04 Inductance Practice Problems
INDUCTANCE | PRACTICE PROBLEMS Complete the following to reinforce your understanding of the concept covered in this module.
PROBLEM 1: What is the energy stored by an inductor with an inductance of 10 H and a current of 5 A?
PROBLEM 2: An inductor is found to generate a voltage of 5.6 mV with a current through its terminals changes at 0.85 A/s. What is the inductance?
PROBLEM 3: The voltage across a 200 mH inductor is given by 𝑣 𝑡 = 3𝑡 ! + 2𝑡 + 4 V for 𝑡 > 0. Determine the current 𝑖 𝑡 through the inductor. Assume that 𝑖 0 = 1 A.
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INDUCTANCE | SOLUTIONS SOLUTION 1: We are solving for the energy stored in the inductor. Since we are given the inductance and current of the circuit, we can simply plug the values into our formula and solve for the energy stored in the inductor. In the problem we are given: • Inductance, 𝐿 = 10 H • Current, 𝐼 = 5 A As we are given the current and inductance of the circuit, we are able to solve for the energy stored in the inductor using the formula for stored energy in an inductor. Energy = 𝐿𝑖!! /2 Plugging in the given inductance and current values, we can calculate the energy stored in the inductor. Energy = 10 H 5 A ! /2 Energy = 125 J
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psst… don’t forget to take notesÎ
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.
SOLUTION 2: In this problem we are solving for the inductance of the inductor, given the voltage and the change in current based on time. In the problem we are given: • Voltage, 𝑣 = 5.6 mV = 5.6×10!! V • Change in current, 𝑑𝑖! /𝑑𝑡 = 0.85 A/s We can use Faraday's Law to solve for the inductance, as we are given the voltage and change in current based on time. 𝑣! (𝑡) = 𝐿(𝑑𝑖! /𝑑𝑡) The formula for FARADAY’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. We can re-arrange Faraday’s Law to solve for the inductance of the inductor
𝐿=
𝑣! 𝑑𝑖! /𝑑𝑡
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psst… don’t forget to take notesÎ
Plugging in the given voltage and change in current over time values, we can the inductance of the inductor.
5.6×10!! V 𝐿= A 0.85 s 𝐿 = 6.6×10!! H = 6.6 mH
SOLUTION 3: In this problem we are given the current as a function of time and are asked to solve for the voltage 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
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!
𝑣! 𝜏 𝑑𝜏 !! !
𝑣! 𝜏 𝑑𝜏 !
psst… don’t forget to take notesÎ
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: • Inductance, 𝐿 = 200 mH = 200×10!! H • Voltage as a function of time, 𝑣! 𝑡 = 3𝑡 ! + 2𝑡 + 4 V, for 𝑡 > 0 The boundary condition for the current at time 𝑡 = 0 is 𝑖! 0 = 1 A We can now plug in the given function for voltage, the value for inductance, and current at time zero into the formula for the voltage-current relationship of an inductor.
1 𝑖! 𝑡 = 𝑖! 0 + 𝐿
!
𝑣! 𝜏 𝑑𝜏 !
1 𝑖! 𝑡 = (1 A) + (200×10!! H)
!
3𝑡 ! + 2𝑡 + 4 V 𝑑𝑡 !
We then evaluate the integral using the properties of integration. 𝑖! 𝑡 = 1 A + 5 𝑡 ! + 𝑡 ! + 4𝑡
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𝑡 0
psst… don’t forget to take notesÎ
We then simplify the function to obtain our function for current as a function of time. 𝑖! 𝑡 = 5𝑡 ! + 5𝑡 ! + 20𝑡 + 1 A
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psst… don’t forget to take notesÎ
PROBLEM 1: What is the energy stored by an inductor with an inductance of 10 H and a current of 5 A?
PROBLEM 2: An inductor is found to generate a voltage of 5.6 mV with a current through its terminals changes at 0.85 A/s. What is the inductance?
PROBLEM 3: The voltage across a 200 mH inductor is given by 𝑣 𝑡 = 3𝑡 ! + 2𝑡 + 4 V for 𝑡 > 0. Determine the current 𝑖 𝑡 through the inductor. Assume that 𝑖 0 = 1 A.
8|PREPINEER.COM
psst… don’t forget to take notesÎ
INDUCTANCE | SOLUTIONS SOLUTION 1: We are solving for the energy stored in the inductor. Since we are given the inductance and current of the circuit, we can simply plug the values into our formula and solve for the energy stored in the inductor. In the problem we are given: • Inductance, 𝐿 = 10 H • Current, 𝐼 = 5 A As we are given the current and inductance of the circuit, we are able to solve for the energy stored in the inductor using the formula for stored energy in an inductor. Energy = 𝐿𝑖!! /2 Plugging in the given inductance and current values, we can calculate the energy stored in the inductor. Energy = 10 H 5 A ! /2 Energy = 125 J
9|PREPINEER.COM
psst… don’t forget to take notesÎ
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.
SOLUTION 2: In this problem we are solving for the inductance of the inductor, given the voltage and the change in current based on time. In the problem we are given: • Voltage, 𝑣 = 5.6 mV = 5.6×10!! V • Change in current, 𝑑𝑖! /𝑑𝑡 = 0.85 A/s We can use Faraday's Law to solve for the inductance, as we are given the voltage and change in current based on time. 𝑣! (𝑡) = 𝐿(𝑑𝑖! /𝑑𝑡) The formula for FARADAY’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. We can re-arrange Faraday’s Law to solve for the inductance of the inductor
𝐿=
𝑣! 𝑑𝑖! /𝑑𝑡
10|PREPINEER.COM
psst… don’t forget to take notesÎ
Plugging in the given voltage and change in current over time values, we can the inductance of the inductor.
5.6×10!! V 𝐿= A 0.85 s 𝐿 = 6.6×10!! H = 6.6 mH
SOLUTION 3: In this problem we are given the current as a function of time and are asked to solve for the voltage 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
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!
𝑣! 𝜏 𝑑𝜏 !! !
𝑣! 𝜏 𝑑𝜏 !
psst… don’t forget to take notesÎ
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: • Inductance, 𝐿 = 200 mH = 200×10!! H • Voltage as a function of time, 𝑣! 𝑡 = 3𝑡 ! + 2𝑡 + 4 V, for 𝑡 > 0 The boundary condition for the current at time 𝑡 = 0 is 𝑖! 0 = 1 A We can now plug in the given function for voltage, the value for inductance, and current at time zero into the formula for the voltage-current relationship of an inductor.
1 𝑖! 𝑡 = 𝑖! 0 + 𝐿
!
𝑣! 𝜏 𝑑𝜏 !
1 𝑖! 𝑡 = (1 A) + (200×10!! H)
!
3𝑡 ! + 2𝑡 + 4 V 𝑑𝑡 !
We then evaluate the integral using the properties of integration. 𝑖! 𝑡 = 1 A + 5 𝑡 ! + 𝑡 ! + 4𝑡
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𝑡 0
psst… don’t forget to take notesÎ
We then simplify the function to obtain our function for current as a function of time. 𝑖! 𝑡 = 5𝑡 ! + 5𝑡 ! + 20𝑡 + 1 A
13|PREPINEER.COM
psst… don’t forget to take notesÎ
