ECET 3000 Electrical Principles

Transcription

1 ECET 3000 Electrical Principles SeriesParallel Circuits Introduction The fundamental concepts and building blocks that form the foundation of basic circuit theory are: Ohm s Law Seriesconnected Resistors Kirchhoff s Voltage Law (KVL) Parallelconnected Resistors Kirchhoff s Current Law (KCL) Series & Parallel Equivalent Resistances Voltage Divider & Current Divider Equations 1

2 Introduction A technique that can be used to solve circuits containing both seriesconnected and parallelconnected elements will now be presented. This technique requires no new rules or laws. Instead, it describes an approach for analyzing more complex circuits using the only the previously presented concepts. SeriesParallel Circuit Example #1 SeriesParallel Circuits A seriesparallel circuit is a circuit that is formed using a combination of seriesconnected and parallelconnected circuit elements. SeriesParallel Circuit Example #1 2

3 Series Elements Two or more elements are connected in series if the current flowing through one of the elements must entirely flow through the other element(s). Note the voltage source is also connected in series with resistors and since the current I1,2 flows through all three elements. I1,2 Series SeriesParallel Circuit Example #1 Parallel Elements Two or more elements are connected in parallel if they are connected across the same two nodes (i.e. the exact same voltage appears across each of the elements). Note although a node is a common point of connection for two or more circuit elements, a section of ideal wire can be considered equivalent to a node since there is no potential difference across an ideal wire. V3 4 parallel SeriesParallel Circuit Example #1 3

4 SeriesParallel Circuits When a circuit contains both series and parallel connected elements, careful inspection is often required in order to identify the correct combinations. For example, in the circuit shown below, resistors and R5 are connected in series since the current I2,5 must flow through both of the resistors. I2,5 Series R5 SeriesParallel Circuit Example #2 SeriesParallel Circuits Careful inspection will also reveal that resistors and are connected in parallel since the ends of both resistors are connected to the same two nodes. parallel R5 SeriesParallel Circuit Example #2 4

5 SeriesParallel Circuits Note that the seriescombination of resistors and R5 may be considered as also being connected in parallel with resistors and. parallel IDC Series R5 SeriesParallel Circuit Example #2 SeriesParallel Circuits Although is connected in series with the battery, it is not connected in series or in parallel with any of the other resistors in the circuit. parallel IDC Series R5 SeriesParallel Circuit Example #2 5

6 SeriesParallel Circuits Take a moment to examine the following circuit in order to determine which resistors are connected in series and which are connected in parallel. In this circuit, there are no series or parallel connected sets of resistors. The techniques required to analyze this type of network will be presented later in the course. R5 R6 SeriesParallel Circuit Example #3 The Reduce & Return Approach Although seriesparallel circuits may be analyzed by a variety of methods, one of the simplest methods is the: Reduce and Return Approach. With this method, the circuit is incrementally reduced in complexity by replacing sets of series or parallel connected elements with their series or parallel equivalents. Each incremental reduction provides a simpler circuit containing fewer elements that can either be further reduced until only a trivial circuit remains. 6

7 The Reduce & Return Approach Once the original circuit is reduced down into a trivial circuit, that circuit can be analyzed in order to determine any unknown voltages or currents. The initiallysolved voltages or currents can then be utilized, stepbystep in reverse order back through the simplified circuits, facilitating the analysis of each incrementally more complex circuit until the desired unknown quantities specified in the original circuit are known. Use the Reduce and Return Approach to solve for the voltage V 4 as specified in the following figure: Step 1 Reduce the Original Circuit Identify a set of seriesconnected or parallelconnected resistors and replace them with a single equivalent resistor. Note if multiple series or parallel combinations exists, the ones furthest from the source are typically reduced first V4 Circuit #1 7

8 Step 1 Reduce the Original Circuit Resistors R 3 and R 4 are connected in series. The seriesequivalent resistance R 3,4 that may be used to replace the series R 3 and R 4 combination is: R R , Series 100 V4 Circuit #1 Step 1 Reduce the Original Circuit When replacing a combination of resistors with a single equivalent resistance, the rest of the circuit remains unchanged. Note it is often useful to highlight the elements that are being reduced in order to help keep track of the circuit changes V4,4 240 Circuit #1 8

9 Step 1a Identify New Variables in the Reduced Circuit Redraw the circuit with the seriesequivalent resistor in place and then analyze the reduced circuit, identifying any new variables that may be required during the completion of the overall problem. 160,4 240 Circuit #2 Step 1a Identify New Variables in the Reduced Circuit Since the rest of the circuit is unaffected by the change, a voltage V 3,4 and a current I 3,4 can be defined in both the simplified circuit (#2) and the original circuit (#1) such that their values will be the same in both circuits. I3,4 I3, , Circuit #2 Circuit #1 9

10 Step 1a Identify New Variables in the Reduced Circuit It is important to understand that the voltage V 4 does not appear in the simplified circuit (#2). But, if the voltage V 3,4 can be determined, we can use that information to solve for V 4 in the original circuit. I3,4 I3, , Circuit #2 Circuit #1 Steps 2? Repeat the Process until a trivial circuit remains. Note although the circuit can be reduced down until only a single resistor remains, circuits are typically reduced to the point where only a single set of either seriesconnected or parallelconnected resistors remain. I3,4 160,4 240 Circuit #2 10

11 Step 2 Repeat the Reduction Process Examine the new circuit to determine if there are any series or parallel connected elements that need to be replaced by their respective equivalents Looking at the new circuit (#2), it can be seen that resistors R 2 and R 3,4 are connected in parallel. I3,4 160, parallel,4 240 Circuit #2 Circuit #2 Step 2 Repeat the Reduction Process Thus, a parallelequivalent resistance R 2 (3,4) can be used to replace the R 2 and R 3,4 combination, where: R 2 (3,4) 1 1 R 3, ,4 240 (3,4) 96 Circuit #2 11

12 Step 2 Repeat the Reduction Process The newly reduced circuit appears as shown below: Once again, the circuit should be analyzed in order to identify any new variables that may be required during the completion of the overall problem. (3,4) 96 Circuit #3 Step 2 Repeat the Reduction Process The voltage V 3,4 across the parallelequivalent resistance and the current I 1 flowing through the parallelequivalent resistance can be defined in both the new circuit (#3) and the previous circuit (#2). I1 I1 (3,4) ,4 240 Circuit #3 Circuit #2 12

13 Step 2 Repeat the Reduction Process Since the circuit has now been reduced down to only a single pair of series connected resistors, the reduction part of this problem is complete. I1 Circuit #3 (3,4) 96 Step 3 Analyze the Trivial Circuit (#3) All that remains is a simple circuit (#3) consisting of two seriesconnected resistors. Thus, we can now easily determine the unknown quantities I 1 and V 2 (3,4) and then begin working backwards through the reduced circuits to find the desired quantity V 4. I1 Circuit #3 (3,4) 96 13

14 Step 3 Analyze the Trivial Circuit (#3) Solving for I 1 and V 2 (3,4) : V DC 12 I 0. 1 amps 1 R R (3,4) R 2 (3,4) 96 V2 (3,4) V DC volts R 2 (3,4) I1 Circuit #3 (3,4) 96 Step 3 Analyze the Trivial Circuit (#3) Note Although we found both I 1 and V 2 (3,4), only one of those quantities is actually needed to complete this problem. I1= 0.1A Circuit #3 (3,4) 96 14

15 Step 3 Analyze the Trivial Circuit (#3) We will use the value of V 2 (3,4) to complete the problem. R 2 (3,4) 96 V2 (3,4) volts R 2 (3,4) Note for practice, you may want to rework the remainder of this problem by applying the value for I 1 to a currentdivider based solution for this problem. I1= 0.1A Circuit #3 (3,4) 96 Step 4 Return to the Previous Circuits Take the results obtained from the final circuit (#3) and apply them incrementally to the morecomplex circuits (#3#2#1), solving for the voltages and/or currents in the circuits that will eventually allow for the solution of the originally desired quantity, V 4. I3,4 (3,4) , V4 Circuit #3 Circuit #2 Circuit #1 15

16 Step 4 Return to the Previous Circuit (#2) As stated earlier, V 2 (3,4) in the final circuit (#3) is equivalent to V 2 (3,4) as defined in the incrementally more complex circuit #2. (3,4) ,4 240 Circuit #3 Circuit #2 Step 4 Return to the Previous Circuit (#2) And, since R 2 is connected in parallel with R 3,4 in that circuit, V 2 (3,4) is also equivalent to V (3,4) as also defined in circuit #2. Since we know V (3,4), we can return to the original circuit (#1) in order to solve V 4. (3,4) ,4 240 Circuit #3 Circuit #2 16

17 Step 5 Return to the Original Circuit (#1) Apply the solution from circuit #2 to the original circuit (#1) and then use the result to solve for the voltage V 4. We can do this because V 3,4 in circuit #2 is equivalent to V 3,4 in the original circuit (#1), which is the total voltage across the seriesconnected resistors R 3 and R ,4 240 Circuit #2 Circuit # V4 Step 5 Return to the Original Circuit (#1) Thus, V 4 can be solved using a voltage divider equation as follows: R V volts R Circuit #1 100 V4 17