09.1 Inductors in Parallel and Series Practice Problems
INDUCTORS IN PARALLEL AND SERIES | PRACTICE PROBLEMS Complete the following problems to reinforce your understanding of the concept covered in this module.
PROBLEM 1: What is the total equivalent inductance for the circuit shown below?
12|PREPINEER.COM
psst…don’t forget to take notesÎ
PROBLEM 2: What is the total equivalent inductance for the circuit shown below?
PROBLEM 3: Two inductors of 10 H and 20 H are connected in series; two other inductors of 30 H and 40 H are also connected in series. What is the total equivalent inductance of the circuit if these series combinations are connected in parallel on a circuit?
13|PREPINEER.COM
psst…don’t forget to take notesÎ
INDUCTORS IN PARALLEL AND SERIES | SOLUTIONS
SOLUTION 1: 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 capacitor or total equivalent inductance.
First, we will simplify the
L 3 =4 H and the L 4 =4 H inductors that 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 value:
LP
1 1 1 + L1 L2
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 4 H and 4 H, we calculate the first equivalent inductance value: 14|PREPINEER.COM
psst…don’t forget to take notesÎ
1 L EQ,1
=
1 1 + 4H 4H
L EQ,1 =2 H
Hint: Use the equation solve function on your calculator to avoid having to do various algebra calculations and fraction manipulations by hand. This is helpful for saving time and avoiding simple algebra calculation errors. 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. Since we converted all of the inductors that were in parallel within the circuit, the remaining inductors are all in series. We will use for the formula for inductors in series to calculate the second and total equivalent inductance value of the circuit:
L EQ,2 = L1 + L 2 + L EQ,1 + L 5 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.
15|PREPINEER.COM
psst…don’t forget to take notesÎ
Given the inductor values of 6 H, 2 H, 5 H and the first calculated equivalent inductance value of 2 H, we can calculate the second and total equivalent inductance value of the circuit using the formula for the equivalent inductance of inductors in series:
L EQ,2 =6 H+2 H+2 H+5 H L EQ,2 =15 H
SOLUTION 2: 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 capacitor or total equivalent inductance. We will begin the problem with the 20 H, 12 H, and 10 H inductors that are in series with each other. We will use for the formula for inductors in series to calculate the first equivalent inductance value.
L S = L1 + L 2 + L 3
16|PREPINEER.COM
psst…don’t forget to take notesÎ
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 20 H, 12 H, 10 H, we can calculate the first equivalent inductance value using the formula for the equivalent inductance of inductors in series:
L EQ,1 =20 H+12 H+10 H L EQ,1 =42 H Next, we will simplify the first inductance equivalence we just calculated and the 7 H inductor that is parallel with it. We will use for the formula for inductors in parallel to calculate the second equivalent inductance value.
LP
1 1 1 + L1 L2
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 value of 7 H and calculated first equivalent inductance value of 42 H, we can calculate the second equivalent inductance value: 17|PREPINEER.COM
psst…don’t forget to take notesÎ
1 L EQ,2
=
1 1 + 42 H 7 H
L EQ,2 =6 H
Hint: Use the equation solve function on your calculator to avoid having to do various algebra calculations and fraction manipulations by hand. This is helpful for saving time and avoiding simple algebra errors. Note: We could have also used the formula for the product-over-the-rule to calculate the equivalent inductance value. 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. Since we converted all the inductors that were in parallel in the circuit, the remaining inductors are all in series. The remaining inductors that are in series are: We will use for the formula for inductors in series to calculate the third and total equivalent inductance value of the circuit:
L S = L1 + L 2 + L 3 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.
18|PREPINEER.COM
psst…don’t forget to take notesÎ
Given the inductor values of 4 H, 8 H and the second calculated equivalent inductance value of 6 H, we can calculate the third and total equivalent inductance value:
L EQ,3 =4 H+6 H+8 H L EQ,3 =18 H
SOLUTION 3: 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 capacitor or total equivalent inductance. We will begin the problem by simplifying the 10 H and the 20 H inductors that are in series with each other. We will use for the formula for inductors in series to calculate the first equivalent inductance value:
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. 19|PREPINEER.COM
psst…don’t forget to take notesÎ
Given the inductor values of 10 H and 20 H , we can calculate the first equivalent inductance value:
L EQ,1 =10 H+20 H L EQ,1 =30 H Next, we will simplify the first inductance equivalence we just calculated and the 40 H inductor that is in series with it. We will use for the formula for inductors in series to calculate the second equivalent inductance value:
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 30 H and 40 H , we can calculate the second equivalent inductance value:
L EQ,2 =30 H+40 H L EQ,2 =70 H
20|PREPINEER.COM
psst…don’t forget to take notesÎ
Lastly, we will simplify the first and second equivalent inductances we calculated. These inductors are in parallel with each other. We will use for the formula for inductors in parallel to calculate the third and total equivalent inductance value of the circuit:
LP
1 1 1 + L1 L2
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 the calculated equivalent inductance values of 30 H and 70 H, we calculate the third and total equivalent inductance value of the circuit:
1 L EQ,3
=
1 1 + 30 H 70 H
L EQ,3 =21 H
Hint: Use the equation solve function on your calculator to avoid having to do various algebra calculations and fraction manipulations by hand. This is helpful for saving time and avoiding simple algebra errors. 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 21|PREPINEER.COM
psst…don’t forget to take notesÎ
recommend using both formulas when studying so you are able to efficiently check your work on the exam.
22|PREPINEER.COM
psst…don’t forget to take notesÎ
