Series, Parallel, and Series-Parallel DC Circuits

12-34 a guard. A B detail regarding the values. With these values, we can now begin to learn more about the nature of the circuit. In this configuration, there is a 12-volt DC source in series with two resistors, R1 = 10 Ω and R2 = 30 Ω. For resistors in a series configuration, the total resistance of the circuit is equal to the sum of the individual resistors. The basic formula is: R T = R1 + R2 + R3 + ………RN For Figure 12-82, this will be: R T = 10 Ω + 30 Ω R T = 40 Ω Now that the total resistance of the circuit is known, the current for the circuit can be determined. In a series circuit, the current cannot be different at different points within the circuit. The current through a series circuit is always the same through each element and at any point. Therefore, the current in the simple circuit can now be determined using Ohm’s Law: Formula, E = I (R) Solve for current, E RI = The variables, E = 12 V and RT = 40 Ω Substitute variables, 12 V 40 ΩI = Current in circuits, I = 0.3 A Ohm’s Law describes a relationship between the variables of voltage, current, and resistance that is linear and easy to illustrate with a few extra calculations. First is the act of changing the total resistance of the circuit while the other two remain constant. In this example, the R T of the circuit in Figure 12-82 is doubled.

The effects on the total current in the circuit are: Formula, E = I (R) Solve for current, E RI = The variables, E = 12 V and RT = 80 Ω Substitute variables, 12 V 80 ΩI = Current in circuits, I = 0.15 A It can be seen quantitatively and intuitively that when the resistance of the circuit is doubled, the current is reduced by half the original value. Next, reduce the RT of the circuit in Figure 12-82 to half of its original value. The effects on the total current are: 12-35 Multiple poles Schematic Symbol for a Relay Relay Illustration A2 B2 A3 A1 B3 B1 C C X1 X2 Normally closed (NC) Normally open (NO) Magnetic field Common (C) or Arm Electromagnet pulls contact down NC NO NC NO X1 X2 X1 X2 (+) (−) N S + − + 12V 30Ω R2 10Ω R1 − Formula, E = I (R) Solve for current, E RI = The variables, E = 12 V and RT = 20 Ω Substitute variables, 12 V 20 ΩI = Current in circuits, I = 0.6 A Voltage Drops & Further Application of Ohm’s Law The example circuit in Figure 12-83 is used to illustrate the idea of voltage drop. It is important to differentiate between voltage and voltage drop when discussing series circuits.

V oltage drop refers to the loss in electrical pressure or emf caused by forcing electrons through a resistor. Because there are two resistors in the example, there are separate voltage drops. Each drop is associated with each individual resistor. The amount of electrical pressure required to force a given number of electrons through a resistance is proportional to the size of the resistor. In Figure 12-83 , the values used to illustrate the idea of voltage drop are: Current, I = 1 mA R 1 = 1 kΩ R 2 = 3 kΩ R 3 = 5 kΩ The voltage drop across each resistor is calculated using Ohm’s Law. The drop for each resistor is the product of each resistance and the total current in the circuit. Keep in mind that the same current flows through series resistor.

Formula: E = I (R) V oltage across R1: E1 = IT (R1) E1 = 1 mA (1 kΩ) = 1 volt V oltage across R2: E2 = IT (R2) E2 = 1 mA (3 kΩ) = 3 volt V oltage across R3: E3 = IT (R3) E3 = 1 mA (5 kΩ) = 5 volt The source voltage can now be determined, which can then be used to confirm the calculations for each voltage drop. Using Ohm’s Law: Formula: E = I (R) Source voltage = current × the total resistance ES = I (RT) 12-36 + 5 kΩ R3 1 kΩ R1IT 1 mA 3 kΩ R2 − RT = 1 kΩ + 3 kΩ + 5 kΩ RT = 9 kΩ Now: ES = I (RT) Substitute ES = 1 mA (9 kΩ) ES = 9 volts Simple checks to confirm the calculation and to illustrate the concept of the voltage drop add up the individual values of the voltage drops and compare them to the results of the above calculation.

1 volt + 3 volts + 5 volts = 9 volts Voltage Sources in Series A voltage source is an energy source that provides a constant voltage to a load. Two or more of these sources in series equals the algebraic sum of all the sources connected in series. The significance of pointing out the algebraic sum is to indicate that the polarity of the sources must be considered when adding up the sources. The polarity is indicated by a plus or minus sign depending on the source’s position in the circuit. In Figure 12-84, all of the sources are in the same direction in terms of their polarity. All of the voltages have the same sign when added up. In the case of Figure 12-84, three cells of a value of 1.5 volts are in series with the polarity in the same direction. The addition is simple enough: ET = 1.5v + 1.5v + 1.5v = +4.5 volts However, in Figure 12-85, one of the three sources has been turned around, and the polarity opposes the other two sources.

Again the addition is simple: ET = + 1.5v − 1.5v + 1.5v = +1.5 volts Kirchhoff’s Voltage Law A law of basic importance to the analysis of an electrical circuit is Kirchhoff’s V oltage Law. This law simply states that the algebraic sum of all voltages around a closed path or loop is zero. Another way of saying it: the sum of all the voltage drops equals the total source voltage. A simplified formula showing this law is shown below: With three resistors in the circuit: E S – E1 – E2 – E3 … –EN = 0 volts Notice that the sign of the source is opposite that of the individual voltage drops. Therefore, the algebraic sum equals zero. Written another way: E S = E1 + E2 + E3 … +EN The source voltage equals the sum of the voltage drops. The polarity of the voltage drop is determined by the direction of the current flow. When going around the circuit, notice that the polarity of the resistor is opposite that of the source voltage. The positive on the resistor is facing the positive on the source, and the negative on the resistor is facing the negative on the source.

V oltage Law. There are two resistors in this example. One has a drop of 14 volts and the other has a drop of 10 volts. The source voltage must equal the sum of the voltage drops around the circuit. By inspection, it is easy to determine the source voltage as 24 volts. drops and one voltage source rated at 24 volts. Two of the voltage drops are known. However, the third is not known. Using Kirchhoff’s V oltage Law, the third voltage drop can be determined. With three resistors in the circuit: E S – E1 – E2 – E3 = 0 volts Substitute the known values: 24v – 12v – 10v − E3 = 0 Collect known values: 2v – E3 = 0 Solve for the unknown: E3 = 2 volts Determine the value of E4 in Figure 12-88. For this example, I = 200mA.

First, the voltage drop across each of the individual resistors must be determined. E1 = I (R1) E1 = (200 mA) (10 Ω) 12-37 + 1.5V ' +'+' 1.5V 1.5V + 1.5V − +−+− 1.5V 1.5V + ES=24V − 14V 10V +− + 10V E224VES − + − + 12V E1 − + ? E3 − reversed. V oltage drop across R1 E1 = 2 volts E2 = I (R2) E2 = (200 mA) (50 Ω) V oltage drop across R2 E2 = 10 volts E3 = I (R3) E3 = (200 mA) (100 Ω) V oltage drop across R3 E3 = 20 volts Kirchhoff’s V oltage Law is now employed to determine the voltage drop across E4. With four resistors in the circuit E S – E1 – E2 – E3 – E4 = 0 volts Substituting values: 100v – 2v – 10v – 20v – E4 = 0 Combine: 68v – E4 = 0 Solve for unknown: E4 = 68v Using Ohm’s Law and substituting in E 4, the value for R 4 can now be determined.

Ohm’s Law: E IR = Specific application: E4 IR4 = Substitute values: 68 V 200 mAR4 = Value for R4: R 4 = 340 Ω Voltage Dividers V oltage dividers are devices that make it possible to obtain more than one voltage from a single power source. A voltage divider usually consists of a resistor, or resistors connected in series, with fixed or movable contacts and two fixed terminal contacts. As current flows through the resistor, different voltages can be obtained between the contacts. Series circuits are used for voltage dividers. The voltage divider rule allows the technician to calculate the voltage across one or a combination of series resistors without having to first calculate the current in the circuit. [Figure 12-89] Because the current flows through each resistor, the voltage drops are proportional to the ohmic values of the constituent resistors.

To understand how a voltage divider works, examine Each load draws a given amount of current: I 1, I 2, I 3. In 12-38 + 100 Ω R3 10 Ω R1 50 Ω R2 − + − +− ? R4 +− 100 VES C B A R1 R2R3 addition to the load currents, some bleeder current (IB) flows. The current (IT) is drawn from the power source and is equal to the sum of all currents. The voltage at each point is measured with respect to a common point. Note that the common point is the point at which the total current (IT) divides into separate currents (I1, I2, I3). Each part of the voltage divider has a different current flowing in it. The current distribution is as follows: Through R1 — bleeder current (IB) Through R2 — IB + I1 Through R3 — IB + I1, + I2 The voltage across each resistor of the voltage divider is: 90 volts across R1 60 volts across R2 50 volts across R3 The voltage divider circuit discussed up to this point has had one side of the power supply (battery) at ground potential. In has been moved to a different point on the voltage divider.

The voltage drop across R1 is 20 volts; however, since tap A is connected to a point in the circuit that is at the same potential as the negative side of the battery, the voltage between tap A and the reference point is a negative (−) 20 volts. Since resistors R2 and R3 are connected to the positive side of the battery, the voltages between the reference point and tap B or C are positive. The following rules provide a simple method of determining negative and positive voltages: (1) If current enters a resistance flowing away from the reference point, the voltage drop across that resistance is positive in respect to the reference point; (2) if current flows out of a resistance toward the reference point, the voltage drop across that resistance is negative in respect to the reference point. It is the location of the reference point that determines whether a voltage is negative or positive.

Tracing the current flow provides a means for determining the voltage polarity. Figure 12-92 shows the same circuit with the polarities of the voltage drops and the direction of current flow indicated. The current flows from the negative side of the battery to R1. Tap A is at the same potential as the negative terminal of the battery since the slight voltage drop caused by the resistance of the conductor is disregarded; however, 20 volts of the source voltage are required to force the current through R 1 and this 20-volt drop has the polarity indicated. Stated another way, there are only 80 volts of electrical pressure left in the circuit on the ground side of R1.

When the current reaches tap B, 30 more volts have been used to move the electrons through R 2, and in a similar manner the remaining 50 volts are used for R 3. But the voltages across R2 and R3 are positive voltages, since they are above ground potential. voltage drops across the resistances are the same; however, the reference point (ground) has been changed. The voltage between ground and tap A is now a negative 100 volts, or the applied voltage. The voltage between ground and tap B is a negative 80 volts, and the voltage between ground and tap C is a negative 50 volts. Determining the Voltage Divider Formula a voltage source. With a few simple calculations, a formula for determining the voltage divisions in a series circuit can be determined.

The voltage drop across any particular resistor shall be called EX, where the subscript x is the value of a particular resistor (1, 2, 3, or 4). Using Ohm’s Law, the voltage drop across any 12-39 C B A 100 V R3=100 Ω R2=60 Ω R1=40 Ω 50 V 30 V 20 V C B A 100 V R3 R2 R1 +50 V + − + − + − +30 V −20 V 200v 150v 90v IT IB I1 I2 I3 R1 R2E R3 Load Load Load C B A 100 V R3 R2 R1 50 V 30 V 20 V resistor can be determined. Ohm’s Law: EX = I (RX) As seen earlier in the handbook, the current is equal to the source voltage divided by the total resistance of the series circuit. Current: ES RT I = The current equation can now be substituted into the equation for Ohm’s Law.

Substitute: ES RT EX = ( )(RX) Algebraic rearrange: RX RT EX = ( )(ES) This equation is the general voltage divider formula. The explanation of this formula is that the voltage drop across any resistor or combination of resistors in a series circuit is equal to the ratio of the resistance value to the total resistance, divided by the value of the source voltage. Figure 12-95 illustrates this with a network of three resistors and one voltage source. RX RT EX = ( ) ES R T = 100 Ω + 300 Ω + 600 Ω = 1,000 Ω E S = 10 V V oltage drop over 100 Ω resistor is: 100 Ω 1,000 ΩEX = ( ) 100 V E 100Ω = 10 V V oltage drop over 300 Ω resistor is: 300 Ω 1,000 ΩEX = ( ) 100 V E 100 Ω = 30 V V oltage drop over 600 Ω resistor is: 600 Ω 1,000 ΩEX = ( ) 100 V E 100Ω = 60 V 12-40 + − + − + − + − + − R1 R2 ES R3 R4 E3 E4 E2 E1 100 Ω 300 Ω 600 Ω + − + − + − + − R1 R2ES100 v R3 Checking work E T = 10 V + 30 V + 60 V = 100 V

Parallel DC Circuits

A circuit in which two of more electrical resistances or loads are connected across the same voltage source is called a parallel circuit. The primary difference between the series circuit and the parallel circuit is that more than one path is provided for the current in the parallel circuit. Each of these parallel paths is called a branch. The minimum requirements for a parallel circuit are the following: • A power source • Conductors • A resistance or load for each current path • Two or more paths for current flow flowing out of the source divides at point A in the diagram and goes through R1 and R2. As more branches are added to the circuit, more paths for the source current are provided.

Voltage Drops The first point to understand is that the voltage across any branch is equal to the voltage across all of the other branches. Total Parallel Resistance The parallel circuit consists of two or more resistors connected in such a way as to allow current flow to pass through all of the resistors at once. This eliminates the need for current to pass one resistor before passing through the next. When resistors are connected in parallel, the total resistance of the circuit decreases. The total resistance of a parallel combination is always less than the value of the smallest resistor in the circuit.

In the series circuit, the current has to pass through the resistors one at a time. This gave a resistance to the current equal the sum of all the resistors. In the parallel circuit, the current has several resistors that it can pass through, actually reducing the total resistance of the circuit in relation to any one resistor value. The amount of current passing through each resistor varies according to its individual resistance. The total current of the circuit is the sum of the current in all branches. It can be determined by inspection that the total current is greater than that of any given branch. Using Ohm’s Law to calculate the total resistance based on the applied voltage and the total current, it can be determined that the total resistance is less than any branch.

An example of this is if there was a circuit with a 100 Ω resistor and a 5 Ω resistor; while the exact value must be calculated, it still can be said that the combined resistance between the two is less than the 5 Ω. Resistors in Parallel The formula for the total parallel resistance is as follows: 1 RT = + + + ... 1 R1 1 R2 1 R3 1 RN If the reciprocal of both sides is taken, then the general formula for the total parallel resistance is: RT = + + + ... 1 R1 1 R2 1 R3 1 RN 1 Two Resistors in Parallel Typically, it is more convenient to consider only two resistors at a time because this setup occurs in common practice. Any number of resistors in a circuit can be broken down into pairs.

Therefore, the most common method is to use the formula 12-41 + − + − + − R1 A B R2 ES for two resistors in parallel. RT = + 1 R1 1 R2 1 Combining the terms in the denominator and rewriting: RT = R1 + R2 R1R2 Put in words, this states that the total resistance for two resistors in parallel is equal to the product of both resistors divided by the sum of the two resistors. In the formula below, calculate the total resistance. General formula RT = R1 + R2 R1R2 Known values R1 = 500 Ω R2 = 400 Ω RT = 500 Ω + 400 Ω 500 Ω 400 Ω RT = 900 Ω 200,000 Ω RT = 222.22 Ω Current Source A current source is an energy source that provides a constant value of current to a load even when the load changes in resistive value. The general rule to remember is that the total current produced by current sources in parallel is equal to the algebraic sum of the individual sources.

Kirchhoff’s Current Law Kirchhoff’s Current Law can be stated as: the sum of the currents into a junction or node is equal to the sum of the currents flowing out of that same junction or node. A junction can be defined as a point in the circuit where two or more circuit paths come together. In the case of the parallel circuit, it is the point in the circuit where the individual branches join. General formula IT = I1 + I2 + I3 Refer to Figure 12-97 for an example. Point A and point B represent two junctions or nodes in the circuit with three resistive branches in between. The voltage source provides a total current IT into node A. At this point, the current must divide, flowing out of node A into each of the branches according to the resistive value of each branch. Kirchhoff’s Current Law states that the current going in must equal that going out. Following the current through the three branches and back into node B, the total current I T entering node B and leaving node B is the same as that which entered node A. The current then continues back to the voltage source.

I 1 = 5 mA I 2 = 12 mA The total current flow into the node A equals the sum of the branch currents, which is: IT = I1 + I2 Substitute IT = 5 mA + 12 mA IT = 17 mA The total current entering node B is also the same. in one branch. Note that the total current into a junction of the three branches is known. Two of the branch currents are known. By rearranging the general formula, the current in branch two can be determined. General formula IT = I1 + I2 + I3 Substitute 75 mA = 30 mA + I2 + 20 mA Solve I2 I2 = 75 mA – 30 mA – 20 mA I2 = 25 mA Current Dividers It can now be easily seen that the parallel circuit is a current divider. As shown in Figure 12-96, there is a current through each of the two resistors. Because the same voltage is applied across both resistors in parallel, the branch currents are inversely proportional to the ohmic values of the resistors.

Branches with higher resistance have less current than those with lower resistance. For example, if the resistive value of R2 is twice as high as that of R1, the current in R2 is half of that of R1. All of this can be determined with Ohm’s Law. By Ohm’s Law, the current through any one of the branches can be written as: 12-42 + − + − + − + − R1 R2 A B R3 ES IT I1 I3 + − + − + − + − A B ES I1=30 mA I3=20 mA I2=? IT=75 mA + − + − + − A B ES I2=12 mA R2R1 IT=17 mA I1=5 mA I X = ES/RX The voltage source appears across each of the parallel resistors and RX represents any one the resistors. The source voltage is equal to the total current times the total parallel resistance.

E S = ITRT Substituting ITRT for ES ITRT RX IX = Rearranging RT RX IX = ( ) IT R2 RT I2 = ( ) IT And R1 RT I1 = ( ) IT This formula is the general current divider formula. The current through any branch equals the total parallel resistance divided by the individual branch resistance, multiplied by the total current.

Series-Parallel DC Circuits

Most of the circuits that the technician encounters will not be a simple series or parallel circuit. Circuits are usually a combination of both, known as series-parallel circuits, which are groups consisting of resistors in parallel and in series. An example of this type of circuit can be seen in Figure 12-100. While the series-parallel circuit can initially appear to be complex, the same rules that have been used for the series and parallel circuits can be applied to these circuits. The voltage source provides a current out to resistor R 1, then to the group of resistors R2 and R3 and then to the next resistor R4 before returning to the voltage source. The first step in the simplification process is to isolate the group R 2 and R3 and recognize that they are a parallel network that can be reduced to an equivalent resistor. Using the formula for parallel resistance, R23 = R2 + R3 R2R3 R2 and R3 can be reduced to R23. Figure 12-101 now shows an equivalent circuit with three series connected resistors. The total resistance of the circuit can now be simply determined by adding up the values of resistors R1, R23, and R4.

Determining the Total Resistance A more quantitative example for determining total resistance and the current in each branch in a combination circuit is shown in the following example. [Figure 12-102] The first step is to determine the current at junction A, leading into the parallel branch. To determine the IT, the total