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DC Circuits Part 3

Combination Circuits We have learned about the series circuit and the parallel circuit. Both circuits can be examined by the use of Ohm’s Law. Let’s now combine the two circuits to build a more complex circuit. This type of circuit will be called a series/parallel circuit or a combination circuit. Ohm’s Law can also be used to examine this type of circuit. A series/parallel circuit is a circuit in which the electrical components are connected in series and parallel groups.

In the electrical industry you hardly ever run across a circuit that is only a series or a parallel circuit. Most all circuits are combination circuits that contain both series elements and parallel elements. The illustration above is a simple representation of a combination circuit. Many such circuits found in the industry are quite complex and contain many amounts of both series and parallel elements. Even with all of the complexity of these circuits the basic laws still apply; i.e. Kirchhoff’s Laws and Ohm’s Law. You will need a sharp eye to decipher the elements contained in complex circuits.

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Notice that in the circuit below, the total current of the circuit will pass through the series section of the circuit and the current will branch out in the parallel portion of the circuit according to the resistances within the parallel section.

The voltage drop across the components in a combination circuit act according across each group of components as if they were in an individual series or parallel circuits. That is the voltages drop across each resistor in the series portion of a combination circuit and the voltage to the common points of the parallel portion of a combination circuit is the same at each common point. Let’s look at a simple combination circuit to visualize this concept.

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In the previous lessons on series and parallel we solved for total resistance of the circuit or an equivalent resistance. When dealing with combination circuits, in order to solve for unknown values, an equivalent circuit can be derived. The Equivalent circuit is a circuit that has been simplified from many components down to only one equivalent component and the source. In the series circuit we simply added the values of the resistances which then equated the sum to one single large valued resistor as an equivalent circuit. In the parallel circuit we had several methods to equate the equivalent resistance depending upon the values of the resistor branches. In the combination circuit to, find the equivalent circuit; you will be using the practices taught for both series and parallel circuits. Again we will use the same simple combination circuit so we may solve for the following values; (1) total resistance, (2) total current, (3) current through the parallel branches, (4) voltage drop across each series component and (5) total power including power dissipated across each component.

There are different methods in solving for the unknown values, but with practice you will be able to choose the most logical methods. We have all of the resistances labeled therefore we can proceed to simplify the circuit to achieve an equivalent or total resistance. We will tackle the parallel section first. We learned several methods to solve for parallel resistors in parallel, but the most logical method for two unequal resistors is the product over the sum.

3 4

3 4

R R =R + R50×25 =

50 + 251250 16.=

75667Ω

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The circuit can now be simplified by replacing the parallel components with an equivalent resistance of 16.667Ω. For ease of understanding, a new schematic will be used.

As the circuit suggests, the parallel section is replaced with an equivalent resistance of 16.667Ω. The circuit is now simplified to just a series circuit in which the three resistances need only to be summed.

1 2 equivalent tR + R + R = R

100Ω + 75Ω + 16.66 191.7Ω = 667Ω Total resistance.

The circuit can now be illustrated with the following diagram.

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We can now solve for the total current within this circuit by using the known total values and Ohm’s law.

or

EI = R

50VI = 191.667Ω

I = 0.2608A 260.8mA

Total current.

The current through the series portion is equal to the total current of the circuit or 260.8mA. Once the current enters the parallel section of the circuit, the current will branch to the two branches according to the resistances within the branches. The total power of this circuit can also be easily ascertained by one of the PIER circle equations.

t

t

P = IEP = 0.2608A × 50VP = 13.04W

Total Power.

The next logical step is to derive the voltage drops across the series resistors and the equivalent resistor.

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The voltage drop across the series components can easily be found by using Ohm’s law. The voltage drop across the parallel components will equal the voltage drop across the equivalent resistor. Since the equivalent resistor is also in series, the same method of deriving the voltage drop can be made.

1 t 1

1

1

V = I RV = 0.2608A × 100ΩV = 26.08V

Voltage across R1

2 t 2

2

2

V = I RV = 0.2608A × 75ΩV = 19.56V

Voltage across R2

The remaining voltage should be the voltage across the parallel components. We can derive the drop by adding the series voltage drops and subtracting the sum from the total voltage (Kirchhoff’s Law) or we can again use Ohms Law.

eq t eq

eq

eq

V = I R

V = 0.2608A × 16.667Ω

V = 4.35V

Voltage across the parallel

Or

( ) ( ) ( )( )

t 1 2 eq

eq

eq

eq or

V + V + V + V = 0

+50V + -26.08V + -19.56V + V = 0

V = -50V + 26.08V + 19.56V

V = -4.36V 4.36V

Using Voltage Kirchhoff’s Law

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Now that the series voltage drops are known, the power dissipated across the series resistors can now easily be determined

We will again refer to the PIER circle to obtain the proper equation to solve for the power dissipation across the series resistors. (The PIER circle is only a tool. Try to memorize as many of the formulas in the circle and remember how to derive the remaining equations from the ones you have stored in your memory. You may not have the luxury of having a copy of the circle when you are working in the field.)

1 t 1

1

1

P = I EP = 0.2608A × 26.08VP = 6.8W

2 t 2

2

2

P = I EP = 0.2608A × 19.56VP = 5.1W

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Power dissipated across R1 & R2 in Watts

We can now look at the parallel components with the data that we have derived from its equivalent circuit. We basically now have a voltage that is common to both branches within the parallel circuit.

By using this Voltage the currents and power can now be derived for these branches. We will use Ohm’s law to find the current in Branch R3.

eq3

3

3

3 or

EI =

R4.35VI = 50Ω

I = 0.087A 87mA

Current of Branch R3

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To get you thinking, let’s explore another way to derive the current in Branch R4. We would normally choose the same method as we did for Branch R3 but this is just a lesson to learn from. Let’s look at the current to resistance ratios. The relationships are inversely proportional.

3 4

4 3

4

4

4 or

R I = R I50Ω I = 25Ω .087A

50 × .087I = 25

I = 0.174A 174mA

Current of Branch R4

We must now prove that the currents are correct. Any ideas on how to do this? Let’s use Kirchhoff’s current law which states: The sum of the currents entering and leaving a point (node) in a circuit equals zero. The mathematical equation for this law looks like this; 1 2 3 ......... 0nI I I I+ + + = Using the currents derived from our equations the results should be:

Now to solve for our last piece of the puzzle is to find the power dissipated across the remaining two resistors that are connected parallel in this circuit. We will use an equation derived from Ohm’s Law.

3 3 eq

3

3 or

P = I E

P = 0.087A × 4.35VP = 0.378W 378mW

Power of branch R3

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It is time again for another experiment. Normally we would solve for the R4 branch using the same method as we did for branch R3. Let’s refer to the PIER circle and grab another equation to solve for power.

2eq

44

2

4

4

4 or

EP =

R

4.35VP = 25Ω

18.923P = 25

P = 0.7 57W 757mW

Power of branch R4

It’s time again for the proof. The sum of the power dissipated across the components equals the total power dissipated in the circuit. Earlier we determined that the power dissipated in the circuit was 13.1 Watts. This was derived from the source voltage times the total current of the circuit. Now all we have to do is to add up the powers found dissipated across each component that we have previously solved for.

1 2 3 4 t

t

t

P + P + P + P = P6.8W + 5.1W + 0.378W + 0.757W =13.035

PW

= P

Due to rounding of numbers, the two totals are not exact yet are basically the same number. Please complete the exercises contained in Practice 13 – Combination Circuit Analysis –

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So far we have worked on a simple combination circuit. More complex circuits may consist of several sections of series and parallel components. In fact there may be series components within one of the branches of a parallel circuit or two or more parallel branches in series. Let’s look at some different circuits and locate the series or parallel sections within the circuits.

As one can see, there are infinite arrangements that a circuit may have. You must be able to identify the different series and parallel sections within a circuit so you can then logically

determine the variables of that circuit. Let’s look at the first circuit and define its sections. As you can see, resistors 1 and 6 can be easily spotted since they are the only connected resistors to the source; that is they are connected in series with the source. Behind resistor 1 the conductor is a common connection to three more resistors. The conductor is highlighted in orange. The common connection defines a parallel section of components connected to the voltage source. Resistor 5 is located in one of the parallel

branches of the circuit yet it is connected in series with another resistor of that branch. You must

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treat the relationship of these two resistors as a series circuit within the larger circuit. The two resistances must be summed to obtain an equivalent series resistance which then becomes the value of one of the parallel resistances. The next circuit is the most complicated. A proper sequence of steps must be met to ensure that the values of the circuit are properly determined.

When you first observe this circuit the large parallel section jumps out at you. For those who can’t see it, the blue circle encompasses it. Within the parallel section of this circuit one branch has three series resistors which can be summed to an equivalent resistance and the other branch has a second parallel section within its branch which must be determined then summed with another resistor (purple) in series within that branch. Once the equivalent resistances of the two branches of the large parallel section are determined (blue circle), the whole section can be summed to an equivalent resistance. There was another parallel section of the circuit located in the lower left corner of the circuit. This section will be independently determined to obtain an equivalent resistance. Once this is done, the circuit is nothing more than a series circuit, two resistors in red and two equivalent resistances, one defining the region within the blue circle and one defined by the red circle in the lower left corner of the circuit. This series circuit can then be easily determined.

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The third example seems to be quite busy but upon further inspection you will find a rather simple circuit. There is a pattern of three resistors connected in parallel connected to another section of three resistors in parallel. This pattern of connections repeats four more times for a total of six parallel groups of resistors connected in series. The six parallel sections are individually determined and the six equivalent quantities are then handled as a series circuit containing six resistors.

The last circuit is a circuit that contains two large strings (series) of resistors connected in parallel. That is, there are only two parallel branches connected to a voltage source. You only have to sum the series connected resistors to obtain an equivalent branch resistance for each branch. The circuit can then be easily solved as a simple parallel circuit. Once you get the hang of it, you will be able to identify the different sections within a circuit and then determine the logical approach to solving for the values that need to be determined within the circuit. Yes, this means practice.

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Now that we can identify the relations components have within a circuit, let’s dive in and begin to solve for specific quantities from given values within a more complicated circuit.

I won’t be able to run through all of the combinations possible with this circuit but we will look at a few switch options to see the affects to the circuit. With all switches open, the circuit consists of only R1 in series with the 130 volt source. With all of the switches closed, a series parallel portion of the circuit is connected in parallel to the resistor R1. Switch 1 can be considered the master switch because no matter what condition the other switches are, switch 1 will ensure that the series parallel portion of the circuit can be disconnected from the source voltage. Switch 2 has a special purpose. If the switch is closed, R2 is bypassed (or shorted) so it is essentially out of the circuit. This switching method removes components from series circuits without turning the circuit completely off. Switch 3 or switch 4 can individually remove either R3 or R4 from the circuit. We will now find the (1) total current, (2) the total resistance, (3) the total power, (4) the current thru the branches, (5) the voltage drops across the resistors, (6) the power dissipated across each resistor for the following switch conditions. Switch 1 open With switch 1 open the only resistor in play is R1. The circuit is a simple series resistor across a 130 volt source. The analysis is as follows: E source 130 volts given R total 2500Ω given

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Current of circuit is;

t

t or

EI = R

130VI = 2500Ω

I = 0.052A 52mA

Total Current

The total power is also the power across the resistor which is:

P = IEP = 0.052 × 130P = 6.76W

Total Power

What’s interesting will be the sudden change of these totals as we manipulate the switches. Switches 1, 3, 4, closed; 2 open Step 1: let’s simplify the parallel section of the circuit. This contains R3 and R4 in parallel. The parallel contains only two branches so we can use the product over the sum method.

3 4eq

3 4

eq

eq

eq

R RR = R + R25 × 125R = 25 + 1253125R = 150

20.R = 833Ω

Step 2: Sum the series resistances of R2 and Req. This equals the total equivalent resistance of the branch.

branch eq 2

branch

branch

R = R + R

R = 20.833Ω + 1170

50.

ΩR = 833Ω

Step 3: Sum R1 and Rbranch as two parallel branches to obtain total resistance of circuit. Since we have used the product over the sum method for parallel resistance, let’s do the reciprocal of the sum of the reciprocals to obtain the parallel equivalent resistance which in this case is the total resistance of the circuit.

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t

1 branch

t

t

t

t

1R = 1 1 + R R

1R = 1 1 + 2500Ω 170.833Ω

1R = 0.0004 + 0.00585366

1R = 0.00625366

159R = .91Ω

Total Resistance

Step 4: The total circuit current can now be determined by Ohm’s Law.

sourcet

t

t

t or

EI = R130VI =

159.91ΩI = 0.8129A 812 .9mA

Total Current

Step 5: Derive the total power of the circuit.

t

t

t

P = IEP = 0.8129A × 130VP = 105.677W

Total power

Let’s stop at this point to analyze these two totals with the first totals when all of the switches were open. All switches open Switch 1,3,4 closed; 2 open Total current 52mA 812.9mA Total power 6.76W 105.68W Total resistance 2500Ω 159.9Ω If you look at both current and power, they both increased by a factor of 15.63. On the other hand the resistance has decreased by a factor of 15.63. So is it true that resistance is inversely proportional to both current and power?

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Of course. Now let’s continue and find the variables of the components in this circuit. Remember that we had a parallel circuit containing R1 and Rbranch. This is where we will begin and we will determine current across these branches. Step 6: Find current of R1 by using Ohm’s Law.

11

1

1 or

EI = R130VI = 2500Ω

I = 0.052A 52mA

Resistance of branch R1

Do notice that this has not changed from the first switch condition when all switches were open! Step 7: Find current of Rbranch by using Ohm’s Law.

branchbranch

branch

branch or

EI = R

130VI = 170.833Ω

I = 0.7609A 760 .9mA

Current of branch Rbranch

To check our work we can now add the two current and they should total the calculated total current as found above. 0.052 + 0.7609 = 0.8129 or 812.9mA………”Ain’t” we good!! Step 8: Find power dissipated across R2. Since R2 is in series in the Rbranch circuit, the total current of the branch will flow through R2. The power dissipated across R2 can be determined as follows:

22 2 2

22

2

P = I R

P = 0.7609 × 150P = 86.84W

Power dissipated across R2

Step 9: Determine voltage drops of R2 and Req. R2 is in series with the parallel components of R3 and R4. An equivalent resistance has been determined to replace the parallel section for computations. That equivalent resistance is 20.833Ω

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2 2

2

2

V = IRV = 0.7609A × 150ΩV = 114.135V

eq eq

eq

eq

V = IR

V = 0.7609A × 20.833Ω

V = 15.85V

Proof: V2 +Veq = 130V 114.135 + 15.85 = 129.985V basically the identical answer. Since the voltage of Veq is 15.85V, this also represents the common voltage of the parallel circuit composing of R3 and R4. Step 10: Determine the current in the branches of R3 and R4. Ohm’s law can be used for both variables.

d3

3

3

3 or

EI = R15.85VI =

25ΩI = 0.634A 634mA

d4

4

4

4 or

EI = R15.85VI = 125Ω

I = 0.1268A 126.8mA

Let’s again check top see if we are on the right path. The sum of all of the current paths should equal the total current that was calculated above. That was 812.9mA. We do remember Kirchhoff’s current law?

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Step 11: For the final step, we only need to find the power dissipated across the resistors R3 and R4. The wattage will be determined as follows;

3 3 d

3

3

P = I EP = 0.634A × 15.85VP = 10.05W

Power dissipated across R3

4 4 d

4

4

P = I EP = 0.1268A × 15.85VP = 2W

Power dissipated across R4

Did we calculate the correct power dissipation across all of the four resistors? To find proof we must sum all of the power across all of the resistors and this should equal the original total power calculated which is 105.68 watts.

1 2 3 4 tP + P + P + P = P6.76W + 86.84 W + 10.05W 1+ 0 2 5W = .65W

Basically the same.

Now it’s your turn. I want you to calculate for all of the values (1) total current, (2) the total resistance, (3) the total power, (4) the current thru the branches, (5) the voltage drops across the resistors, (6) the power dissipated across each resistor for the following switch condition. Switch 1, 2, 3 and 4 all closed

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One more thought before we go on to the next subject. I want you to compare this simple close schematic with the circuit that we have been working on.

You have just analyzed a Power Circuit Breaker close scheme.

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Circuit Re-draws So far we have analyzed circuit schemes that are offered in a format that can be easily defined so that the variables can be determined. Sometimes we are forced to actually trace out a circuit from the actual wire board to obtain a schematic to work from. These tracings are not quite as organized as a manufacturer’s schematic. Let’s look at some typical circuits that don’t meet our standard layout. I will present the drawing in rough form then we will begin our procedure to simplify the drawing to a workable layout. The eventual task is to determine the equivalent series circuit. Drawing 1

Our first step is to define all of the common connection points within the circuit. We will begin at the negative post of the battery. There is a node that three resistors are connected to. Being an electrician is being in a world of color, so we will color code this node yellow and all of the conductors connected to this node.

The four resistors, 10Ω, 15Ω, 25Ω and 20Ω are all connected together to one common point. We will highlight these connections green. There is a third common connection point located on the three resistors, 5Ω, 25Ω, 20Ω and the positive post of the battery. This common point with all conductors will be color highlighted in pink.

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Our next task is to re-draw the circuit and use the color highlighted connection points as a connector bus. This circuit has three busses.

The schematic now seems to have a far better order of appearance to it. It can now be simplified to the series equivalent circuit remembering that was our original intent for this circuit. There are two parallel circuit sections each containing a pair of resistors. (15Ω, 10Ω), (25Ω, 20Ω) These two circuits can be simplified to two equivalent resistors connected in series. The calculations to obtain the two equivalent resistances were performed using the product over the sum method for simplifying parallel circuits.

eq1

eq1

eq1

15Ω × 10ΩR = 15Ω + 10Ω150R = 25

R = 6Ω

eq2

eq2

eq2

25Ω × 20ΩR = 25Ω + 20Ω500R = 45

R = 11.11Ω

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The two equivalent parallel resistors are now in series and can the simply be added to obtain the series resistance of the circuit section. The value of that resistance is 17.11Ω. The last step to obtain the equivalent series circuit is to sum the two remaining parallel branches. One branch consists of the equivalent resistance of 17.11Ω and the other branch is the remaining 5Ω resistor. We can again apply the product over the sum method to solve this parallel circuit.

t

t

t

17.11Ω × 5ΩR = 17.11Ω + 5Ω85.55R = 22.13

1R = .87Ω

Let’s try another circuit but this time we will only simplify it to the common busses. Drawing 2

As you can see, some of these rough circuit drawings can be quite busy. We will now define this circuit by highlighting the common connections associated with the conductors.

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The pink highlight represents a common connection between the battery positive post and resistors 35Ω, 25Ω and 10Ω. The green highlight represents common connections for series components. The yellow highlight represents the common connection point for resistors 20Ω, 15Ω, 30Ω, and 5Ω. The following drawing can now be redrawn to a simpler form using the common busses.

The circuit is now re-drawn to a simpler format so the calculations concerning any variable can now be made. Please complete the exercises contained in Practice 14 – Larger Complex Circuits –

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Review

1. A combination circuit contains elements of a series circuit and elements of a parallel circuit. 2. The six parameters of a series circuit apply to the series portion of the combination circuit.

3. The six parameters of a parallel circuit apply to the parallel portion of a combination circuit.

4. To find values across components of a combination circuit, it may be necessary to simplify the circuit to a

series equivalent circuit to obtain the circuit totals.

5. Sections of the combination circuit must be identified as series sections or parallel sections so the rules involving these types of circuits can be applied.

6. Kirchhoff’s voltage and current laws still apply.

7. Abstract circuit will have to be organized so a proper analysis of the circuit may be made.

You have completed Trade Theory – DC Circuits. You must complete the Final Exam with a score of 75% or better to receive credit for this course. Use the link below to take you to the Final Exam. DCCircuitsQuizFinalExam_v0.09.tbk