Previously in Lesson 4, it was
mentioned that there are two different ways to connect two or more electrical
devices together in a circuit. They can be connected by means of series
connections or by means of parallel connections. When all the devices in a
circuit are connected by series connections, then the circuit is referred to as
a series circuit. When all the devices in a circuit are
connected by parallel connections, then the circuit is referred to as a parallel
circuit. A third type of circuit involves the dual use
of series and parallel connections in a circuit; such circuits are referred to
as compound circuits or combination circuits. The circuit depicted at the right
is an example of the use of both series and parallel connections within the
same circuit. In this case, light bulbs A and B are connected by parallel
connections and light bulbs C and D are connected by series connections. This
is an example of a combination circuit.
When analyzing combination
circuits, it is critically important to have a solid understanding of the
concepts that pertain to both series circuits and parallel
circuits. Since both types of connections are used in
combination circuits, the concepts associated with both types of circuits apply
to the respective parts of the circuit. The main concepts associated with
series and parallel circuits are organized in the table below.
Series Circuits · The current is the same in every
resistor; this current is equal to that in the battery. · The sum of the voltage drops across the
individual resistors is equal to the voltage rating of the battery. · The overall resistance of the
collection of resistors is equal to the sum of the individual resistance
values, Rtot = R1 + R2 + R3 + ... |
Parallel Circuits · The voltage drop is the same across
each parallel branch. · The sum of the current in each
individual branch is equal to the current outside the branches. · The equivalent or overall resistance of
the collection of resistors is given by the equation 1/Req =
1/R1 + 1/R2 + 1/R3 ... |
Each of the above concepts has a mathematical expression.
Combining the mathematical expressions of the above concepts with the Ohm's law
equation (ΔV = I • R) allows one to conduct a complete analysis of a
combination circuit.
The basic strategy for the analysis of combination circuits
involves using the meaning of equivalent resistance for parallel branches to
transform the combination circuit into a series circuit. Once transformed into
a series circuit, the analysis can be conducted in the usual manner. Previously
in Lesson 4, the method for determining the equivalent
resistance of parallel are equal, then the total or equivalent resistance of
those branches is equal to the resistance of one branch divided by the number
of branches.
This method is consistent with the formula
1 / Req = 1 / R1 + 1 / R2 + 1 / R3 + ...
where R1,
R2, and R3 are the resistance values of the individual
resistors that are connected in parallel. If the two or more resistors found in
the parallel branches do not have equal resistance, then the above formula must
be used. An example of this method was presented in a previous
section of Lesson 4.
By applying one's understanding of the equivalent resistance
of parallel branches to a combination circuit, the combination circuit can be
transformed into a series circuit. Then an understanding of the equivalent
resistance of a series circuit can be used to determine the total resistance of
the circuit. Consider the following diagrams below. Diagram A represents a
combination circuit with resistors R2 and R3 placed in
parallel branches. Two 4-Ω resistors in parallel is equivalent to a
resistance of 2 Ω. Thus, the two branches can be replaced by a single
resistor with a resistance of 2 Ω. This is shown in Diagram B. Now
that all resistors are in series, the formula for the total resistance of
series resistors can be used to determine the total resistance of this circuit:
The formula for series resistance is
Rtot = R1 + R2 + R3 + ...
So in Diagram B, the total resistance of the circuit is
10 Ω.
Once the total resistance of the circuit is determined, the
analysis continues using Ohm's law and voltage and resistance values to
determine current values at various locations. The entire method is illustrated
below with two examples.
The first example is the easiest case - the resistors placed
in parallel have the same resistance. The goal of the analysis is to determine
the current in and the voltage drop across each resistor.
As discussed above, the first step is to simplify the circuit
by replacing the two parallel resistors with a single resistor that has an
equivalent resistance. Two 8 Ω resistors in series is equivalent
to a single 4 Ω resistor. Thus, the two branch resistors (R2 and R3)
can be replaced by a single resistor with a resistance of 4 Ω. This
4 Ω resistor is in series with R1 and R4.
Thus, the total resistance is
Rtot = R1 + 4 Ω + R4 = 5 Ω +
4 Ω + 6 Ω
Rtot = 15 Ω
Now the Ohm's law equation (ΔV = I • R) can be used to
determine the total current in the circuit. In doing so, the total resistance
and the total voltage (or battery voltage) will have to be used.
Itot = ΔVtot / Rtot = (60 V) / (15 Ω)
Itot = 4 Amp
The 4 Amp current calculation represents the current at the
battery location. Yet, resistors R1 and R4 are in
series and the current in series-connected resistors is everywhere the same.
Thus,
Itot = I1 = I4 = 4 Amp
For parallel branches, the sum of the current in each
individual branch is equal to the current outside the branches. Thus, I2 + I3 must equal
4 Amp. There are an infinite number of possible values of I2 and I3 that
satisfy this equation. Since the resistance values are equal, the current
values in these two resistors are also equal. Therefore, the current in
resistors 2 and 3 are both equal to 2 Amp.
I2 =
I3 = 2 Amp
Now that the current at each individual resistor location is
known, the Ohm's law equation (ΔV = I • R) can be used to determine the
voltage drop across each resistor. These calculations are shown below.
ΔV1 = I1 • R1 = (4 Amp) • (5 Ω)
ΔV1 = 20 V
ΔV2 = I2 • R2 = (2 Amp)
• (8 Ω)
ΔV2 = 16 V
ΔV3 = I3 • R3 = (2 Amp)
• (8 Ω)
ΔV3 = 16 V
ΔV4 = I4 • R4 = (4 Amp)
• (6 Ω)
ΔV4 = 24 V
The analysis is now complete and the results are summarized
in the diagram below.
The second example is the more difficult case - the resistors
placed in parallel have a different resistance value. The goal of the analysis
is the same - to determine the current in and the voltage drop across each
resistor.
As discussed above, the first step is to simplify the circuit
by replacing the two parallel resistors with a single resistor with an
equivalent resistance. The equivalent resistance of a 4-Ω and
12-Ω resistor placed in parallel can be determined using the usual
formula for equivalent resistance of parallel branches:
1 / Req = 1 / R1 + 1 / R2 + 1 / R3 ...
1 / Req = 1 /
(4 Ω) + 1 / (12 Ω)
1 / Req =
0.333 Ω-1
Req = 1 /
(0.333 Ω-1)
Req =
3.00 Ω
Based on this calculation, it can be said that the two branch
resistors (R2 and R3) can be replaced by a single
resistor with a resistance of 3 Ω. This 3 Ω resistor
is in series with R1 and R4. Thus, the total resistance is
Rtot = R1 + 3 Ω + R4 = 5 Ω +
3 Ω + 8 Ω
Rtot = 16 Ω
Now the Ohm's law equation (ΔV = I • R) can be used to
determine the total current in the circuit. In doing so, the total resistance
and the total voltage (or battery voltage) will have to be used.
Itot = ΔVtot / Rtot = (24 V) / (16 Ω)
Itot = 1.5 Amp
The 1.5 Amp current calculation represents the current at the
battery location. Yet, resistors R1 and R4 are in
series and the current in series-connected resistors is everywhere the same.
Thus,
Itot = I1 = I4 = 1.5 Amp
For parallel branches, the sum of the current in each
individual branch is equal to the current outside the branches. Thus, I2 + I3 must equal
1.5 Amp. There are an infinite possibilities of I2 and I3 values
that satisfy this equation. In the previous example, the two resistors in
parallel had the identical resistance; thus the current was distributed equally
among the two branches. In this example, the unequal current in the two
resistors complicates the analysis. The branch with the least resistance will
have the greatest current. Determining the amount of current will demand that
we use the Ohm's law equation. But to use it, the voltage drop across the
branches must first be known. So the direction that the solution takes in this
example will be slightly different than that of the simpler case illustrated in
the previous example.
To determine the voltage drop across the parallel branches,
the voltage drop across the two series-connected resistors (R1 and R4)
must first be determined. The Ohm's law equation (ΔV = I • R) can be used
to determine the voltage drop across each resistor. These calculations are
shown below.
ΔV1 = I1 • R1 = (1.5 Amp) • (5 Ω)
ΔV1 = 7.5 V
ΔV4 = I4 • R4 = (1.5
Amp) • (8 Ω)
ΔV4 = 12 V
This circuit is powered by a 24-volt source. Thus, the
cumulative voltage drop of a charge traversing a loop about the circuit is 24
volts. There will be a 19.5 V drop (7.5 V + 12 V) resulting from passage
through the two series-connected resistors (R1 and R4).
The voltage drop across the branches must be 4.5 volts to make up the
difference between the 24 volt total and the 19.5-volt drop across R1 and R4.
Thus,
ΔV2 = V3 = 4.5 V
Knowing the voltage drop across the parallel-connected
resistors (R1 and R4) allows one to use the Ohm's
law equation (ΔV = I • R) to determine the current in the two branches.
I2 = ΔV2 / R2 = (4.5 V) / (4 Ω)
I2 = 1.125 A
I3 = ΔV3 / R3 = (4.5 V)
/ (12 Ω)
I3 = 0.375 A
The analysis is now complete and the results are summarized
in the diagram below.
The two examples above illustrate an effective concept-centered strategy for analyzing combination
circuits. The approach demanded a firm grasp of the series and parallel
concepts discussed earlier. Such analyses are often conducted in order to
solve a physics problem for a specified unknown. In such situations, the
unknown typically varies from problem to problem. In one problem, the resistor
values may be given and the current in all the branches are the unknown. In
another problem, the current in the battery and a few resistor values may be
stated and the unknown quantity becomes the resistance of one of the resistors.
Different problem situations will obviously require slight alterations in the
approaches. Nonetheless, every problem-solving approach will utilize the same
principles utilized in approaching the two example problems above.
The following suggestions for approaching combination circuit
problems are offered to the beginning student:
· If a
schematic diagram is not provided, take the time to construct one. Use schematic symbols such as those shown in the
example above.
· When
approaching a problem involving a combination circuit, take the time to
organize yourself, writing down known values and equating them with a symbol
such as Itot, I1, R3,
ΔV2, etc. The organization scheme used in the two examples
above is an effective starting point.
· Know and
use the appropriate formulae for the equivalent resistance of series-connected
and parallel-connected resistors. Use of the wrong formulae will guarantee
failure.
· Transform
a combination circuit into a strictly series circuit by replacing (in your
mind) the parallel section with a single resistor having a resistance value
equal to the equivalent resistance of the parallel section.
· Use the
Ohm's law equation (ΔV = I • R) often and appropriately. Most answers will
be determined using this equation. When using it, it is important to substitute
the appropriate values into the equation. For instance, if calculating I2,
it is important to substitute the ΔV2 and the R2 values into the equation.
For further practice analyzing combination
circuits, consider analyzing the problems
in the Check Your Understanding section below
1. A combination circuit is shown in the diagram at the right. Use the
diagram to answer the following questions.
a. The current at location A is _____ (greater than, equal
to, less than) the current at location B.
b. The current at location B is _____ (greater than, equal
to, less than) the current at location E.
c. The current at location G is _____ (greater than, equal
to, less than) the current at location F.
d. The current at location E is _____ (greater than, equal
to, less than) the current at location G.
e. The current at location B is _____ (greater than, equal
to, less than) the current at location F.
f. The current at location A is _____ (greater than, equal
to, less than) the current at location L.
g. The current at location H is _____ (greater than, equal
to, less than) the current at location I.
The current outside the branches of a combination circuit is
everywhere the same. The current inside of the branches is always less than
that outside of the branches. When comparing the current of two
parallel-connected resistors, the resistor with the least resistance will have
the greatest current. The current within a single branch will be the same above and below the
resistor.
a. The current at location
A is equal
to the
current at location B.
b. The current at location B is greater than the current at
location E.
c. The current at location G is less than the current at
location F.
d. The current at location E is greater than the current at
location G.
e. The current at location B is greater than the current at
location F.
f. The current at location A is equal to the current at
location L.
g. The current at location H is less than the current at
location I.
2. Consider the combination circuit in the diagram at the right. Use the
diagram to answer the following questions. (Assume that the voltage drops in
the wires themselves in negligibly small.)
a. The electric potential difference (voltage drop) between
points B and C is _____ (greater than, equal to, less than) the electric
potential difference (voltage drop) between points J and K.
b. The electric potential difference (voltage drop) between
points B and K is _____ (greater than, equal to, less than) the electric
potential difference (voltage drop) between points D and I.
c. The electric potential difference (voltage drop) between points
E and F is _____ (greater than, equal to, less than) the electric potential
difference (voltage drop) between points G and H.
d. The electric potential difference (voltage drop) between
points E and F is _____ (greater than, equal to, less than) the electric
potential difference (voltage drop) between points D and I.
e. The electric potential difference (voltage drop) between
points J and K is _____ (greater than, equal to, less than) the electric
potential difference (voltage drop) between points D and I.
f. The electric potential difference between points L and A
is _____ (greater than, equal to, less than) the electric potential difference
(voltage drop) between points B and K.
The voltage drop across a resistor is dependent upon the current
in the resistor and the resistance of the resistor. In situations in which the
current is the same for both resistors (such as for series-connected
resistors), the resistor with the greatest resistance will have the greatest
voltage drop.
a. The electric potential
difference (voltage drop) between points B and C is greater than the electric potential
difference (voltage drop) between points J and K.
b. The electric potential difference (voltage drop) between points
B and K is greater than the electric potential difference (voltage drop) between
points D and I.
c. The electric potential difference (voltage drop) between points
E and F is equal to the electric potential difference (voltage drop) between
points G and H.
d. The electric potential difference (voltage drop) between points
E and F is equal to the electric potential difference (voltage drop) between
points D and I.
e. The electric potential difference (voltage drop) between points
J and K is greater than the electric potential difference (voltage drop) between
points D and I.
f. The electric
potential difference between points L and A is equal to the electric potential
difference (voltage drop) between points B and K.
close
3. Use the concept of equivalent resistance to determine the
unknown resistance of the identified resistor that would make the circuits equivalent.
For parallel-connected resistors:
1/Req = 1/R1 + 1/R2 =
1 / (6 )
+ 1 / (6 )
= 2 / (6 )
Req = 3
For series-connected resistors:
Req = R1 + R2 + R3 = 3 + 3 + 5
Req = 11
For parallel-connected resistors:
1/Req = 1/R1 + 1/R2 =
1 / (12 )
+ 1 / (6 )
= 3 / (12 )
Req = 4
For series-connected resistors:
Req = R1 + R2 + R3 = 9 + 4 + 5
Req = 18
For parallel-connected resistors:
1/Req = 1/R1 + 1/R2 =
1 / (12 Ω) + 1 / (6 Ω) = 3 / (12 Ω)
Req = 4 Ω
For series-connected resistors:
Req = R1 +
R2 + R3 + R4 =
3 Ω + 6 Ω + 4 Ω + 5 Ω
Req = 18 Ω
4. Analyze the following
circuit and determine the values of the total resistance, total current, and
the current at and voltage drops across each individual resistor.
The first step is to simplify the circuit by replacing the two
parallel resistors with a single resistor with an equivalent resistance. The
equivalent resistance of a 4 Ω and 6 Ω resistor
placed in parallel can be determined using the usual formula for equivalent
resistance of parallel branches:
1 / Req = 1 / R1 +
1 / R2 + 1 / R3 ...
1 / Req = 1 / (3
Ω) + 1 / (6 Ω)
1 / Req =
0.500 Ω-1
Req = 1 /
(0.500 Ω-1)
Req = 2.00 Ω
Based on this calculation, it can be said that the two branch
resistors (R2 and R3) can be replaced by a single
resistor with a resistance of 2 Ω. This 2 Ω resistor
is in series with R1 and R4. Thus, the total
resistance is
Rtot = R1 +
2 Ω + R4 = 2 Ω +
2 Ω + 4 Ω
Rtot = 8 Ω
Now the Ohm's law equation (ΔV = I • R) can be used to
determine the total current in the circuit. In doing so, the total resistance
and the total voltage (or battery voltage) will have to be used.
Itot = ΔVtot / Rtot =
(24 V) / (8 )
Itot = 3.0 Amp
The 3.0 Amp current calculation represents the current at the
battery location. Yet, resistors R1 and R4 are
in series and the current in series-connected resistors is everywhere the same.
Thus,
Itot = I1 =
I4 = 3.0 Amp
For parallel branches, the sum of the current in each individual
branch is equal to the current outside the branches. Thus, I2 +
I3 must equal 3.0 Amp. There are an infinite possibilities of I2 and
I3 values which satisfy this equation. Determining the amount
of current in either branch will demand that we use the Ohm's law equation. But
to use it, the voltage drop across the branches must first be known. To
determine the voltage drop across the parallel branches, the voltage drop
across the two series-connected resistors (R1 and R4)
must first be determined. The Ohm's law equation (ΔV = I • R) can be used
to determine the voltage drop across each resistor. These calculations are
shown below.
ΔV1 =
I1 • R1 = (3.0 Amp) • (2 )
= 6.0 V
Δ
V4 =
I4 • R4 = (3.0 Amp) • (4 )
= 12 V
This circuit is powered by a 24-volt source. Thus, the
cumulative voltage drop of a charge traversing a loop about the circuit is 24
volts. There will be a 18.0 V drop (6.0 V +
12.0 V) resulting from passage through the two series-connected resistors (R1 and
R4). The voltage drop across the branches must be 6.0 volts to make
up the difference between the 24 volt total and the 18.0 volt drop across R1 and
R4. Knowing the voltage drop across the parallel-connected resistors
(R1 and R4) allows one to use the Ohm's law equation
(ΔV = I • R) to determine the current in the two branches.
I2 = ΔV2 /
R2 = (6.0 V) / (3 )
= 2.0 A
I3 = ΔV3 /
R3 = (6.0 V) / (6 )
= 1.0 A
5. Referring to the diagram in question #4, determine the ...
a.
... power rating of resistor 4.
b. ... rate at which energy is consumed by resistor 3.
The power rating of a resistor can be determined from any one of
the following formulae:
P = I • ΔV |
P = I2 • R |
P = ΔV 2 /
R |
Since all three variables (I, ΔV and R) are known,
either one of the equations can be used to calculate power.
a. P4 =
I4 • ΔV4 = (3 A) • (12 V) = 36.0 Watt
b. P3 =
I3 • ΔV3 = (1 A) • (6 V) = 6.0 Watt