09 Inductors in Parallel and Series FINAL
INDUCTORS IN PARALLEL AND SERIES | CONCEPT OVERVIEW The topic of INDUCTORS IN PARALLEL AND SERIES can be referenced on page 201 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. Capacitors, inductors, and resistors can all be present within the same circuit. Similar to resistors and capacitors, inductors can be combined together to calculate an equivalent inductance. Inductors that are in PARALLEL can be combined together, using the formula that is similar to calculating the equivalent resistance of resistors in parallel.
To calculate the equivalent inductance of a circuit with inductors in parallel, we would use the formula:
LP
1 1 1 1 1 + + + L1 L2 L 3 L 4
1|PREPINEER.COM
psst…don’t forget to take notesÎ
The formula for INDUCTORS IN PARALLEL can be referenced under the topic of CAPACITORS AND INDUCTORS IN SERIES AND PARALLEL on page 201 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. If there are only 2 inductors that are in parallel, we can calculate the equivalent inductance by using the formula for the PRODUCT-OVER-THE-SUM-RULE. This formula is helpful in saving time and reducing errors as it is easier to follow and is more concise.
LT=
L1 x L2 L 1 +L 2
Where: • L 1 and L 2 are the inductance values of the two inductors in the circuit that are in parallel.
It is important to remember that we can use this formula multiple times within a circuit to calculate multiple equivalences, as long as we are able to group 2 inductors that are in parallel at a time. This formula will not work if we are trying to find the equivalent inductance of more than 2 inductors. If you are trying to calculate the equivalence of more than 2 inductors, you must use the formula for inductors in parallel from the Reference Handbook. Note: This formula is not provided in the Reference Handbook.
2|PREPINEER.COM
psst…don’t forget to take notesÎ
Inductors that are in SERIES can be summed together to calculate the equivalent inductance of a circuit. This would be similar to calculating the equivalent resistance of resistors in series.
To calculate the equivalent inductance of a circuit with inductors in series, we would use the formula:
L S = L1 + L 2 + L 3 + L 4 The formula for INDUCTORS IN SERIES can be referenced under the topic of CAPACITORS AND INDUCTORS IN SERIES AND PARALLEL on page 201 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing.
3|PREPINEER.COM
psst…don’t forget to take notesÎ
CONCEPT EXAMPLE: What is the equivalent inductance for the circuit shown below?
SOLUTION: In this problem we are looking to calculate the equivalent or total inductance of the circuit. We will need to identify pairs or groups of inductors that we can combine and simplify until we are able to simplify the circuit to have one inductor or total equivalent inductance. When working with like circuits like this one, it works best if you choose a side of the circuit to start on, and work your way from that reference point. For this problem, we are going to start on the right side of the circuit and work our way left.
4|PREPINEER.COM
psst…don’t forget to take notesÎ
We will begin the problem with the 14 H inductor on the right side of the circuit. The 14 H inductor and 6 H inductor are in parallel with each other. We will use the formula for inductors that are in parallel with each other to calculate the first equivalent inductance:
LP
1 1 1 1 1 + + + L1 L2 L 3 L 4
The formula for INDUCTORS IN PARALLEL can be referenced under the topic of CAPACITORS AND INDUCTORS IN SERIES AND PARALLEL on page 201 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. Given inductor values of 14 H and 6 H, we calculate the first equivalent inductance value:
1 L EQ,1
=
1 1 + 14 H 6 H
L EQ,1 =4.2 H
Hint: Use the equation solve function on your calculator to avoid using having to do various algebra calculations and fraction manipulations by hand. This is helpful for saving time and avoiding simple algebra errors.
5|PREPINEER.COM
psst…don’t forget to take notesÎ
Note: We could have also used the formula for the product-over-the-rule to calculate the equivalent inductance. Either way will get you the correct answer, but we recommend using both formulas when studying so you are able to efficiently check your work on the exam. For the rest of this problem, we will be using the product-over-the-sum-rule formula for inductors in parallel, but, again, you can use either formula. Next, we will use the calculated inductance equivalence and the 15.8 H inductor to further simplify the circuit. These inductors are in series, so we will use the following formula:
L S = L1 + L 2 The formula for INDUCTORS IN SERIES can be referenced under the topic of CAPACITORS AND INDUCTORS IN SERIES AND PARALLEL on page 201 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. Given the inductor values of 4.2 H and the first equivalent inductance value of 15.8 H, we can calculate the second equivalent inductance value:
L EQ,2 =15.8 H+4.2 H
L EQ,2 =20 H
6|PREPINEER.COM
psst…don’t forget to take notesÎ
Next, we will simplify the second inductance equivalence we just calculated and the 60 H inductor that is parallel with it. As there are 2 inductors that are parallel with each other, we can calculate the third equivalent inductance value of the circuit by using the PRODUCT-OVER-THE-SUMRULE.
LT=
L1 x L2 L 1 +L 2
Note: This formula is not provided in the Reference Handbook. Given the inductor values of 60 H and the second equivalent inductance value of 20 H, we can calculate the third equivalent inductance of the circuit:
L EQ,3 =
20 H x 60 H 20 H + 60 H
L EQ,3 =15 H
Next, we will simplify the third inductance equivalence we just calculated and the 5 H inductor that is in series with it. We will use for the formula for inductors in series to calculate the fourth equivalent inductance value.
7|PREPINEER.COM
psst…don’t forget to take notesÎ
L S = L1 + L 2
The formula for INDUCTORS IN SERIES can be referenced under the topic of CAPACITORS AND INDUCTORS IN SERIES AND PARALLEL on page 201 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. Given the inductor values of 5 H and the third equivalent inductance value of 15 H, we can calculate the fourth equivalent inductance:
L EQ,4 =15 H+5 H
L EQ,4 =20 H
Next, we will simplify the fourth inductance equivalence we just calculated and the 80 H inductor that is in parallel with it. As the 2 inductors are in parallel with each other, we can calculate the fifth equivalent inductance value of the circuit by using the PRODUCT-OVER-THE-SUM-RULE.
LT=
L1 x L2 L 1 +L 2
Note: This formula is not provided in the Reference Handbook. Given the inductor values of 20 H and 80 H, we can calculate the fifth equivalent inductance value: 8|PREPINEER.COM
psst…don’t forget to take notesÎ
L EQ,5 =
20 H x 80 H 20 H + 80 H
L EQ,5 =16 H
Next, we will simplify the fifth inductance equivalence we just calculated and the 24 H inductor that is in series with it. We will use for the formula for inductors in series to calculate the sixth equivalent inductance.
L S = L1 + L 2 The formula for INDUCTORS IN SERIES can be referenced under the topic of CAPACITORS AND INDUCTORS IN SERIES AND PARALLEL on page 201 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. Given the inductor values of 16 H and the fifth equivalent inductance value of 24 H, we can calculate the sixth equivalent inductance:
L EQ,6 =16 H+24 H
L EQ,6 =40 H
9|PREPINEER.COM
psst…don’t forget to take notesÎ
Next, we will simplify the sixth inductance equivalence we just calculated and the 10 H inductor that is in parallel with it. As the 2 inductors are in parallel with each other, we can calculate the seventh equivalent inductance value of the circuit by using the PRODUCT-OVER-THE-SUMRULE.
LT=
L1 x L2 L 1 +L 2
Note: This formula is not provided in the Reference Handbook. Given the inductor values of 20 H and 80 H, we calculate the seventh equivalent inductance of the circuit:
L EQ,7 =
40 H x 10 H 40 H+10 H
L EQ,7 =8 H Lastly, we will simplify the seventh inductance equivalence we just calculated and the 12 H inductor that is in series with it. We will use for the formula for inductors in series to calculate the eighth and total equivalent inductance of the circuit.
10|PREPINEER.COM
psst…don’t forget to take notesÎ
L S = L1 + L 2
The formula for INDUCTORS IN SERIES can be referenced under the topic of CAPACITORS AND INDUCTORS IN SERIES AND PARALLEL on page 201 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. Given the inductor values of 12 H and the seventh equivalent inductance value of 8 H, we can calculate the eighth and final equivalent inductance value:
L EQ,8 =12 H+8 H L EQ,8 =20 H
11|PREPINEER.COM
psst…don’t forget to take notesÎ
