41

DC Circuits Part II

The Parallel Circuit Let’s now look at the parallel DC circuit. Instead of the circuit components being connected in series, they are connected on one side by one common point and the other side to another common point, or parallel.

We can now add a voltage source to make a complete circuit. Visualize that in this circuit that the “gob” of resistors in parallel can be replaced by one equivalent resistor in a series circuit.

42

A more usable schematic depicting a parallel circuit is illustrated below. This schematic is more easily understood than the example showing a clutter of resistors connected to one point. This example is uniform, spread out and assessable to interpretation.

Two Equal Resistors We will now look to see how voltage affects a parallel circuit. Let’s again look to see how voltage affects a series circuit. We will use a very simple circuit using only one resistor. In this circuit we will use a 10 volt battery as a source. The voltage across the battery is also the voltage across the resistor.

Notice the polarity of the battery. The red conductor is connected to the negative terminal therefore the negative polarity extends to the top of the resistor as negative. The positive terminal of the battery is connected to the resistor with the black conductor extending the positive polarity to the bottom of the resistor. If the resistor had a resistance of 100 Ω, what would the current be limited to? By using Ohm’s Law, we can solve for the current with the following formula.

volt

EI = R10I = 100

I = 0.10 1or 00mAΩ

Let’s look at the exact setup as this series circuit except we will substitute the single series resistor with two resistors of the same value in parallel. Again the voltage across the two resistors connected to the same common point is equal to the voltage across the battery. There is now a path across the two equal resistors in which current will flow. The current will split across the two paths. We have the same voltage, same value resistors; yet two paths, therefore the same magnitude of current should flow in each path.

43

Current

Since the numbers are the same as the series circuit mentioned above, the current through each leg of the parallel circuit is 100mA. The two 100mA currents will join at the bottom of the parallel circuit summing to 200mA. That means a total of 200mA circulates through this circuit. We know that there are 10 total volts in this circuit and a total current of 200ma. With those totals, a total resistance can now be calculated.

100.250

ERI

R

R

=

=

= Ω

The equivalent circuit.

Here’s the rule to follow: For the same value resistors connected in parallel, divide the value of one resistor by the total number of resistors of the same value. This means that if I had three 100Ω resistors in parallel, I would divide the value of the one (100Ω) by 3 (the total resistors in parallel of the same value) 100

3 33.3Ω = Ω This is only true for resistors of equal value.

Please complete the exercises contained in Practice 8 - Parallel Branches of Equal Resistors

Current

44

In parallel circuits each parallel path in which current may flow is called a branch (or leg). In our simple parallel schemes in which we are now solving, only one resistor exist and we can define each branch by a resistor number i.e. R1, R2, R3 etc. But in more complicated circuits the branches may contain more than one component; therefore that portion of the circuit, one which a current may divide into, is considered a branch.

Two or More Unequal Resistors If the resistors are of unequal values, the current will be different through the two resistors. The rules that we so far have discovered are;

Each load component (leg/branch) operates at the same voltage. Each load component is connected separately to the source. The current divides proportionally down each leg or branch.

Another Law to consider is Kirchhoff’s Current Law. This law states; The algebraic sum of the currents entering and leaving any junction of connectors is equal to zero.

45

A junction is a connection point within a circuit which current can divide its total into branch currents. Another name for a junction which is quite commonly used is the node.

In order for the law to be functional you must first label the currents so that any current entering the junction is a positive current and any current exiting the junction is a negative current.

46

Let’s consider the current through the branches containing two unequal resistors. Again the voltage at the common connection is the same for both branches, but the different resistances will determine the magnitudes of current that will flow in each branch. We will use an example of a parallel circuit which contains two branches. Branch 1 contains a 200Ω resistor and branch 2 contains a 300Ω resistor. This circuit is connected to a 50 volt DC source.

We can begin to analyze the circuit by first using Ohm’s law and solving for the currents in each branch. Then by adding the two currents, we can derive the total current of the circuit. And the total circuit resistance can be determined by the quotient of the total voltage divided by the total current. Branch 1:

volts

1

1 or

50I = 200

I = 0.25 250mAΩ

Branch 2:

volts

2

2 or

50I = 300

I = 0.167 167mAΩ

Notice that the branch with the greater resistance has a lower magnitude of current passing through it. To establish the total current, we just add the two currents together. That gives us a total of: or0.250 0.167 0.417 417mA+ =

47

The total resistance of the circuit can now be established.

totaltotal

total

voltstotal

amps

t or

ER = I

50R = 0.417

R 119.9 120= Ω

The circuit’s total resistance is also equal to the series equivalent circuit. Did you notice that the total resistance in the previous example was smaller than the smaller resistance found in the branches? This will always be true. If you calculate a total resistance, that total will always be smaller than the smallest valued branch resistance. If not, you miscalculated. Let’s look at another method to determine the total resistance of a parallel circuit. We will concentrate just on the parallel resistances in the branches and no other values. In fact we will use the same resistors so we may compare methods. Method 1 The reciprocal of the sum of the reciprocals of the resistances in parallel equals the total or equivalent resistance of that circuit. In the analysis of the circuit above; 1 2 total orR + R R 200Ω + 300Ω 120Ω≠ ≠ . Yet when we applied Ohm’s law to the two branches we came up with two currents which when totaled and applied to the total current, an equivalent resistance (or total resistance) was derived as 120Ω. What actually has taken place was a sum of the reciprocals of the branches equaled the reciprocal of the total. That is;

1 2

1 1 1

1 1 1200 300

3 2 15600 600 600

1 1200.00833

t

t

R R R

R

+ =

+ =Ω Ω

+ =

= Ω

This is; the reciprocal of the total equals the sum of the reciprocals.

48

If you notice that a manipulation of fractions in finding the least common denominator must be exercised in order to achieve an answer to this equation. Let’s change this equation to obtain an easier equation to work from.

( ) ( ) t

1 2

1 2 t

t1 2

1 2

1 2

t

2

t

1

1

t

2

t Multiply both sides by R

1 1Divide both sides by +

R R

1 1 1 1 + + R R R R

1 1 1 + = R R R

1 1R + = 1 R R

1 1 + R R 1R =

1R =

R R

1 1 + R R

This new formula now becomes the reciprocal of the sum of the reciprocals of parallel resistances equals the equivalent or total resistance. In looking at the two different equations, one would say, “What’s the difference”? The difference is that with the later equation we can now work with decimals instead of fractions.

49

Let’s look at the following example (all three branches) and calculate the total or equivalent resistance of the circuit. You will now find that by using the new equation, it is will be much easier to solve.

total

branch 1 branch 2 branch 3

total

totalor

1 R 1 1 1R R R

1 R1 1 1200 300 500

1 1 R0.005 0.00333 0.002 0.01033

1 0.01 96.0 833

=+ +

=+ +

=+ +

÷ = Ω

50

In circuits with many resistive branches the method above is easier to use in calculating equivalent resistances. Let’s do one more example of this. We have a circuit which has five resistors in parallel. The resistances are as follows,

With the following rule we may begin the calculation. The reciprocal of the sum of the reciprocals of the resistances in parallel equals the total or equivalent resistance of that circuit.

total

1 2 3 4 5

total

total

total

total

1 R1 1 1 1 1R R R R R

1 R1 1 1 1 115 65 25 30 55

1 R0.0667 0.0154 0.04 0.0333 0.0181

1

5

R

.7

0.1735

R 6

=+ + + +

=+ + + +

=+ + + +

=

= Ω

Well, I didn’t say, “it was easy”, I said, “it was easier” Please complete the exercises contained in Practice 9 –Unequal Parallel Resistance -- Method 2: Product over the sum

51

The product of the resistances divided by the sum of the resistances equals the total resistance.

We will again begin with that equation; 1 2 t

1 1 1 + = R R R

.

This equation we have already proved to be true. We can now manipulate this equation to a new equation to solve for the equivalent resistance for a parallel circuit containing only two branches. By the described equation we added the two resistive components;

1 2

1 1 + R R

To add these fractions, we must find the least common denominator which is, 1 2R R .

We can now multiply the first fraction, 1

1R

by 1 or 2

2

RR

The product of this multiplication is, 2

1 2

RR R

We can repeat this process on the second fraction and multiply by 1 or 1

1

RR

The product of this multiplication is, 1

1 2

RR R

With these new products we can now re-build the equation to form a new equation.

2 1 1 2

1 2 1 2 1 2

R R R + R + = R R R R R R

But we must remember that the true sum is the reciprocal of this sum or

t1 2

1 2

1 = RR + RR R

We must now invert and multiply. Since we will multiply by only 1, the final equation will be;

1 2t

1 2

R R = RR + R

This is what we call the product over the sum. We will again use the following circuit scheme so we may compare the equations to obtain the same answer.

52

But remember this equation will only work for circuits with only two branches. There is an exception later explained.

Branch 1 = 200Ω Branch 2 = 300Ω

total

total

total

20

12

0 300R200 30060,000R

0500

R

Ω× Ω=

Ω + Ω

=

= Ω

This limits the calculation to circuits with only two branches. If your circuit contains three branches, you may calculate two of the three branches by this method and then use that quotient as a resistance and the third branch resistance as the other, then calculate again as if they are two branches. For example we will throw in a third branch of 500Ω.

53

We will then use the Rtotal of the first two branches, label it as R1+2 then again calculate.

1+2 3total

1+2 3

total

total

total or

R RR R + R120 500R 120 50060,000R

620R 9 96.86.77

×=

Ω× Ω=

Ω + Ω

=

= Ω

Again notice that the total resistance is smaller than the calculated R1+2 and 500Ω of branch 3. You may continue this method over again for any number of branches that your circuit has. The problem is this may become quite cumbersome with many parallel branches. Therefore I will recommend an easier method to solve for multiple branches (more that two). Please complete the exercises contained in Practice 10 – Parallel Product/Sum--

54

Alternate methods to solve for parallel resistors The following rule can also be used, but be careful and follow the rules involving fractions. The sum of the reciprocals of each of the resistances (branches) in a parallel circuit is equal to the reciprocal of the total resistance of that circuit. Let’s use the following circuit to prove this method.

The reciprocals of the following resistances are:

1 1 15 50 25

and andΩ Ω Ω

The sum of these reciprocals equals the reciprocal of the total resistance. That is;

total

total

total

total

1 1 1 1 find the common denominator5 50 25 R

10 1 2 13 1 invert to remove reciprocal50 50 50 50 R50 R13

3.85R

+ + =Ω Ω Ω

+ + = =

=

= Ω

This may be easier for some, but you must work in fractions and not decimals.

55

The final method is instead of calculating resistance, you may calculate the conductance, which (by the way) is the reciprocal of resistance. Well, we actually did this method by the sum of the reciprocals of the resistances as stated prior. Be aware of this fact because I may plan to trick you in a couple of questions when I describe a resistor not by its resistance (ohms, Ω) but by its conductance (siemens, G) Described below is the equation to derive total conductance in a parallel circuit. 1 2 3 ......t nG G G G G= + + + Do notice that even though the circuit is parallel you add the conductance’s to obtain a total (like resistors in series). To convert resistance to conductance and conductance to resistance the following equations are used.

totaltotal

and1 1 1 R= and RR

GG G

= =

Let’s solve the following circuit that contains the following conductances.

total 1 2 3

total

total

0.2 0.02 0.040.26

G G G GG G G GG G

= + += + +=

therefore

total

total

total

1R

1R0.26

R 3.85

tG

G

=

=

= Ω

Please complete the exercises contained in Practice 11 –Parallel Alt. Methods

56

Power in a Parallel Circuit The total power dissipated in a parallel circuit is the sum of the individual power dissipated across each branch of the circuit. That means: total 1 2 3 nP P +P +P +.......P= Let’s look at an example. A circuit consists of three resistors connected to a voltage source in parallel. The voltage source has a voltage of 100 volts. The resistive values of the resistors are as follows:

Remember that the voltage is the same at all branches in the circuit. The red line represents an equal voltage. Find the total power and the power across each branch in this circuit. We can first find the total resistance of the circuit and then determine the total current. Total current times the total voltage equals the total power.

t

t

t

t

1R 1 1 115 45 30

1R0.0667 0.0222 0.033

8.18

31R

0.1222R

=+ +

=+ +

=

= Ω

and

tt

t

t

t

VIR100VI8.112.22 m

8ΩsI a p

=

=

=

The total power can now be derived.

or

P=IEP=12.22A×100VP= 1,222W 1.22kW

57

The power across the branches of this circuit can now be determined. We can look at the PIER circle and find a power equation which contains a voltage and a resistance. That equation is

2EP = R

.

Branch 1: Branch 2:

2

2

EP = R

100P =15

10,

6

000P =

66.67 1

w5

P = atts

2

2

EP = R

100P =45

10,

2

000P =

22.22 4

w5

P = atts

Branch 3:

2

2

EP = R

100P =30

10,

3

000P =

33.33 3

w0

P = atts

If our calculations are correct, we can sum the individual powers and they should equal the total which was calculated prior. Please note that we usually do this to check or correct our calculations in our working environment. We do not want to make a mistake when we are working on actual circuits, so get used to this practice.

total 1 2 3

total

total

P P + P + PP = 666.67 + 222.22 + 333.33P = 1,222 watts

=

The two calculations come out the same proving that our answer is correct. Please complete the exercises contained in Practice 12 – Power in Parallel –

58

Assumed Voltage Another tool to solve for resistive quantities of a parallel circuit is to assume a voltage applied to the circuit. By calculating the branch currents from the assumed voltage a total current can then be assumed (since the voltage was assumed in the first place). The ratio of assumed voltage divided by the assumed current will actually produce the proper equivalent of total resistance. In order to avoid small decimal quantities the assumed voltage should be greater than the largest impedance of any parallel branch. Choose a voltage in the power of ten so any multiplication or divisions can be made quite easily. Let’s look at an example that we will first solve by the reciprocal of the sum of the branch reciprocals. Then we will solve it by assuming a voltage. We will use the following schematic circuit.

Method 1: Reciprocal of the sum of the reciprocals

t

1 2 3

t

t

t

t

1R = 1 1 1 + + R R R

1R = 1 1 110 6 40

1R = 0.1 + 0.1667 + 0.025

1R = 0.2917

R 2 = 3.428

+ +

Ω

59

Method 2: We will assume a voltage of 100 volts across the circuit

1 amps1

2 amps2

3 amps3

total 1 2 3

total

total amps

totaltotal

E 100I = = = 10R 10E 100I = = = 16.667

R 6E 100I = = = 2.5R 40

I = I + I + II = 10 + 16.667 + 2.5I = 29.167

E 100R = = = 3I 29.

.4167

285Ω

The two methods came out with the same answer proving that both methods are valid and we performed the correct computations. At your first glance you would probably say that the assumed voltage method requires just as much computation as the conventional method. This is not necessarily the case when polar and rectangular conversions are required along with division in AC circuit solutions. Now for you crafty people if you assumed a voltage of one volt, how much difference would this be to method one. To some people the voltage assumption may be easier to visualize.

60

Special Equations If you have a parallel circuit situation to where you have three of the following four values, you may choose to solve using the following equations. Given: Any three of the four values for I1, I2, R1, R2

It is known that the voltage across both resistors in parallel is the same.

1 1 1 2 2 2

1 2

1 1 2 2

and

also

therefore

R I = E R I = EE = E

R I = R I

By the relationships stated above the following equations can be formed to solve for each unknown value of the four values.

2 2 1 11 2

1 2

2 2 1 11 2

1 2

R I R II = I = R R

R I R IR = R = I I

These equations are derived by substituting into the original equations of:

and E EI = R = R I

Notice that you did not have to know the voltage value for these calculations.

61

Let’s look at another instance in which requires special consideration. A circuit has two resistors of a known value connected in parallel which a total current of a known value is supplied to this circuit.

Using Ohm’s Law we can state the obvious relationships as equations. Relationship 1: 1 1 1 2 2 2andE = R I E = R I Relationship 2: 1 2 1 1 2 2thereforeE = E R I = R I The following two equations were derived from the first special consideration prior to this instance. 2 2 1 1

1 21 2

and R I R II = I =

R R

We must first derive a branch current so that we may then establish the voltage supplied to this circuit. The following equations will derive the final equation which is used to solve for the branch circuit. We will begin with the branch which contains R1. Solving for I1 the following relationships hold true: total 1 2 1 total 2 2 total 1and and I - I - I = 0 I = I - I I = I - I

62

By substitution we can then say:

( )

( ) ( )

( ) ( ) ( )

( )

( ) ( )

( )

1 1t 1

2

2 t 1 1 1

2 t 2 1 1 1

2 t 2 1 1 1

2 t 1 2 1

1

R I1 I - I = R

2 R I - I = R I

3 R I - R I R I

4 R I = R I + R I

5 R I = I R + R

R6 I =

therefore

therefore

therefore

therefore

therefore

=

2 t

2 1

IR + R

And I2 is solved by

( )

( ) ( )

( ) ( ) ( )

( )

( ) ( )

( )

2 2t 2

1

1 t 2 2 2

1 t 1 2 2 2

1 t 1 2 2 2

1 t 2 1 2

2

R I1 I - I = R

2 R I - I = R I

3 R I - R I R I

4 R I = R I + R I

5 R I = I R + R

R6 I =

therefore

therefore

therefore

therefore

therefore

=

1 t

1 2

IR + R

The above equations are rarely but may be used depending upon the known values that your circuit has. Many types of equipment will give you an ammeter which displays just the total current. Look again at the process of deriving these two final equations and appreciate your math skills.

63

Parallel Circuit Parameters The following statements represent the parameters for parallel circuits. These parameters are the rules in which we have just covered in the previous paragraphs concerning parallel circuits. They can be considered a recap of what we have learned about parallel circuits. Just like the series circuits, there are six parameters listed below; 1. The same voltage exists across each branch of a

parallel circuit and is equal to the source voltage. 2. The current through a branch of a parallel

network is inversely proportional to the amount of resistance of the branch 3. The total current of a parallel circuit is equal to the sum of the currents of the individual

branches of the circuit. 4. The total resistance of a parallel circuit is equal to the reciprocal of the sum of the reciprocals

of the individual branch resistances of the circuit. 5. The total power dissipated in a parallel circuit is equal to the sum of the power consumption

of the individual branch resistances. 6. Ohm’s Law applies to each component and to the entire parallel circuit. Study these parameters and commit them to memory.

64

Shorts and Opens Involving Parallel Circuits Let’s look at a typical parallel circuit which includes four resistors connected in parallel. We will make the values of the resistors all the same, 100Ω. This will make for easy computation when we decipher the outcomes of opens and shorts. This circuit will have a voltage supply of 25 volts.

The voltage across each branch is the same, 25 volts. The total resistance is calculated by the following equation:

tresistors

t

100ΩR = 4

R = 25Ω

The total current is calculated as follows:

tt

volt

t mp

t

a

EI = R25I = 25Ω

1I =

The current across each branch is calculated:

branchbranch

voltbranch

branch or

EI = R25I = 100Ω

I = 0.25 250mA

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These are the values of this circuit in normal operation. Let’s open one of the resistor branches and determine what should happen. Any guesses? If I remove one resistor in parallel will the total resistance go up or go down? What will happen to the total voltage? Does the total current in the circuit go up or go down?

There are now three resistors in parallel instead of four. The equation states that the total resistance of equal resistors in parallel is determined by the value of one resistor divided by the total amount of equal value resistors. That will be 100Ω divided by the three resistors. The new total resistance is 33.3Ω. With fewer resistors, the resistance becomes greater. The total current should become less because the total resistance, opposition to current, went up. Remember that we stated that the total current was the sum of all of the branch currents. If we removed a branch, we also removed a current source from the total. The currents across the branches remain the same because the voltage will never change in a parallel circuit and the resistances of the remaining resistors remains the same. Since we have 250mA across each branch in the circuit, the remaining three branches should sum to 750mA for the total current in this circuit. Let’s prove this by calculating the total current by the voltage divided by the total resistance.

tt

t

t or

EI = R

25VI = 33.33Ω

I = 0.75 750mA

Another open in this circuit will raise the total resistance and reduce the total current of the circuit.

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We will now switch gears and determine the outcome of a parallel circuit with a short across a branch.

We once stated that the total resistance of a parallel circuit will be lesser than the least valued resistor. We essentially added another branch to the circuit. This branch, being a short, has a resistance of zero. The total resistance of the circuit is now zero and if there is no resistance to impede current flow, the current will flow to its maximum potential.

67

Loading of Parallel Circuits

Let’s look at the schematic above. It is early morning and the store is closed and the factory is not producing at this time. We essentially have three branches, the houses, in a parallel circuit. The voltage supplied to any and every will remain the same. This is a parameter of a parallel circuit, 120 volts supplied to every branch no matter what the branches load is. We have learned that the total current in the circuit is the sum of all of the currents within the branches of a parallel circuit. As seen in the drawing only the three houses are in the circuit. Let’s calculate the total current. House 1 House 2 House 3

house 11

house 1

house 1

EI = R120I = 100

I = 1.2A

house 21

house 2

house 2

EI = R120I = 120

I = 1A

house 31

house 3

house 3

EI = R120I = 300

I = 0.40A

t 1 2 3

t

t

I = I + I + II = 1.2A + 1A + 0.40AI = 2.6A

The total resistance of the circuit is:

tt

t

t

ER = I120VR = 2.6A

R = 46.15Ω

68

The morning passes and people are at work and the stores open. Additional load picks up with the opening of these businesses. Let’s look at the circuit now.

The circuit now has five branches all being supplied with the 120 volt source. The voltage will not change because the components are connected in parallel. We now have two new loads to include into the total. The first three loads were unchanged so we only have to calculate and add the new load to obtain the total load. Food Mart Factory

food martfood mart

food mart

food mart

EI = R120VI = 50

I A= 2.4Ω

factoryfactory

factory

factory

EI = R

120VI = 2

I A0

= 6Ω

The total current can now be found.

t 1 2 3 4 5

t

t

I = I + I + I I + II = 1.2A + 1A + 0.40A + 2.4A + 6AI A = 11

+

As we stated before the total resistance is lower than the least branch resistance. The factory branch has the least amount of resistance, 20Ω. The total resistance should be less so let’s calculate it.

tt

t

t

ER = I120VR = 11Ω

R = 10.9Ω

The total resistance is indeed lower than the least branch resistance.

69

Conclusion: If we were to design a power system to supply electricity to residential and commercial customers, we would design it to put all of the loads in parallel. By doing so we can guarantee the supply voltage will be unchanged no matter what variances happen to the different loads. Our only caution is that if we insert too many loads in parallel, the total resistance of the system will become very low and the current may rise to an unacceptable level or overload.

70

Review of Parallel Circuits:

1. A parallel circuit has more than one current path connected to a common voltage source. 2. Each load component operates at the same voltage.

3. The current paths split proportionally to the reciprocal of the branch/leg resistances

4. The sum of the branch currents equals the total current.

5. The sum of the power dissipated in the branches equals the total power consumption.

6. The more branches (or legs) the parallel circuit has the greater the current will become (add another branch

and the current will become still greater).

7. As more resistive branches are added, the total or equivalent resistance decreases.

8. The total resistance will always be lower than the lowest branch resistance.

9. Kirchhoff’s current law states: The algebraic sum of the currents entering or leaving any one point will always equal zero.

10. The total or equivalent resistance can be calculated by the following methods

a. For equal resistance branches, the value of one branch divided by the total number of branches. b. For two branches you may use the sum over the product.

c. The sum of the reciprocals of the branch resistances equals the reciprocal of the total resistance.

d. The reciprocal of the sum of the reciprocals of the branch resistances equals the total resistance

e. The total conductance in a parallel circuit is the sum of the branch conductance’s. The total

resistance is the reciprocal of the total conductance.

f. The total resistance in a parallel circuit segment can be calculated by assuming a voltage

11. The six parameters of a parallel circuit are as follows;

1. The same voltage exists across each branch of a parallel circuit and is equal to the source voltage.

2. The current through a branch of a parallel network is inversely proportional to the amount of resistance of the branch

3. The total current of a parallel circuit is equal to the sum of the currents of the individual branches

of the circuit.

4. The total resistance of a parallel circuit is equal to the reciprocal of the sum of the reciprocals of the individual branch resistances of the circuit.

5. The total power dissipated in a parallel circuit is equal to the sum of the power consumption of the

individual branch resistances.

6. Ohm’s Law applies to each component and to the entire parallel circuit.

71

12. The more load (resistance) you place in a parallel circuit, the total resistance decreases and more current is allowed to flow within the circuit. Yet, voltage remains the same no matter how many or how few parallel components are added. This makes a parallel a good circuit to supply a steady voltage source to different loads. The only caution to this is the current magnitude may become too high with additional loading.

13. Any open within a parallel circuit will remove the component from the circuit, lowering the total current

flow.

14. Any short within a parallel circuit will remove all of the components from the circuit.