04 Inductance Review
INDUCTANCE | CONCEPT OVERVIEW The topic of INDUCTANCE can be referenced on page 201 of the NCEES Supplied Reference Handbook, 9.3 version for Computer Based Testing. INDUCTANCE (L) represents the property of an inductor resisting change in current flow through the coil. INDUCTANCE is given in units of HENRIES (H).
INDUCTORS store energy in MAGNETIC FIELDS, in comparison to capacitors that store energy in electric fields. The current through an inductor cannot change abruptly. When the current is CHANGING, a voltage is present, but when the current is CONSTANT, a voltage is not present. FARADAY’S LAW governs the electrical behavior of an inductor, relating VOLTAGE across the terminals with the RATE OF CHANGE of the current flowing through the inductor. 1|PREPINEER.COM
psst… don’t forget to take notesÎ
𝑣! 𝑡 = 𝐿 𝑑𝑖! /𝑑𝑡 The formula for FARADY’S LAW can be referenced under the topic of CAPACITORS AND INDUCTORS on page 201 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. When given a time interval, we can solve for current through integration using the formula for the VOLTAGE-CURRENT RELATIONSHIP OF FARADAY’S LAW.
1 𝑖! 𝑡 = 𝑖! 0 + 𝐿
!
𝑣! 𝜏 𝑑𝜏 !
where • 𝑣! 𝑡 = inductor voltage, V • 𝐿 = inductance, which is a constant of proportionality and is in henrys, H • 𝑖! 𝑡 = inductor current in amperes, A • 𝑖! 0 = the current at time zero, and is called a boundary condition The formula for the VOLTAGE-CURRENT RELATIONSHIP OF AN INDUCTOR can be referenced under the topic of CAPACITORS AND INDUCTORS on page 201 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. It is important to note that we may need to use INTEGRATION when evaluating these formulas as they contain variables and equations as a function of time. 2|PREPINEER.COM
psst… don’t forget to take notesÎ
CONCEPT EXAMPLE: A 5 mH inductor has a current 𝑖(𝑡) as a function of time into its positive terminal, and a voltage
𝑣(𝑡)
as
a
function
of
time
across
its
terminals.
Find
𝑖(𝑡)
if
𝑣 𝑡 = 400 cos 200𝑡 V and 𝑖(0) = 400 A.
SOLUTION: In this problem we are given the voltage as a function of time and are asked to solve for the current. We can use the formula for the voltage-current relationship of an inductor to solve for the current, as we can integrate the voltage functions and obtain the current as a function of time.
𝑖! 𝑡 = 𝑖! 𝑡!
1 + 𝐿
1 𝑖! 𝑡 = 𝑖! 0 + 𝐿
!
𝑣! 𝜏 𝑑𝜏 !! !
𝑣! 𝜏 𝑑𝜏 !
The formula for the VOLTAGE-CURRENT RELATIONSHIP OF AN INDUCTOR can be referenced under the topic of CAPACITORS AND INDUCTORS on page 201 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. In the problem we are given: 3|PREPINEER.COM
psst… don’t forget to take notesÎ
• Inductance, 𝐿 = 5 mH = 5×10! H • The voltage as a function of time, 𝑣 𝑡 = 400 cos 200𝑡 V • The boundary condition for the current at time 𝑡 = 0, 𝑖(0) = 400 A We now plug in the function for voltage as function of time, the value for inductance, and the current at time zero into the formula for the voltage-current relationship of an inductor.
1 𝑖 𝑡 = 400 A + 5×10! H
!
400 cos 200𝑡 𝑑𝑡 !
We then evaluate the integral using the properties of integration. 𝑖 𝑡 = 400 A + 400 cos 200𝑡 |!! The next step is to evaluate the integral with the specified boundary conditions for 0 to 𝑡. 𝑡 = 400 A + 400 cos 200𝑡 − 0 We then simplify the function to obtain our function for current as a function of time. 𝑡 = 400 + 400 cos 200𝑡 A
4|PREPINEER.COM
psst… don’t forget to take notesÎ
STORED ENERGY OF AN INDUCTOR The topic of INDUCTORS can be referenced on page 201 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. As current passes through an inductor, a MAGNETIC FIELD is created that stores energy. As the current increases, energy is stored in the magnetic field, and as the current decreases the stored energy is released adding to the current. Energy is NOT DISSIPATED by an inductor, but only stored. In comparison, a capacitor stored and dissipates energy. The energy stored in the inductor is given by Energy = 𝐿𝑖!! /2 Where • The energy stored in an inductor given in units of Joules, J • 𝐿 = the inductance given in Henrys, H • 𝑖! = the current given in amperes, A The formula for the ENERGY STORED IN AN INDUCTOR can be referenced under the topic of CAPACITORS AND INDUCTORS on page 201 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing.
5|PREPINEER.COM
psst… don’t forget to take notesÎ
CONCEPT EXAMPLE: For a circuit where a battery provides a voltage source of 150 volts, a resistor provides a resistance of 10 𝛺, and an inductor provides an inductance of 330 mH, how much energy is stored in the inductor?
SOLUTION: In this problem we are solving for the amount of energy stored in the magnetic field of the inductor given the resistance and inductance of an inductor. Given the voltage and resistance, we are able to solve for the current of the circuit using Ohm’s Law. 𝑉 = 𝐼𝑅
The formula for the OHM’S LAW can be referenced under the topic of RESITIVITY on page 200 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing.
Rearrange the equation and solve for the current.
𝐼=
𝑉 𝑅
6|PREPINEER.COM
psst… don’t forget to take notesÎ
𝐼=
150 V 10 Ω
𝐼 = 15 A
Now that we have the current and inductance of the circuit, we are able to solve for the energy stored in the inductor. Energy = 𝐿𝑖!! /2 Plugging in the inductance and current values, we can calculate the energy stored in the inductor. Energy = 0.33 H 15 A ! /2 Energy = 37 J The formula for the ENERGY STORED IN AN INDUCTOR can be referenced under the topic of CAPACITORS AND INDUCTORS on page 201 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing.
7|PREPINEER.COM
psst… don’t forget to take notesÎ
INDUCTORS store energy in MAGNETIC FIELDS, in comparison to capacitors that store energy in electric fields. The current through an inductor cannot change abruptly. When the current is CHANGING, a voltage is present, but when the current is CONSTANT, a voltage is not present. FARADAY’S LAW governs the electrical behavior of an inductor, relating VOLTAGE across the terminals with the RATE OF CHANGE of the current flowing through the inductor. 1|PREPINEER.COM
psst… don’t forget to take notesÎ
𝑣! 𝑡 = 𝐿 𝑑𝑖! /𝑑𝑡 The formula for FARADY’S LAW can be referenced under the topic of CAPACITORS AND INDUCTORS on page 201 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. When given a time interval, we can solve for current through integration using the formula for the VOLTAGE-CURRENT RELATIONSHIP OF FARADAY’S LAW.
1 𝑖! 𝑡 = 𝑖! 0 + 𝐿
!
𝑣! 𝜏 𝑑𝜏 !
where • 𝑣! 𝑡 = inductor voltage, V • 𝐿 = inductance, which is a constant of proportionality and is in henrys, H • 𝑖! 𝑡 = inductor current in amperes, A • 𝑖! 0 = the current at time zero, and is called a boundary condition The formula for the VOLTAGE-CURRENT RELATIONSHIP OF AN INDUCTOR can be referenced under the topic of CAPACITORS AND INDUCTORS on page 201 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. It is important to note that we may need to use INTEGRATION when evaluating these formulas as they contain variables and equations as a function of time. 2|PREPINEER.COM
psst… don’t forget to take notesÎ
CONCEPT EXAMPLE: A 5 mH inductor has a current 𝑖(𝑡) as a function of time into its positive terminal, and a voltage
𝑣(𝑡)
as
a
function
of
time
across
its
terminals.
Find
𝑖(𝑡)
if
𝑣 𝑡 = 400 cos 200𝑡 V and 𝑖(0) = 400 A.
SOLUTION: In this problem we are given the voltage as a function of time and are asked to solve for the current. We can use the formula for the voltage-current relationship of an inductor to solve for the current, as we can integrate the voltage functions and obtain the current as a function of time.
𝑖! 𝑡 = 𝑖! 𝑡!
1 + 𝐿
1 𝑖! 𝑡 = 𝑖! 0 + 𝐿
!
𝑣! 𝜏 𝑑𝜏 !! !
𝑣! 𝜏 𝑑𝜏 !
The formula for the VOLTAGE-CURRENT RELATIONSHIP OF AN INDUCTOR can be referenced under the topic of CAPACITORS AND INDUCTORS on page 201 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. In the problem we are given: 3|PREPINEER.COM
psst… don’t forget to take notesÎ
• Inductance, 𝐿 = 5 mH = 5×10! H • The voltage as a function of time, 𝑣 𝑡 = 400 cos 200𝑡 V • The boundary condition for the current at time 𝑡 = 0, 𝑖(0) = 400 A We now plug in the function for voltage as function of time, the value for inductance, and the current at time zero into the formula for the voltage-current relationship of an inductor.
1 𝑖 𝑡 = 400 A + 5×10! H
!
400 cos 200𝑡 𝑑𝑡 !
We then evaluate the integral using the properties of integration. 𝑖 𝑡 = 400 A + 400 cos 200𝑡 |!! The next step is to evaluate the integral with the specified boundary conditions for 0 to 𝑡. 𝑡 = 400 A + 400 cos 200𝑡 − 0 We then simplify the function to obtain our function for current as a function of time. 𝑡 = 400 + 400 cos 200𝑡 A
4|PREPINEER.COM
psst… don’t forget to take notesÎ
STORED ENERGY OF AN INDUCTOR The topic of INDUCTORS can be referenced on page 201 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. As current passes through an inductor, a MAGNETIC FIELD is created that stores energy. As the current increases, energy is stored in the magnetic field, and as the current decreases the stored energy is released adding to the current. Energy is NOT DISSIPATED by an inductor, but only stored. In comparison, a capacitor stored and dissipates energy. The energy stored in the inductor is given by Energy = 𝐿𝑖!! /2 Where • The energy stored in an inductor given in units of Joules, J • 𝐿 = the inductance given in Henrys, H • 𝑖! = the current given in amperes, A The formula for the ENERGY STORED IN AN INDUCTOR can be referenced under the topic of CAPACITORS AND INDUCTORS on page 201 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing.
5|PREPINEER.COM
psst… don’t forget to take notesÎ
CONCEPT EXAMPLE: For a circuit where a battery provides a voltage source of 150 volts, a resistor provides a resistance of 10 𝛺, and an inductor provides an inductance of 330 mH, how much energy is stored in the inductor?
SOLUTION: In this problem we are solving for the amount of energy stored in the magnetic field of the inductor given the resistance and inductance of an inductor. Given the voltage and resistance, we are able to solve for the current of the circuit using Ohm’s Law. 𝑉 = 𝐼𝑅
The formula for the OHM’S LAW can be referenced under the topic of RESITIVITY on page 200 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing.
Rearrange the equation and solve for the current.
𝐼=
𝑉 𝑅
6|PREPINEER.COM
psst… don’t forget to take notesÎ
𝐼=
150 V 10 Ω
𝐼 = 15 A
Now that we have the current and inductance of the circuit, we are able to solve for the energy stored in the inductor. Energy = 𝐿𝑖!! /2 Plugging in the inductance and current values, we can calculate the energy stored in the inductor. Energy = 0.33 H 15 A ! /2 Energy = 37 J The formula for the ENERGY STORED IN AN INDUCTOR can be referenced under the topic of CAPACITORS AND INDUCTORS on page 201 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing.
7|PREPINEER.COM
psst… don’t forget to take notesÎ
