Electromagnetism - Lesson 2 – Applications of Induced Currents Part 2

Self-Inductance

• EMF induced by changing magnetic field (external or from the wire itself).
• Changing current in a coil induces an EMF opposing the change, reducing potential difference and net current.
• Self-inductance: Property of a wire to create an induced EMF opposing the change in potential difference.
• Self-inductance is the property of the current- carrying coil that resists or opposes the change of current flowing through it.

Transformers

• Devices that increase or decrease potential differences (AC only).
• Two coils (primary and secondary) electrically insulated but wound around the same iron core.
• Changing current in primary coil creates a changing magnetic field, inducing a varying EMF in the secondary coil.
• Mutual inductance: EMF and current in one coil due to changing current in another.
• Step-up transformer: Secondary voltage > primary voltage.
• Step-down transformer: Secondary voltage < primary voltage.
• Isolation transformers: Primary and secondary coils have the same number of turns, with identical input and output potential differences, used for safety reasons.
• Ideal transformer: $P<em>p = P</em>s$
• Transformer equation relates number of turns, currents, and potential differences: $\frac{V<em>s}{V</em>p} = \frac{N<em>s}{N</em>p}$
• In real transformers, some electrical energy is converted to thermal energy, reducing efficiency.
• Efficiency is the ration of the output power to the input power.

Practical Applications of Transformers

• Long-distance transmission uses high potential differences to reduce current and energy loss.
• Step-up transformers increase voltage at power sources (e.g., up to 480,000 V).
• Step-down transformers reduce voltage for consumer use (e.g., 120 V in homes).

Transmission of Electrical Energy

• High voltage, low current minimizes power loss during long-distance transmission.

Transformer Problems

• Step-up transformer example: Primary coil (200 turns, 90.0 V), secondary coil (3000 turns).
  • Secondary voltage: $V<em>s = \frac{N</em>s}{N<em>p}V</em>p = \frac{(3000)(90.0 V)}{200} = 1350 V$
  • Secondary current is 2.0 A, primary current: $I<em>p = \frac{V</em>sI<em>s}{V</em>p} = \frac{(1350 V)(2.0 A)}{90.0 V} = 3.0 \times 10^1 A$