University:
Kentucky State University
Course:
PHY 212 | General Physics II
Academic year:

2024
Views:

374

Pages:

38
Author:

Nicole Wang

**Chapter #13: Magnetically Coupled Circuits** * Oersted has discovered that when (I) passes through a current carrying conductor, a magnetic field is created. [Image of a current-carrying conductor with magnetic field lines around it] To know the direction of the magnetic field (right hand rule): Thumb is in the direction of I, other fingers with the magnetic field. **Faraday's Law:** [Image of a coil with a changing magnetic field inducing a voltage] Ei induced voltage **Creating a variable magnetic field:** [Image of a coil connected to a voltage source, creating a variable magnetic field] * To determine the magnetic field direction, we use the right-hand rule: Thumb with the magnetic field while other fingers with the current. i(t) V₂(t): Induced Voltage V₁(t) = L₁ di₁(t)/dt L₁ is the self-inductance [H] If i was a DC current, V₁ = L₁ * 0 = 0 & the magnetic field will be constant. M₁₂: Mutual Inductance [H] V₂(t) = M₁₂ di₁(t)/dt V₁(t) = L₁ di₁(t)/dt V₂(t) = M₁₂ di₁(t)/dt Self-induced Voltage Mutual induced Voltage **Example:** Find I₁, I₂, I₃: [Image of a circuit diagram with three loops and current labels I₁, I₂, I₃] **Solution:** -100∠0° + j5ω(I₁-I₂) + j2ω(I₁-I₃) + 6I₁ = 0 j4ω(I₂-I₁) + j3ω(I₂-I₃) + 6I₂ + jω(I₃-I₂) + 5I₃ = 0 ⇒ 6I₂ + j4ωI₂ - j4ωI₁ + j3ωI₂ - j3ωI₃ + jωI₃ - jωI₂ + 5I₃ = 0 ⇒ -j4ωI₁ + (6+j4ω)I₂ - j2ωI₃ = 0 6(I₃-I₁) + j6ω(I₃-I₂) + j2ωI₂ + jω(I₁-I₂) + 5I₃ = 0 ⇒ -6I₁ - jωI₁ + j4ωI₂ + (11+j6ω)I₃ = 0 **Energy stored & equality of M12 & M21** @ zero initial conditions. [Image of a circuit diagram with two inductors L1 and L2, and a mutual inductance M] **Experiment 1:** Let i₂ = 0; open circuit. [Image of a circuit diagram with one inductor L1, a current source, and an open circuit] a) + increase i₁ from 0 → I₁ from t = 0 to t = t₁ Instantaneous power: P(t) = v₁(t) i₁(t) = L₁ di₁(t)/dt * i₁(t) P₂(t) = 0 Energy = ∫power dt E₁ = ∫₀ᵗ¹ L₁ i₁(t) di₁(t) = 1/2 L₁ I₁² E₂,₁ = 0 b) now, hold i₁(t) @ I₁ Increase i₂ from 0 → I₂ from t₁ to t₂ [Image of a circuit diagram with two inductors L1 and L2, and a current source] E₂'₁ = ∫(t₁ to t₂) v₁ dt = L₁ ∫(t₁ to t₂) di₁ = 0 E₁₂ = ∫(t₁ to t₂) v₂ dt = M₁₂ ∫(t₁ to t₂) di₂ = M₁₂ I₂ = M₁₂ I₁ I₂