Faraday’S Law Of Induction- Mutual And Self-Inductance

Slide 1

Faraday’s Law of Induction

Explains the relationship between a changing magnetic field and induced emf
States that the induced emf in a circuit is directly proportional to the rate of change of magnetic flux through the circuit

Equation:

Induced emf (ε) = -dΦ/dt

Key Points:

Magnetic flux (Φ) is a measure of the magnetic field passing through a given area.

The negative sign in the equation indicates the direction of the induced emf.

The unit for induced emf is volts (V).

Example:

When a magnetic field of 0.5 T cuts across a coil of 100 turns at a rate of 5000 Wb/s, calculate the induced emf.

Slide 2

Mutual Inductance

Refers to the interaction of two or more coils with respect to each other
Induced emf is produced in one coil due to the changing current in another nearby coil

Key Points:

The magnetic field produced by one coil influences the magnetic field in another coil.

Mutual inductance is denoted by the symbol M and measured in henries (H).

Induced emf (ε) in one coil is given by the equation ε = -M(dI/dt), where dI/dt is the rate of change of current.

Example:

Two coils have a mutual inductance of 0.2 H. If the current in one coil changes at a rate of 5 A/s, find the induced emf.

Equation:

ε = -M(dI/dt)

Slide 3

Self-Inductance

Refers to the induction of emf in a single coil due to its own changing current
When current changes in a coil, a changing magnetic field is produced, inducing an emf in the same coil

Key Points:

Self-inductance is denoted by the symbol L and measured in henries (H).

The induced emf in a coil is given by the equation ε = -L(dI/dt), where dI/dt is the rate of change of current.

Example:

A coil with a self-inductance of 0.1 H has a current changing at a rate of 2 A/s. Determine the induced emf.

Equation:

ε = -L(dI/dt)

Slide 4

Inductive Reactance

Opposes the flow of alternating current through an inductor
Similar to resistance, but depends on frequency and inductance

Key Points:

Inductive reactance is denoted by the symbol XL and measured in ohms (Ω).

Inductive reactance depends on the frequency of the alternating current and the inductance of the coil.

The higher the frequency, the larger the inductive reactance.

Equation:

XL = 2πfL

Example:

An inductor with inductance 0.02 H is connected to a current source with frequency 50 Hz. Calculate the inductive reactance.

Slide 5

RL Circuits

Composed of a resistor (R) and an inductor (L) connected in series
Behaves differently for direct current (DC) and alternating current (AC)

Key Points:

In a DC RL circuit, the inductor acts as a short circuit once it is fully energized.

For an AC RL circuit, the inductor impedes the flow of current at the beginning due to inductive reactance.

The time constant (τ) of an RL circuit is given by the equation τ = L/R.

Example:

In a 2 H inductor and a 10 Ω resistor connected in series, calculate the time constant of the RL circuit.

Equation:

τ = L/R

Slide 6

RC Circuits

Composed of a resistor (R) and a capacitor (C) connected in series
Behaves differently for direct current (DC) and alternating current (AC)

Key Points:

In a DC RC circuit, the capacitor acts as an open circuit once it is fully charged.

For an AC RC circuit, the capacitor charges and discharges continuously, providing a phase shift of the current.

The time constant (τ) of an RC circuit is given by the equation τ = RC.

Example:

In a 100 μF capacitor and a 5 kΩ resistor connected in series, calculate the time constant of the RC circuit.

Equation:

τ = RC

Slide 7

LC Circuits

Composed of an inductor (L) and a capacitor (C) connected in parallel
Oscillates at a natural frequency known as the resonant frequency

Key Points:

An LC circuit stores energy in both the magnetic field of the inductor and the electric field of the capacitor.

The resonant frequency (ω0) of an LC circuit is given by the equation ω0 = 1/√(LC).

At resonance, the reactance of the inductor cancels out the reactance of the capacitor, resulting in a purely resistive circuit.

Example:

In an LC circuit with inductance 2 H and capacitance 20 μF, calculate the resonant frequency.

Equation:

ω0 = 1/√(LC)

Slide 8

Impedance

The total opposition to the flow of alternating current in a circuit
Combines resistance, inductive reactance, and capacitive reactance

Key Points:

Impedance (Z) is denoted by a complex number consisting of two parts: real (R) and imaginary (jX).

The magnitude of impedance, |Z|, is given by the equation |Z| = √(R^2 + X^2).

The net effect of resistance, inductive reactance, and capacitive reactance determines the behavior of the circuit.

Equation:

Z = R + jX

Example:

A circuit has a resistance of 10 Ω, an inductive reactance of 5 Ω, and a capacitive reactance of -3 Ω. Calculate the impedance.

Slide 9

AC Power in Inductive and Capacitive Circuits

Power in an AC circuit varies due to complex impedance

Key Points:

In an inductive circuit, the power factor (cosθ) lags behind the voltage, leading to lower power consumption.

In a capacitive circuit, the power factor leads the voltage, resulting in improved power efficiency.

The power dissipation in an AC circuit is given by the equation P = IV cosθ, where I is the current, V is the voltage, and cosθ is the power factor.

Example:

A circuit has a current of 5 A, a voltage of 220 V, and a power factor of 0.8. Calculate the power dissipation.

Equation:

P = IV cosθ

Slide 10

Transformers

Electrical devices used to change the voltage of alternating current
Consist of two separate coils, primary and secondary, wound around a common iron core

Key Points:

The primary coil is connected to the input voltage source, while the secondary coil is connected to the load.

Transformers work based on the principle of mutual inductance, where a changing current induces emf in the secondary coil.

The ratio of the number of turns in the primary coil to the number of turns in the secondary coil determines the voltage ratio.

Example:

A transformer has a primary coil with 500 turns and a secondary coil with 250 turns. Determine the voltage ratio.

Slide 11

Faraday’s Law of Induction

Explains the relationship between a changing magnetic field and induced emf
States that the induced emf in a circuit is directly proportional to the rate of change of magnetic flux through the circuit Equation:
Induced emf (ε) = -dΦ/dt Key Points:
Magnetic flux (Φ) is a measure of the magnetic field passing through a given area.
The negative sign in the equation indicates the direction of the induced emf.
The unit for induced emf is volts (V). Example:
When a magnetic field of 0.5 T cuts across a coil of 100 turns at a rate of 5000 Wb/s, calculate the induced emf.

Slide 12

Mutual Inductance

Refers to the interaction of two or more coils with respect to each other
Induced emf is produced in one coil due to the changing current in another nearby coil Key Points:
The magnetic field produced by one coil influences the magnetic field in another coil.
Mutual inductance is denoted by the symbol M and measured in henries (H).
Induced emf (ε) in one coil is given by the equation ε = -M(dI/dt), where dI/dt is the rate of change of current. Example:
Two coils have a mutual inductance of 0.2 H. If the current in one coil changes at a rate of 5 A/s, find the induced emf. Equation:
ε = -M(dI/dt)

Slide 13

Self-Inductance

Refers to the induction of emf in a single coil due to its own changing current
When current changes in a coil, a changing magnetic field is produced, inducing an emf in the same coil Key Points:
Self-inductance is denoted by the symbol L and measured in henries (H).
The induced emf in a coil is given by the equation ε = -L(dI/dt), where dI/dt is the rate of change of current. Example:
A coil with a self-inductance of 0.1 H has a current changing at a rate of 2 A/s. Determine the induced emf. Equation:
ε = -L(dI/dt)

Slide 14

Inductive Reactance

Opposes the flow of alternating current through an inductor
Similar to resistance, but depends on frequency and inductance Key Points:
Inductive reactance is denoted by the symbol XL and measured in ohms (Ω).
Inductive reactance depends on the frequency of the alternating current and the inductance of the coil.
The higher the frequency, the larger the inductive reactance. Equation:
XL = 2πfL Example:
An inductor with inductance 0.02 H is connected to a current source with frequency 50 Hz. Calculate the inductive reactance.

Slide 15

RL Circuits

Composed of a resistor (R) and an inductor (L) connected in series
Behaves differently for direct current (DC) and alternating current (AC) Key Points:
In a DC RL circuit, the inductor acts as a short circuit once it is fully energized.
For an AC RL circuit, the inductor impedes the flow of current at the beginning due to inductive reactance.
The time constant (τ) of an RL circuit is given by the equation τ = L/R. Example:
In a 2 H inductor and a 10 Ω resistor connected in series, calculate the time constant of the RL circuit. Equation:
τ = L/R

Slide 16

RC Circuits

Composed of a resistor (R) and a capacitor (C) connected in series
Behaves differently for direct current (DC) and alternating current (AC) Key Points:
In a DC RC circuit, the capacitor acts as an open circuit once it is fully charged.
For an AC RC circuit, the capacitor charges and discharges continuously, providing a phase shift of the current.
The time constant (τ) of an RC circuit is given by the equation τ = RC. Example:
In a 100 μF capacitor and a 5 kΩ resistor connected in series, calculate the time constant of the RC circuit. Equation:
τ = RC

Slide 17

LC Circuits

Composed of an inductor (L) and a capacitor (C) connected in parallel
Oscillates at a natural frequency known as the resonant frequency Key Points:
An LC circuit stores energy in both the magnetic field of the inductor and the electric field of the capacitor.
The resonant frequency (ω0) of an LC circuit is given by the equation ω0 = 1/√(LC).
At resonance, the reactance of the inductor cancels out the reactance of the capacitor, resulting in a purely resistive circuit. Example:
In an LC circuit with inductance 2 H and capacitance 20 μF, calculate the resonant frequency. Equation:
ω0 = 1/√(LC)

Slide 18

Impedance

The total opposition to the flow of alternating current in a circuit
Combines resistance, inductive reactance, and capacitive reactance Key Points:
Impedance (Z) is denoted by a complex number consisting of two parts: real (R) and imaginary (jX).
The magnitude of impedance, |Z|, is given by the equation |Z| = √(R^2 + X^2).
The net effect of resistance, inductive reactance, and capacitive reactance determines the behavior of the circuit. Equation:
Z = R + jX Example:
A circuit has a resistance of 10 Ω, an inductive reactance of 5 Ω, and a capacitive reactance of -3 Ω. Calculate the impedance.

Slide 19

AC Power in Inductive and Capacitive Circuits

Power in an AC circuit varies due to complex impedance Key Points:
In an inductive circuit, the power factor (cosθ) lags behind the voltage, leading to lower power consumption.
In a capacitive circuit, the power factor leads the voltage, resulting in improved power efficiency.
The power dissipation in an AC circuit is given by the equation P = IV cosθ, where I is the current, V is the voltage, and cosθ is the power factor. Example:
A circuit has a current of 5 A, a voltage of 220 V, and a power factor of 0.8. Calculate the power dissipation. Equation:
P = IV cosθ

Slide 20

Transformers

Electrical devices used to change the voltage of alternating current
Consist of two separate coils, primary and secondary, wound around a common iron core Key Points:
The primary coil is connected to the input voltage source, while the secondary coil is connected to the load.
Transformers work based on the principle of mutual inductance, where a changing current induces emf in the secondary coil.
The ratio of the number of turns in the primary coil to the number of turns in the secondary coil determines the voltage ratio. Example:
A transformer has a primary coil with 500 turns and a secondary coil with 250 turns. Determine the voltage ratio.

Slide 21

Faraday’s Law of Induction - Mutual and Self-Inductance - Example 1 (Motional Emf)

Consider a conducting rod of length L moving perpendicular to a magnetic field with a constant velocity v. Key Points:
The rod experiences a motional emf, given by the equation ε = Bvl, where B is the magnetic field strength, v is the velocity, and l is the length of the rod.
The motional emf is a result of the changing magnetic field as the rod moves across it.
The induced current in the rod can be determined using Ohm’s law: I = ε/R, where R is the resistance of the rod. Example:
A conducting rod of length 0.5 m moves across a magnetic field of 0.3 T with a velocity of 2 m/s. Calculate the motional emf and the current when the resistance of the rod is 2 Ω. Equation:
ε = Bvl

Slide 22

Transformer Efficiency Key Points:

The efficiency of a transformer is the ratio of output power to input power, given by the equation Efficiency = output power/input power.
Transformer losses include copper losses (I^2R losses) and iron losses (hysteresis and eddy current losses).
Copper losses are due to resistance in the windings, while iron losses are due to energy lost in the core material.
The efficiency of a transformer can be improved by using materials with low resistivity and reducing core losses. Example:
A transformer has an input power of 1000 W and an output power of 950 W. Calculate the efficiency of the transformer. Equation:
Efficiency = output power/input power

Slide 23

Power Factor Key Points:

Power factor is a measure of how effectively an electrical device converts electrical power into useful work.
It is the ratio of real power (P) to apparent power (S), given by the equation Power factor = P/S.
Power factor is a value between 0 and 1, with a higher value indicating better power efficiency.
Power factor is influenced by the presence of reactive components such as inductors and capacitors in the circuit. Example:
A circuit has a real power of 500 W and an apparent power of 600 VA. Calculate the power factor of the circuit. Equation:
Power factor = P/S

Slide 24

Transformers in AC Circuits Key Points:

Transformers are commonly used in AC circuits to step up or step down voltage levels.
The turns ratio of a transformer determines the ratio of primary and secondary voltages: Vp/Vs = Np/Ns.
For a step-up transformer, the secondary voltage is higher than the primary voltage. For a step-down transformer, it is lower.
Transformers work based on the principle of electromagnetic induction and the conservation of energy. Example:
A transformer has a turns ratio of 1:10. If the primary voltage is 220 V, calculate the secondary voltage. Equation:
Vp/Vs = Np/Ns

Slide 25

Resonance in LC Circuits Key Points: