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 µA:
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.
- 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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