The discussion of electrical circuits presented up to this point has been concerned with series circuits in which there is only one path for current. There is another basic type of circuit known as the PARALLEL CIRCUIT with which you must become familiar. Where the series circuit has only one path for current, the parallel circuit has more than one path for current.
Ohm’s law and Kirchhoff’s law apply to all electrical circuits, but the characteristics of a parallel dc circuit are different than those of a series dc circuit.
1. Parallel Circuit Characteristics
A PARALLEL CIRCUIT is defined as one having more than one current path connected to a common voltage source. Parallel circuits, therefore, must contain two or more resistances which are not connected in series. An example of a basic parallel circuit is shown in Figure 1.
Start at the voltage source (E_{s}) and trace counterclockwise around the circuit. Two complete and separate paths can be identified in which current can flow. One path is traced from the source, through resistance R_{1}, and back to the source. The other path is from the source, through resistance R_{2}, and back to the source.
1.1. Voltage in a Parallel Circuit
You have seen that the source voltage in a series circuit divides proportionately across each resistor in the circuit. IN A PARALLEL CIRCUIT, THE SAME VOLTAGE IS PRESENT IN EACH BRANCH. (A branch is a section of a circuit that has a complete path for current.) In Figure 1 this voltage is equal to the applied voltage (E_{s}). This can be expressed in equation form as:
E_{S} = E_{R1} = E_{R2}
Voltage measurements taken across the resistors of a parallel circuit, as illustrated by Figure 2 verify this equation. Each meter indicates the same amount of voltage. Notice that the voltage across each resistor is the same as the applied voltage.
Example: Assume that the current through a resistor of a parallel circuit is known to be 4.5 milliamperes (4.5 mA) and the value of the resistor is 30,000 ohms (30 kΩ). Determine the source voltage. The circuit is shown in Figure 3.
Given:
Solution:
Since the source voltage is equal to the voltage of a branch:
To simplify the math operation, the values can be expressed in powers of ten as follows:
If you are not familiar with the use of the powers of 10 or would like to brush up on it, Mathematics, Vol. 1, NAVEDTRA 10069C, will be of great help to you.
1.2. 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 of current is determined by the total resistance of the circuit and the applied voltage. In a parallel circuit the source current divides among the available paths.
The behavior of current in parallel circuits will be shown by a series of illustrations using example circuits with different values of resistance for a given value of applied voltage.
Part (A) of Figure 4 shows a basic series circuit. Here, the total current must pass through the single resistor. The amount of current can be determined.
Given:
Solution:
Part (B) of Figure 4 shows the same resistor (R_{1}) with a second resistor (R_{2}) of 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).
Given:
Solution:
It is apparent that if there is 5 amperes of current through each of the two resistors, there must be a TOTAL CURRENT of 10 amperes drawn from the source.
The total current of 10 amperes, as illustrated in Figure 4(B), leaves the negative terminal of 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 of 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 of the source. The source supplies a total current of 10 amperes and each of the two equal resistors carries onehalf the total current.
Each individual current path in the circuit of Figure 4(B) is referred to as a BRANCH. Each branch carries a current that is a portion of the total current. Two or more branches form a NETWORK.
From the previous explanation, the characteristics of current in a parallel circuit can be expressed in terms of the following general equation:
I_{T} = I_{1} + I_{2} + . . . I_{n}
Compare part (A) of Figure 5 with part (B) of the circuit in Figure 4. Notice that doubling the value of the second branch resistor (R_{2}) has no effect on the current in the first branch (IR_{1}), but does reduce the second branch current (IR_{2}) to onehalf its original value. The total circuit current drops to a value equal to the sum of the branch currents. These facts are verified by the following equations.
Given:
Solution:
The amount of current flow in the branch circuits and the total current in the circuit shown in Figure 5(B) are determined by the following computations.
Given:
Solution:
Notice that the sum of the ohmic values in each circuit shown in Figure 5 is equal (30 ohms), and that the applied voltage is the same (50 volts). However, the total current in 341(B) (15 amps) is twice the amount in 341(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 of current in a parallel network follows a definite pattern. This pattern is described by KIRCHHOFF’S CURRENT LAW which states:
"The algebraic sum of the currents entering and leaving any junction of conductors is equal to zero."
This law can be stated mathematically as:
I_{a} + l_{b} + . . . I_{n} + 0
where: I_{a}, I_{b}, 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 SIGNS ATTACHED.
Example: Solve for the value of I3 in Figure 6.
Given:
Solution:
I_{a} + l_{b} + . . . I_{a} + 0
The currents are placed into the equation with the proper signs.
I_{3} has a value of 2 amperes, and the negative sign shows it to be a current LEAVING the junction.
Example. Using Figure 7, solve for the magnitude and direction of I_{3}.
Given:
Solution:
I_{3} is 2 amperes and its positive sign shows it to be a current entering the junction.
1.3. Resistance in a Parallel Circuit
In the example diagram, Figure 8, there are two resistors connected in parallel across a 5volt battery. Each has a resistance value of 10 ohms. A complete circuit consisting of two parallel paths is formed and current flows as shown.
Computing the individual currents shows that there is onehalf of an ampere of current through each resistance. The total current flowing from the battery to the junction of the resistors, and returning from the resistors to the battery, is equal to 1 ampere.
The total resistance of the circuit can be calculated by using the values of total voltage (ET) and total current (I_{T}).
Note

From this point on the abbreviations and symbology for electrical quantities will be used in example problems. 
Given:
Solution:
This computation shows the total resistance to be 5 ohms; onehalf the value of either of the two resistors.
Since the total resistance of a parallel circuit is smaller than any of the individual resistors, total resistance of a parallel circuit is not the sum of the individual resistor values as was the case in a series circuit. The total resistance of resistors in parallel is also referred to as EQUIVALENT RESISTANCE (R_{eq}). The terms total resistance and equivalent resistance are used interchangeably.
There are several methods used to determine the equivalent resistance of parallel circuits. The best method for a given circuit depends on the number and value of the resistors. For the circuit described above, where all resistors have the same value, the following simple equation is used:
This equation is valid for any number of parallel resistors of EQUAL VALUE.
Example: Four 40ohm resistors are connected in parallel. What is their equivalent resistance?
Given:
Solution:
Figure 9 shows two resistors of unequal value in parallel. Since the total current is shown, the equivalent resistance can be calculated.
Given:
Solution:
The equivalent resistance of the circuit shown in Figure 9 is smaller than either of the two resistors (R_{1}, R_{2}). An important point to remember is that the equivalent resistance of a parallel circuit is always less than the resistance of any branch.
Equivalent resistance can be found if you know the individual resistance values and the source voltage. By calculating each branch current, adding the branch currents to calculate total current, and dividing the source voltage by the total current, the total can be found. This method, while effective, is somewhat lengthy. A quicker method of finding equivalent resistance is to use the general formula for resistors in parallel:
If you apply the general formula to the circuit shown in Figure 9 you will get the same value for equivalent resistance (2Ω) as was obtained in the previous calculation that used source voltage and total current.
Given:
Solution:
Convert the fractions to a common denominator.
Since both sides are reciprocals (divided into one), disregard the reciprocal function.
The formula you were given for equal resistors in parallel
is a simplification of the general formula for resistors in parallel
There are other simplifications of the general formula for resistors in parallel which can be used to calculate the total or equivalent resistance in a parallel circuit.
RECIPROCAL METHOD.—This method is based upon taking the reciprocal of each side of the equation. This presents the general formula for resistors in parallel as:
This formula is used to solve for the equivalent resistance of a number of unequal parallel resistors. You must find the lowest common denominator in solving these problems. If you are a little hazy on finding the lowest common denominator, brush up on it in Mathematics Volume 1, NAVEDTRA 10069 (Series).
Example: Three resistors are connected in parallel as shown in Figure 10. The resistor values are: R_{1} = 20 ohms, R_{2} = 30 ohms, R_{3} = 40 ohms. What is the equivalent resistance? (Use the reciprocal method.)
Given:
Solution:
PRODUCT OVER THE SUM METHOD.—A convenient method for finding the equivalent, or total, resistance of two parallel resistors is by using the following formula.
This equation, called the product over the sum formula, is used so frequently it should be committed to memory.
Example: What is the equivalent resistance of a 20ohm and a 30ohm resistor connected in parallel, as in Figure 11?
Given:
Solution:
1.4. Power in a Parallel Circuit
Power computations in a parallel circuit are essentially the same as those used for the series circuit. Since power dissipation in resistors consists of a heat loss, power dissipations are additive regardless of how the resistors are connected in the circuit. The total power is equal to the sum of the power dissipated by the individual resistors. Like the series circuit, the total power consumed by the parallel circuit is:
Example: Find the total power consumed by the circuit in Figure 12.
Given:
Solution:
Since the total current and source voltage are known, the total power can also be computed by:
Given:
Solution:
1.5. Equivalent Circuits
In the study of electricity, it is often necessary to reduce a complex circuit into a simpler form. Any complex circuit consisting of resistances can be redrawn (reduced) to a basic equivalent circuit containing the voltage source and a single resistor representing total resistance. This process is called reduction to an EQUIVALENT CIRCUIT.
Figure 13 shows a parallel circuit with three resistors of equal value and the redrawn equivalent circuit. The parallel circuit shown in part A shows the original circuit. To create the equivalent circuit, you must first calculate the equivalent resistance.
Given:
Solution:
Once the equivalent resistance is known, a new circuit is drawn consisting of a single resistor (to represent the equivalent resistance) and the voltage source, as shown in part B.
1.6. Rules for Parallel DC Circuits

The same voltage exists across each branch of a parallel circuit and is equal to the source voltage.

The current through a branch of a parallel network is inversely proportional to the amount of resistance of the branch.

The total current of a parallel circuit is equal to the sum of the individual branch currents of the circuit.

The total resistance of a parallel circuit is found by the general formula: or one of the formulas derived from this general formula.

The total power consumed in a parallel circuit is equal to the sum of the power consumptions of the individual resistances.
2. Solving Parallel Circuit Problems
Problems involving the determination of resistance, voltage, current, and power in a parallel circuit are solved as simply as in a series circuit. The procedure is the same — (1) draw the circuit diagram, (2) state the values given and the values to be found, (3) select the equations to be used in solving for the unknown quantities based upon the known quantities, and (4) substitute the known values in the equation you have selected and solve for the unknown value.
Example: A parallel circuit consists of five resistors. The value of each resistor is known and the current through R_{1} is known. You are asked to calculate the value for total resistance, total power, total current, source voltage, the power used by each resistor, and the current through resistors R_{2}, R_{3}, R_{4}, and R_{5}.
Given:
Find:
This may appear to be a large amount of mathematical manipulation. However, if you use the step bystep approach, the circuit will fall apart quite easily.
The first step in solving this problem is for you to draw the circuit and indicate the known values as shown in Figure 14.
There are several ways to approach this problem. With the values you have been given, you could first solve for R_{T}, the power used by R_{1}, or the voltage across R_{1}, which you know is equal to the source voltage and the voltage across each of the other resistors. Solving for R_{T} or the power used by R_{1} will not help in solving for the other unknown values.
Once the voltage across R_{1} is known, this value will help you calculate other unknowns. Therefore the logical unknown to solve for is the source voltage (the voltage across R_{1}).
Given:
Solution:
Now that source voltage is known, you can solve for current in each branch.
Given:
Solution:
Since R_{3} = R_{4} = R_{5} and the voltage across each branch is the same:
Solving for total resistance.
Given:
Solution:
An alternate method for solving for R_{T} can be used. By observation, you can see that R_{3}, R_{4}, and R_{5} are of equal ohmic value. Therefore an equivalent resistor can be substituted for these three resistors in solving for total resistance.
Given:
Solution:
The circuit can now be redrawn using a resistor labeled Req1 in place of R_{3}, R_{4}, and R_{5} as shown in Figure 15.
An equivalent resistor can be calculated and substituted for R_{1} and R_{2} by use of the product over the sum formula.
Given:
Solution:
The circuit is now redrawn again using a resistor labeled Req2 in place of R_{1} and R_{2} as shown in Figure 16.
You are now left with two resistors in parallel. The product over the sum method can now be used to solve for total resistance.
Given:
Solution:
This agrees with the solution found by using the general formula for solving for resistors in parallel.
The circuit can now be redrawn as shown in Figure 17 and total current can be calculated.
Given:
Solution:
This solution can be checked by using the values already calculated for the branch currents.
Given:
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Solution: 1601 dry run mount hope road
Now that total current is known, the next logical step is to find total power.
Given:
Solution:
Solving for the power in each branch.
Given:
Solution:
Since I_{R3} = I_{R4} = I_{R5} then, P_{R3} = P_{R4} = P_{R5} = 1800 W. The previous calculation for total power can now be checked.
Given:
Solution: