then be multiplied by
the resistance of each resistor to get the voltage * across * that
resistor. Note the marked voltage polarities and the current direction.
I think these are * very * important! With conventional current,
the current flows * downhill * (from higher voltage to lower), so
that is the way the arrow was drawn. If the arrow had been drawn the
other way, we would get a negative current. A negative current going to
the right is equivalent to a positive current going to the left. Further,
with a positive conventional current, the voltage * across * a resistor
will be positive on the end the current enters and negative on the end
the current leaves.

We can verify our calculations by using Kirchoff's Voltage Law
(the sum of the voltage rises or drops around any closed loop is zero).
If we add the voltage rises, starting at the bottom of the left battery,
we find we have 5V + 172.4mV + 1.035V + 459.8mV - 6.667V = 200uV, which
is * close * to zero. The only reason it's not * exactly *
zero volts is due to rounding errors in our calculations. Close enough!

To determine the voltage at any point in the circuit, we can start
at a point where we know the voltage (with respect to ground) and "wind"
our way through the circuit accumulating voltage rises and drops. For
example, to determine the voltage a point A, we can start at ground on
the left side of figure 5. Ground is zero volts. We "go up" 5 volts
as we go through V12, since we are "coming out" the positive side of V12
as we work our way towards point A. We go up an additional 172.4 mV as
we go through R12, since we are also coming out the positive side of R12.
This makes V_{A} 5.1724 volts.

We could have determined V_{A} starting below the right
voltage source. In that case, we start at zero volts below the source,
go up 6.667 volts going through V34, go down 459.8mV as we go through R34
(since we are coming out the negative end), then go down an additional
1.035 volts as we go through R5. In this case,
V_{A} = 6.667V - 459.8mV - 1.035V = 5.1722 volts. The minor
difference is, again, due to rounding.

We have a couple ways of finding V_{B}. Let's just say
it's 1.035 volts above V_{A} or 6.207 volts. These voltages are
substituted back in the original circuit in figure 6. Once these voltages
are known, all node voltages in the circuit are known. Differential voltages
can be easily determined by subtracting
(V_{AB}=V_{A}-V_{B}). Current through any component
can be determined by "differential Ohm's law". By making these calculations,
we can "analyze the circuit to death". As practice, try showing that the
sum of the currents * into * the node at the junction of R1, R2, and R5
is zero (Kirchoff's current law). I get 4.828mA + (-5.172mA) + 345uA = 1uA.
Pretty close!

Next time we'll look at Norton equivalents, followed by Millman's Theorem. Ten extra points to anyone who sends me biographical information on Norton or Millman.