Navy Electricity and Electronics Training Series (NEETS)
1—Introduction to Matter, Energy, and Direct Current
Pages 3-51 through 3-60
Module 1—Introduction to Matter, Energy, and Direct Current
i - ix
, 1-1 to 1-10
1-11 to 1-20
, 1-21 to 1-30
1-41 to 1-50
,1-51 to 1-60
1-61 to 1-65
, 2-1 to 2-10
2-11 to 2-20
, 2-21 to 2-29
3-1 to 3-10
, 3-11 to 3-20
3-21 to 3-30
, 3-31 to 3-40
3-41 to 3-50
,3-51 to 3-60,
3-61 to 3-70
, 3-71 to 3-80
3-81 to 3-90
, 3-91 to 3-100
3-101 to 110
, 3-111 to 3-120
3-121 to 3-126
Since the source
voltage is equal to the voltage a branch:
To simplify the math operation, the values can be expressed in powers ten as follows:
If you are not familiar with the use the powers 10 or would like to brush up on it, Mathematics, Vol.
1, NAVEDTRA 10069-C, will be great help to you.
Q33. What would the source voltage (ES) in figure 3-39 be
if the current through R2 were 2 milliamps?
Current in a Parallel Circuit
Ohm’s law states that the current in a circuit is inversely proportional to the circuit resistance. This fact
is true in both series and parallel circuits.
There is a single path for current in a series circuit. The
amount current is determined by the total resistance the circuit and the applied voltage. In a parallel circuit
the source current divides among the available paths.
The behavior current in parallel circuits will be
shown by a series illustrations using example circuits with different values resistance for a given value applied
Part (A) figure 3-40 shows a basic series circuit. Here, the total current must pass through the single
resistor. The amount current can be determined.
Figure 3-40.—Analysis current in parallel circuit.
Part (B) figure 3-40 shows the same resistor (R1) with a second resistor (R2) equal value connected in
parallel across the voltage source. When Ohm’s law is applied, the current flow through each resistor is found to
be the same as the current through the single resistor in part (A).
It is apparent that if there is 5 amperes current through each the two resistors, there must be a TOTAL
CURRENT 10 amperes drawn from the source.
The total current 10 amperes, as illustrated in figure 3-40(B),
leaves the negative terminal the battery and flows to point a. Since point a is a connecting point for the two
resistors, it is called a JUNCTION. At junction a, the total current divides into two currents 5 amperes each.
These two currents flow through their respective resistors and rejoin at junction b. The total current then flows
from junction b back to the positive terminal the source. The source supplies a total current 10 amperes and each
the two equal resistors carries one-half the total current.
Each individual current path in the circuit
figure 3-40(B) is referred to as a BRANCH. Each branch carries a current that is a portion the total current. Two
or more branches form a NETWORK.
From the previous explanation, the characteristics current in a parallel circuit can be expressed in terms
the following general equation:
IT = I1 + I 2 + . . . In
Compare part (A) figure 3-41 with part (B) the circuit in figure 3-40. Notice that doubling the value
the second branch resistor (R2) has no effect on the current in the first branch (IR1), but does reduce the second
branch current (IR2) to one-half its original value. The total circuit current drops to a value equal to the sum
the branch currents. These facts are verified by the following equations.
Figure 3-41.—Current behavior in parallel circuits.
The amount current flow in the branch circuits and the total current in the circuit shown in figure
3-41(B) are determined by the following computations.
Notice that the sum the ohmic values in each circuit shown in figure 3-41 is equal (30 ohms), and that
the applied voltage is the same (50 volts). However, the total current in 3-41(B) (15 amps) is twice the amount in
3-41(A) (7.5 amps). It is apparent, therefore, that the manner in which resistors are connected in a circuit, as
well as their actual ohmic values, affect the total current.
The division current in a parallel network
follows a definite pattern. This pattern is described by Kirchhoff's CURRENT LAW which states:
“The algebraic sum the currents entering and leaving any junction conductors is equal to zero.“ This
law can be stated mathematically as:
Ia + Ib + . . . I n + O
where: Ia, Ib, etc., are the currents entering and leaving the junction. Currents ENTERING the junction
are considered to be POSITIVE and currents LEAVING the junction are considered to be NEGATIVE. When solving a
problem using Kirchhoff's current law, the currents must be placed into the equation WITH THE PROPER POLARITY
Example: Solve for the value I3 in figure 3-42. Given:
Ia + lb + . . . I a + O
Figure 3-42.—Circuit for example problem.
The currents are placed into the equation with the proper signs.
13 has a value 2 amperes, and the negative sign shows it to be a current LEAV1NG the junction. Example.
Using figure 3-43, solve for the magnitude and direction 13.
Figure 3-43.—Circuit for example problem.
I3 is 2 amperes and its positive sign shows it to be a current entering the junction.
There is a relationship between total current and current through the individual components in a circuit. What is
this relationship in a series circuit and a parallel circuit?
Q35. In applying Kirchhoff's current law,
what does the polarity the current indicate?
Resistance in a Parallel Circuit
In the example diagram, figure 3-44, there are two resistors connected in parallel across a 5-volt battery.
Each has a resistance value 10 ohms. A complete circuit consisting two parallel paths is formed and current flows
Figure 3-44.—Two equal resistors connected in parallel.
Computing the individual currents shows that there is one-half an ampere current through each
resistance. The total current flowing from the battery to the junction the resistors, and returning from the
resistors to the battery, is equal to 1 ampere.
The total resistance the circuit can be calculated by
using the values total voltage (ET) and total current (IT).
NOTE: From this point on the abbreviations and
symbology for electrical quantities will be used in example problems.
This computation shows the total resistance to be 5 ohms; one-half the value either the two resistors.
Since the total resistance a parallel circuit is smaller than any the individual resistors, total resistance a
parallel circuit is not the sum the individual resistor values as was the case in a series circuit. The total
resistance resistors in parallel is also referred to as EQUIVALENT RESISTANCE (Req). The terms total resistance
and equivalent resistance are used interchangeably.
There are several methods used to determine the
equivalent resistance parallel circuits. The best method for a given circuit depends on the number and value the
resistors. For the circuit described above, where all resistors have the same value, the following simple equation
This equation is valid for any number parallel resistors EQUAL VALUE.
Example: Four 40-ohm
resistors are connected in parallel. What is their equivalent resistance? Given:
Figure 3-45 shows two resistors unequal value in parallel. Since the total current is shown, the
equivalent resistance can be calculated.
Introduction to Matter, Energy, and Direct Current,
to Alternating Current and Transformers, Introduction to Circuit Protection,
Control, and Measurement
, Introduction to Electrical Conductors, Wiring Techniques,
and Schematic Reading
, Introduction to Generators and Motors
Introduction to Electronic Emission, Tubes, and Power Supplies,
Introduction to Solid-State Devices and Power Supplies
Introduction to Amplifiers, Introduction to
Wave-Generation and Wave-Shaping Circuits
, Introduction to Wave Propagation, Transmission
Lines, and Antennas
, Microwave Principles,
, Introduction to Number Systems and Logic Circuits, Introduction
to Microelectronics, Principles of Synchros, Servos, and Gyros
Introduction to Test Equipment
, Radar Principles,
The Technician's Handbook,
Master Glossary, Test Methods and Practices,
Introduction to Digital Computers,
Magnetic Recording, Introduction to Fiber Optics