Electricity - Basic Navy Training Courses
NAVPERS 10622 - Chapter 8

Chapter 8
the SERIES CIRCUIT
WHY KNOW CIRCUITS?

You have a good chance to outrank the bull in the China shop - just try fooling around with circuits before you know what you're doing! After burning up a few instruments at one hundred dollars per copy, you'll probably wind up electrocuting yourself or your shipmates!
Electrical circuits are the power carriers of the electrical system. When they are connected properly, they are efficient and trouble free. BUT, foul them up by improper connections and they'll put your whole system out of commission.
The few circuits you have been made acquainted with are SIMPLE circuits - only a source, its load, and the connecting wires. Few practical circuits can be so simple. For example, it would be a terrific waste if EVERY electrical load had its OWN generator and its OWN feeders. Imagine the confusion of cables, lines, and wires if every light, every motor, every telephone, every heater, had its own separate feeder lines directly from the dynamo room. For reasons of ECONOMY and EFFICIENCY, therefore, most circuits are, actually VERY COMPLEX. They are designed so that ONE generator can feed MANY electrical loads. But no matter how complex any particular circuit becomes, it is one of three general types - the SERIES, the PARALLEL, or the SERIES-PARALLEL.

Every electrical load is designed to contain a specific resistance and operate at a certain rated voltage. The resistance of the road controls the amount of current at the rated voltage. The proper type of circuit connection insures the load of its rated voltage and current. Imagine the fireworks if a searchlight got 200 volts instead of 95 volts. Or - you'd wait a long time for chow from a 220 volt galley stove connected on a 110 volt line. Both these things could happen by using incorrect connections.

VOLTAGE IN SERIES CIRCUITS

The first type-the series circuit is a ONE-PATH circuit. You can always recognize a series connection by two facts-it will NEVER HAVE MORE THAN ONE CONDUCTOR CONNECTED TO ANY TERMINAL, and you will find only ONE PATH from source to load (or loads) and back to source.
Figure 37 shows a simple series circuit. In A, a voltmeter is connected across the total resistance. It reads the TOTAL voltage drop of the TOTAL resistance - in this case, 6 volts. notICE that the voltage drop which occurs in the wires is ignored. This is standard practice for short wires because the resistance is very small. But for longer wires, connecting load and source, the resistance is large and the voltage drop is appreciable. It must be considered in the circuit. In B, the voltmeter is connected across only HALF the resistance. In this case, the voltmeter reads one-half of 6 volts or 3 volts, indicating that HALF voltage is used for HALF the resistance. Now connect two voltmeters-each one across half the resistance - as in C. Each voltmeter reads 3 volts. This gives you the law for voltage drops in a series circuit. the TOTAL VOLTAGE IN ANY SERIES CIRCUIT IS the SUM OF ALL the VOLTAGE DROPS. Mathematically the law says -

E_t = E₁ + E₂ + E₃, etc.

when -
      E_t = the total voltage;
      E₁ = the voltage drop of the first load;
      E₂ = the voltage drop of the second load;
      E₃ =the voltage drop of the third load.
In figure 37, E₃ is 0; hence,
     E_t = E₁ + E₂ + E₃, etc.
     6 v. = 3 v. + 3 v. + 0

Figure 37. - Voltage in a series circuit.

Two more examples of series circuits are shown in figure 38. In A, three equal resistors, of 4 ohms each, are connected in series. The voltmeters indicate that it requires 2 volts to force the current through each resistor. notice that the total voltage is equal to the sum of all the voltage drops -
 
      E_t = E₁ + E₂ + E₃, etc.
      E_t = 2 + 2 + 2 = 6 v.

Figure 38. - Voltages across separate loads.

In B of figure 38, two UNEQUAL resistances are connected in series. In this case, the voltage drops are not equal - THEY ARE PROPORTIONAL TO the resistance. Again -
      E_t = E₁ + E₂ + E₃, etc.
     12 v.= 3 v. + 9v.
The searchlight circuit is a good example of this principle of a series circuit. The ship's voltage is usually around 120 volts. But the standard voltage for the Navy's largest searchlight is only 80 volts. In order to reduce the 120 volts to the operating standard of 80 volts, a resistor is placed in series with the light. This resistor uses up about 40 volts, leaving a drop of 80 volts for the light. (120 v. = 40 v. + 80 v.).

current IN SERIES CIRCUITS

There is only one path for current in a series circuit. And the amount of current passing any point in the circuit is the same as the amount of current passing any other point in the circuit. This gives you the law for current in a series circuit. the current IN A SERIES CIRCUIT IS the SAME IN ALL PARTS. Or mathematically -

       I_t = I₁ = I₂= I₃, etc.
when -
      I_t = total current;
      I₁ = current through the first load;
      I₂ = current through the second load;
      I₃ = current through the third load.
Suppose you connect a lamp, a switch, and a motor in series as in figure 39. The current through each part of the circuit is the same.

Figure 39. - Control in a series circuit.

If the switch is opened the current through the switch becomes zero - open circuit. Likewise, the current through both the motor and the lamp be come zero. The opened switch "shut-off" the loads. Switches CONTROLLING electrical loads are ALWAYS in series with the loads. In the same circuit, imagine that the lamp is broken. A broken lamp is also an open circuit and would stop the current through the motor and the switch. For this reason, two loads which are intended to be INDEPENDENT of each other, are NEVER connected in series.
 

resistance IN SERIES CIRCUITS

Figure 40. - resistance in series.

Two resistances are connected in series in figure 40. Tracing the circuit, note that the current passes first through one resistor 'and then through the other. This means that the current is opposed by the force of both resistances. In figure 40, each load has a resistance of 10 ohms, but since the current passes through BOTH loads, the total opposition to current is 20 ohms. This gives you the law for resistances connected in series - the TOTAL resistance IN ANY SERIES CIRCUIT IS EQUAL TO the SUM OF ALL the INDIVIDUAL resistanceS. Mathematically this says -

      R_t = R₁ + R₂ + R₃, etc.
when -
      R_t = total resistance,
      R₁ = resistance of the first load;
      R₂ = resistance of the second load;
      R₃ = resistance of the third load.
In the problem of figure 40 -
      R_t=R₁+R₂
      20 Ω=10 Ω+10 Ω
This principle is employed in limiting the amount of current through a load by inserting resistors in series with the load. For example, the rated amperage through the lamp in figure 41 is 2 amperes. The lamp itself has 40 ohms resistance; but on a 120 volt line, 40 ohms will permit 3 amperes to pass -
      I = E/R = 120/40 = 3 amps.

Figure 41. - Controlling current by series resistance.

This means that, as long as the lamp is operated on 120 volts, it will pass 3 amperes. And 3 amperes in a 2 ampere lamp will melt the filament (burn it out) . You can see that more. than the lamp's 40 ohms of resistance is required to reduce the current to its safe value of 2 amperes. Using Ohm's law to calculate the total resistance required to limit the current to 2 amperes -
      R = E/I = 120/2 = 60 Ω
This means that a 20 ohm resistor would have to be added in series to the 40 ohms of the lamp. Then the total resistance is 60 ohms -
      R_t = R₁ + R₂
      60 Ω = 40 Ω + 20 Ω
Now the current is limited to 2 amperes.

KEEP OFF the ROCKS

Here are a few simple rules that will help keep you off the rocks -
1. Draw a schematic of your problem.
2. Label each value on the schematic that is known.
3. Determine What it is that you want to know. Is it the current, VOLTAGE, resistance, POWER, EFFICIENCY, INPUT OR OUTPUT?
4. Select the equation for your solution and use the EASIEST form of the equation. If you are looking for current, use I = E/R. If you are looking for VOLTAGE, use E = IR. If you are looking for resistance, use R = E/I.
5. (IMPORTANT) Apply the law to either a PART or the WHOLE circuit. DoN'T MIX UP YOUR VALUES. If you are looking for the resistance of a PART of a circuit, use the values or current and voltage for that PART. If you are looking for the TOTAL resistance, use the values of current and voltage for the TOTAL circuit.
6. Substitute the numerical values in your equation and solve.
You've been given a pretty big dose of formulas and equations all at once. They're hard to take - so, here's a "crutch" to help you over the rough spots. When you have a problem in current, voltage, or resistance sketch a "pie" and divide it into three pieces. Label each piece as in figure 42. Now, using your finger, cover up the quantity you want to know. What's left is the formula for solving for that quantity.
For example, if you want to know current, cover I; E/R remains, and I = E/R. If you want to know the voltage, cover E; IR remains, and E = IR. Likewise, if you want to know resistance, cover R; E/I remains, and R = E/I. You can make the same kind of "pie" for the power equation. It's shown in figure 43.

Figure 42. - Ohm's law.                    Figure 43. - Power equation.

Simple, isn't it? Let's try it on a few problems.
 

PROBLEMS IN SERIES CIRCUITS

The BEST (not always the easiest) way to get a clear understanding of circuits is to work circuit problems. You will learn a lot by going through each of the examples which follow. Each example has been selected to illustrate a practical circuit problem.

EXAMPLE 1 -
Three lamps are connected in series. Each has a resistance of 60 ohms. A generator producing 120 v. of EMF powers the circuit. What is the total current in the circuit, and what is the current through each lamp?
Rule 1. Sketch the circuit - figure 44.

Figure 44. - Example 1.

Rule 2. Label values.
Rule 3. What is wanted? I_t, I₁, I₂ and I₃.

Rule 4. Use I = E/R
Rule 5. Use I = E/R first, for the TOTAL current, and then, for the current of EACH LAMP.
Rule 6. Use I_t = E_t/R_t = 120/180 = 2/3 amp.
(for I_t)                           (R_t = R₁ + R₂ + R₃)
(for I₁, I₂ and I₃) E₁/R₁ = 40/60 = 2/3 amp. (Voltage divides in a series circuit)
Is it strange that the answers are the same? No - I_t = I₁ = I₂ = I₃ in a series circuit. Therefore, the total current is equal to the part currents. You will learn to make use of short cuts like this. But until you are sure of what you're doing, it's. best to be complete AND CORRECT. Suppose you had slipped up on Rule 5 (using a combination of a PART and the WHOLE). Say you had used the TOTAL voltage and the resistance of a PART. See what happens -
      I = E/R = 120/40 = 3 amps. ABSOLUTELY WRONG.

EXAMPLE 2 -

Figure 45. - Example 2.

Two lamps are connected in series on a 240-v. line. One lamp has 40 ohms resistance, the other has 80 ohms resistance (see figure 45). What is the total current and power? What is the current and power of each lamp?
    I_t = E_t/R_t = 240/120 = 2 amps.
    I_t = I₁ = I₂ therefore I₁ = I₂ = 2 amps.
    P_t = E_tI_t = 240 x 2 = 480 w.
The power of each of the two lamps can be found in two ways.
      (1) Find the voltage drop across each lamp -  (RF Cafe note: original had E₁ and E₂ reversed)
           E₁ = I₁ x R₁ = 2 x 40 = 80 v.
           E₂ = I₂ x R₂ = 2 x 80 = 160 v.
Then -  (RF Cafe note: original had P₁ and P₂ reversed - thanks to Lou N.)

          P₁ = E₁ x I₁ = 80 x 2 = 160 w.   (RF Cafe note: original had E₂ x I₂₎
          P₂ = E₂ x I₂ = 160 x 2 = 320 w.
OR -
      (2) In the equation P = EI, substitute for E the value of IR.
Then -(RF Cafe note: original had P₁ and P₂ reversed - thanks to Lou N.)

         P = I x IR, or P = I²R
         P₁ = I₁²R₁ = 4 x 40 = 160 w.
         P₂ = I₂²R₂ = 4 x 80 = 320 w.
 

Figure 46. - Example 3.

EXAMPLE 3 -
Figure 46 shows three 1,000 watt heaters connected in series to a 240-volt battery. Each heater has a current of 12.5 amperes. (1) What is the total resistance? (2) resistance of each heater? (-3) Total current? (4) Total power?
      (1) R_t = E_t/I_t = 240/12.5 = 19.2 Ω
      (2) R₁ = P₁/I₁² = 1,000/(12.5 x 12.5) = 1,000/156.25 = 6.4 Ω
           R₁ = R₂ = R₃ (equal power - all are 1,000 w.)
           R₁ = R₂ = R₃ = 6.4 Ω
      (3) I_t = I₁ = I₂ = I₃ = 12.5 amps.
      (4) P₁ = E₁I₁ = 240 x 12.5 = 3,000 w.
           or P_t = P₁ + P₂ + P₃ = 1,000 + 1,000 + 1,000 = 3,000 w.

Chapter 8 Quiz

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