Suppose the water level has just covered wire (a) in
Figure 1. In the circuit, this means (a) will be at 9V and
the circuit will behave as if everything to the right of the
vertical resistor connected to (a) isn’t even present (since
b and c aren’t connected to anything). If you think at least
the 1K resistors should still be in the circuit, you are sort of
right, but we’ll say no to this for now.
The voltmeter attached to Vout draws almost zero
current, so virtually no current will flow through the 1Ks,
and by Ohm’s Law again, the voltage drop across them
will be V = IR. However, with I = 0 (or nearly so), the
voltage drop across either is zero. Thus, the circuit will
resemble Figure 4.
Looking at it carefully, you’ll see that it is similar to the
voltage divider with equal resistors. Thus, Vout will read 4. 5
volts, and we have our first water level to voltage
mapping. In fact, if you read 4. 5 volts, the water level is
low — you might want to add some water to your
reservoir. This is something about voltage dividers worth
remembering: If they are built with two equal resistors, the
voltage at the connection point of the resistors will be half
of the supply voltage.
What if the water is covering both wires (a) and (b)?
The circuit will now look like Figure 5. So, how can we
figure Vout now? Well, notice in Figure 5 that (a) and (b)
are now both connected to 9V (through the water), so
why not rearrange the circuit a bit? We’ll tie (a) and (b)
together; imagine them connected to a single battery
which we’ll draw toward the left, while being careful that
ground is still on the left end of the 2K resistor. The result
is in Figure 6.
Now, again assuming that the voltmeter connected to
Vout draws no current, we see that the 2K and 1K on the
right are in series. We can combine them into a single 3K
resistor, and our circuit becomes Figure 7 (A). Now, we
see the 2K and 3K are in parallel, so they can be
combined into a single 1.2K resistor as in Figure 7 (B).
Lastly, the 1.2K and 2K are in series, and can be combined
into a single 3.2K resistor as in Figure 7 (C).
Why go through all of this? Because the 3.2K resistor
represents the total resistance the R-2R ladder presents to
the battery when both wires (a) and (b) are submerged.
From this, we can find the total current the 9V battery is
supplying to the circuit. From Figure 7 (C), it’ll be (from I
= V/R) I = 9V/3200Ω = 0.0028 A. Going back to Figure 7
28 November 2015
■ FIGURE 6. The circuit in Figure 5, rearranged to
make analyzing it a bit easier.
■ FIGURE 7. (A) is Figure 6 with the rightmost 2K and
1K combined into a single 3K resistor; (B) is circuit (A)
redrawn with the parallel 3K and 2K resistors
combined into a single 1.2K resistor; (C) is circuit (B)
redrawn with the series 1.2K and 2K resistors
combined into a single 3.2K resistor.
■ FIGURE 4. The R-2R ladder from Figure 3, as it
would behave if only wire (a) was connected to 9V.
■ FIGURE 5. The R-2R ladder from Figure 3, as it would
behave if wires (a) and (b) were connected to 9V.