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    Series & Parallel


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    Electricity 4: Series and Parallel Circuits

    In this section we will look at some simple DC series and parallel circuits. The purpose will be to observe the relationships occurring between Voltage, Current, Resistance and power. We will explore how changing the Voltage affects the Current and the fact that Power is a square law function relative to Voltage and Current.   In all of these examples we will assume that there is NO resistance in the wires, ammeter or the battery [Not possible in the real world] .

    Figure 1, below, demonstrates a simple complete circuit with one resistor. It could be viewed as a Series circuit because it is in series with the ammeter, and could also be considered a Parallel circuit because the load resistor is connected directly across the battery.

    Figure 1

     

     

     

    art4_fig1gif.gif

    The  
    answer
     is:

    Example
     1
    P = EI = 10W E = IR if V = ??,   I = 1 Amp, R = 10 W 10V
    Example  
    2
    P = EI = 40W I = E/R if V = 20V, I = ?? Amps, R = 10 W 2 Amps
    Example  
    3
    P = EI = 576W R = E/I if V = 12V, I = 48 Amps, R = ?? W 25 W

    In this circuit the current flows from the Battery through the Ammeter, R1 and back to the Battery. Note the current is equal in all parts of this circuit. In the first example, if the voltage was unknown, we could find it using Ohms Law E = IR.  

    [E] ?? = 1a [I] X 10 ohms [R] = 10 volts  
    The Power [P] is found using the Power formula P=EI  
    P = [E]10V X [I] 1a. = 10 Watts

    In the 2nd example we will change the Battery Voltage to 20v and leave the resistor 10 ohms. Now to find the current this circuit will draw we will use Ohms Law again. I = E/R  

    [I] ?? = [E] 20v / [R] 10 ohms = 2 amps  

    Now if we calculate the Power   P=EI we see that:

    [P] ?? = [E] 20v X [I] 2 a. = 40 watts

    Note that when the Source Voltage is doubled and the Resistance remains constant, The Power increases by 4. This is a "Square Law" function. If the Voltage had been increased by 3 times the Power would have increased by 9 times.

    In the 3rd example we have a 12 Volt Battery and the ammeter is indicating 48 amps of current flowing. What is the value of R1?  

    [R] ?? = [E] 12 v /48a [I] = .25 ohm

     The Power would then be:

     [P] ?? = [E] 12v X [I] 48a = 576 watts

    The purpose of this exercise was to illustrate the effects of changing Voltage, Current and Resistance in a circuit and their relationship to each other and to observe how these changes affect the total Power.

    Figure 2, below, illustrates two resistors connected in series. Note that in this circuit the Current will be the same in all parts of the circuit.

    Figure 2

     

     

     

    art4_fig2.gif

    In this circuit the current path is from the battery through the ammeter through R1 then R2 then back to the battery. The value of the two resistors ADD together when they are in SERIES:

     R1 + R2 = 5 + 15 = 20 ohms  

    With 20 volts applied, E = IR = 20 v = 1a [I] X 20 ohms [R]. Note: The voltage drop across R1 would be 5 V, as shown by:

    [E] 5v = [I] 1a X [R] 5 ohms

    The same would be true for R2. NOTE: The voltage drop in a SERIES circuit is always proportional to the resistance.  

     If one was to assume R2 was a light and that R1 was a piece of wire that had 5 ohms resistance, it is obvious that ¼ of the power would be lost getting to the light. This is how one adapts or utilizes these concepts in practical applications. If you understand what happens in a circuit like this, then when you take a voltmeter and start measuring you can understand what it is telling you and locate the problems.

    Figure 3, below, illustrates a circuit with 2 resistors in parallel, and what happens.

    Figure 3

     

     

     

     

     

     

    art4_fig3.gif

    In this circuit we will not go through all of the E = IR calculations. Note that Both Resistors have 20 volts across them. If we take the Current flowing through R1, which is 1amp, and the Current flowing through R2 and add them together we get 1a + 4a = 5a, or the total circuit current is 5a.

    If we look at E = IR,  then:

    20v @ 5a = 4 ohms.  

    Using P = EI the total power provided by the battery is:

    20v X 5a =100 watts  

    Of these 5 watts are dissipated as heat in R2 and 20 are dissipated in R1.   Note the parallel resistance formula in line 4, calculates to 4 ohms which agrees with the power calculations previously done.   If more than 2 resistors are connected in parallel the formula would be the one shown in the bottom line.  

    Note: IN THIS CIRCUIT WITH 2 RESISTORS IN PARALLEL THE CURRENT THROUGH THE RESISTORS IS INVERSELY PROPORTIONAL TO THE RESISTANCE OF THE RESISTORS.  

    Note also that when we insert Ohms Law [E = IR, and I = E/R] into the Power Formula:

     P = EI,   then P = [IR]I,   therefore P = [I]2R.    

    If we insert P = E[E/R], we get the formula P = [E] 2/R.  

    You can see these calculations used in line 2 and 3 above. This circuit should demonstrate how current flow is inverse to resistance values in parallel circuits and that the current flow is not the same in all parts of the circuit.   Further Note: If a resistor were to replace the ammeter, you would have a series parallel circuit and the 2 parallel resistors R1 and R2 would look like a 4 ohm resistor to the series resistor and the voltage drop across the two would be proportional to added resistor and the 4 ohms for this pair.

    From here we will proceed to some practical troubleshooting and magnetism, not necessarily in that order, but whatever I get finished next.  

    Al...


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