the current that goes through both
resistors by calculating the effective
resistance. Since the 100 ohm
resistor and the one million ohm
resistor are in series, we can simply
add them to determine what the
effective resistance is to the circuit.
The same current will flow through
1,000,000 + 100 = 1,000, 100
Note that this is still very near
one million ohms.
The current would then be:
I = V / R = 10/1,000, 100 =
That’s pretty small. It’s a lot
smaller than 0.1A we had before the
Since the same small current
flows through both resistors, we can
calculate how much voltage drops
across each resistor AND how much
power will be dissipated by each
resistor. Yes, the meter will dissipate
power, just hopefully very little:
V100 = I x R = 0.000009999 x
100 = 0.0009999 volts
Power100 = V x I = 0.000000009998 watts
The voltage across the 100 ohm resistor is now
0.0009999 volts. Not going to zap
V1000000 = I x R = 9.999 volts
Power1000000 = I x R =
The voltage across the one
million ohm resistor is now 9.999
volts. So, because the meter
impedance is so high, the voltage
across it is essentially the voltage
source. We can also see here that a
high resistance can result in the
higher voltage. Sounds weird but it’s
Also note that the total power
used by both resistors is far less than
that used by the single 100 ohm
resistor in the original circuit.
Resistance goes up, voltage goes up,
but power goes way down. Again,
weird but true.
In reality, no voltmeter has an
infinite impedance input. However,
most of them are pretty high. Even a
relatively inexpensive meter (the kind
that are sometimes given free by
certain stores) may have an input
impedance of a million ohms — far
higher than most devices in circuits
we use every day.
Just in case there are some non-believers, here are
some photos of me performing the experiment to show
In Figure 5, I’ve constructed the circuit placing my
May/June 2018 55
Got Two, Get Two Free
With Ohm's Law and the Power Equation, we have
two formulas that provide an insignt as to what is
happening with electrical systems — at least DC
Ohm's Law can be expressed in three ways:
V = I x R, I = V/R, and R = V/I
where V = voltage, R = resistance, and I = current.
If we know any algebra, we can see that each of
these equations can give us the other two simply by
solving for the value that we do not have.
The Power Equation is:
P = V x I
Wherever there is both voltage and current, there
will be power going somewhere.
Using these two equations, we can get any two
values given the other two. For example, assume that
we know the voltage across and resistance of a device.
The device may be anything. Knowing these two values,
we can determine the other two:
I = V/R
Placing the voltage value for V and the resistance
value for R, we have determined I (the current).
Knowing both the current and voltage, we can
determine the power dissipated by the “thing:”
P = V x I, = V x V/R = SQR(V)/R
Learn at least a bit of algebra. Then, these equations
become less memorization and better understood.
FIGURE 5. Measuring across a break
with 1,000 ohms.