Simple series
circuits
Let's start with a series circuit consisting
of three resistors and a single battery:
The first principle to understand about
series circuits is that the amount of current is the same
through any component in the circuit. This is because there
is only one path for electrons to flow in a series circuit,
and because free electrons flow through conductors like
marbles in a tube, the rate of flow (marble speed) at any
point in the circuit (tube) at any specific point in time
must be equal.
From the way that the 9 volt battery is
arranged, we can tell that the electrons in this circuit
will flow in a counter-clockwise direction, from point 4 to
3 to 2 to 1 and back to 4. However, we have one source of
voltage and three resistances. How do we use Ohm's Law here?
An important caveat to Ohm's Law is that all
quantities (voltage, current, resistance, and power) must
relate to each other in terms of the same two points in a
circuit. For instance, with a single-battery,
single-resistor circuit, we could easily calculate any
quantity because they all applied to the same two points in
the circuit:
Since points 1 and 2 are connected together
with wire of negligible resistance, as are points 3 and 4,
we can say that point 1 is electrically common to point 2,
and that point 3 is electrically common to point 4. Since we
know we have 9 volts of electromotive force between points 1
and 4 (directly across the battery), and since point 2 is
common to point 1 and point 3 common to point 4, we must
also have 9 volts between points 2 and 3 (directly across
the resistor). Therefore, we can apply Ohm's Law (I = E/R)
to the current through the resistor, because we know the
voltage (E) across the resistor and the resistance (R) of
that resistor. All terms (E, I, R) apply to the same two
points in the circuit, to that same resistor, so we can use
the Ohm's Law formula with no reservation.
However, in circuits containing more than
one resistor, we must be careful in how we apply Ohm's Law.
In the three-resistor example circuit below, we know that we
have 9 volts between points 1 and 4, which is the amount of
electromotive force trying to push electrons through the
series combination of R1, R2, and R3.
However, we cannot take the value of 9 volts and divide it
by 3k, 10k or 5k Ω to try to find a current value, because
we don't know how much voltage is across any one of those
resistors, individually.
The figure of 9 volts is a total
quantity for the whole circuit, whereas the figures of 3k,
10k, and 5k Ω are individual quantities for
individual resistors. If we were to plug a figure for total
voltage into an Ohm's Law equation with a figure for
individual resistance, the result would not relate
accurately to any quantity in the real circuit.
For R1, Ohm's Law will relate the
amount of voltage across R1 with the current
through R1, given R1's resistance,
3kΩ:
But, since we don't know the voltage across
R1 (only the total voltage supplied by the
battery across the three-resistor series combination) and we
don't know the current through R1, we can't do
any calculations with either formula. The same goes for R2
and R3: we can apply the Ohm's Law equations if
and only if all terms are representative of their respective
quantities between the same two points in the circuit.
So what can we do? We know the voltage of
the source (9 volts) applied across the series combination
of R1, R2, and R3, and we
know the resistances of each resistor, but since those
quantities aren't in the same context, we can't use Ohm's
Law to determine the circuit current. If only we knew what
the total resistance was for the circuit: then we
could calculate total current with our figure for
total voltage (I=E/R).
This brings us to the second principle of
series circuits: the total resistance of any series circuit
is equal to the sum of the individual resistances. This
should make intuitive sense: the more resistors in series
that the electrons must flow through, the more difficult it
will be for those electrons to flow. In the example problem,
we had a 3 kΩ, 10 kΩ, and 5 kΩ resistor in series, giving us
a total resistance of 18 kΩ:
In essence, we've calculated the equivalent
resistance of R1, R2, and R3
combined. Knowing this, we could re-draw the circuit with a
single equivalent resistor representing the series
combination of R1, R2, and R3:
Now we have all the necessary information to
calculate circuit current, because we have the voltage
between points 1 and 4 (9 volts) and the resistance between
points 1 and 4 (18 kΩ):
Knowing that current is equal through all
components of a series circuit (and we just determined the
current through the battery), we can go back to our original
circuit schematic and note the current through each
component:
Now that we know the amount of current
through each resistor, we can use Ohm's Law to determine the
voltage drop across each one (applying Ohm's Law in its
proper context):
Notice the voltage drops across each
resistor, and how the sum of the voltage drops (1.5 + 5 +
2.5) is equal to the battery (supply) voltage: 9 volts. This
is the third principle of series circuits: that the supply
voltage is equal to the sum of the individual voltage drops.
However, the method we just used to analyze
this simple series circuit can be streamlined for better
understanding. By using a table to list all voltages,
currents, and resistances in the circuit, it becomes very
easy to see which of those quantities can be properly
related in any Ohm's Law equation:
The rule with such a table is to apply Ohm's
Law only to the values within each vertical column. For
instance, ER1 only with IR1 and R1;
ER2 only with IR2 and R2;
etc. You begin your analysis by filling in those elements of
the table that are given to you from the beginning:
As you can see from the arrangement of the
data, we can't apply the 9 volts of ET (total
voltage) to any of the resistances (R1, R2,
or R3) in any Ohm's Law formula because they're
in different columns. The 9 volts of battery voltage is
not applied directly across R1, R2,
or R3. However, we can use our "rules" of series
circuits to fill in blank spots on a horizontal row. In this
case, we can use the series rule of resistances to determine
a total resistance from the sum of individual
resistances:
Now, with a value for total resistance
inserted into the rightmost ("Total") column, we can apply
Ohm's Law of I=E/R to total voltage and total resistance to
arrive at a total current of 500 �A:
Then, knowing that the current is shared
equally by all components of a series circuit (another
"rule" of series circuits), we can fill in the currents for
each resistor from the current figure just calculated:
Finally, we can use Ohm's Law to determine
the voltage drop across each resistor, one column at a time:
Just for fun, we can use a computer to
analyze this very same circuit automatically. It will be a
good way to verify our calculations and also become more
familiar with computer analysis. First, we have to describe
the circuit to the computer in a format recognizable by the
software. The SPICE program we'll be using requires that all
electrically unique points in a circuit be numbered, and
component placement is understood by which of those numbered
points, or "nodes," they share. For clarity, I numbered the
four corners of our example circuit 1 through 4. SPICE,
however, demands that there be a node zero somewhere in the
circuit, so I'll re-draw the circuit, changing the numbering
scheme slightly:
All I've done here is re-numbered the
lower-left corner of the circuit 0 instead of 4. Now, I can
enter several lines of text into a computer file describing
the circuit in terms SPICE will understand, complete with a
couple of extra lines of code directing the program to
display voltage and current data for our viewing pleasure.
This computer file is known as the netlist in SPICE
terminology:
series circuit
v1 1 0
r1 1 2 3k
r2 2 3 10k
r3 3 0 5k
.dc v1 9 9 1
.print dc v(1,2) v(2,3) v(3,0)
.end
Now, all I have to do is run the SPICE
program to process the netlist and output the results:
v1 v(1,2) v(2,3) v(3) i(v1)
9.000E+00 1.500E+00 5.000E+00 2.500E+00 -5.000E-04
This printout is telling us the battery
voltage is 9 volts, and the voltage drops across R1,
R2, and R3 are 1.5 volts, 5 volts, and
2.5 volts, respectively. Voltage drops across any component
in SPICE are referenced by the node numbers the component
lies between, so v(1,2) is referencing the voltage between
nodes 1 and 2 in the circuit, which are the points between
which R1 is located. The order of node numbers is
important: when SPICE outputs a figure for v(1,2), it
regards the polarity the same way as if we were holding a
voltmeter with the red test lead on node 1 and the black
test lead on node 2.
We also have a display showing current
(albeit with a negative value) at 0.5 milliamps, or 500
microamps. So our mathematical analysis has been vindicated
by the computer. This figure appears as a negative number in
the SPICE analysis, due to a quirk in the way SPICE handles
current calculations.
In summary, a series circuit is defined as
having only one path for electrons to flow. From this
definition, three rules of series circuits follow: all
components share the same current; resistances add to equal
a larger, total resistance; and voltage drops add to equal a
larger, total voltage. All of these rules find root in the
definition of a series circuit. If you understand that
definition fully, then the rules are nothing more than
footnotes to the definition.
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REVIEW:
-
Components in a series circuit share the
same current: ITotal = I1 = I2
= . . . In
-
Total resistance in a series circuit is
equal to the sum of the individual resistances: RTotal
= R1 + R2 + . . . Rn
-
Total voltage in a series circuit is equal
to the sum of the individual voltage drops: ETotal
= E1 + E2 + . . . En
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