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Magnetic Fields and Forces
 Lecture number:
 11
 Pages:
 4
 Type:
 Lecture Note
 School:
 The University of Vermont
 Course:
 Phys 012  Elementary Physics
 Edition:
 1
Documents in this Packet

6 pages

2 pages

Lecture 38 : Length Contraction
1 pages

Lecture 37 : Time Dilation Practice Problems
2 pages

Lecture 36 : Special Relativity
2 pages

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Lecture 34 : Quantized Electron Orbitals
3 pages

Lecture 33 : WaveParticle Duality of Light and Electrons
4 pages

Lecture 32 : Photoelectric and Compton Effects
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Lecture 30 : Particles and Waves
2 pages

Lecture 29 : Single Slit Diffraction and Resolving Power
2 pages

Lecture 28 : Thin Film Interference and Diffraction
2 pages

Lecture 27 : Diffraction and Thin Film Interference
3 pages

Lecture 26 : Diffraction and Interference
4 pages

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Lecture 24 : Total Internal Reflection
4 pages

Lecture 23 : Refraction of Light
4 pages

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Lecture 21 : Light Rays and Reflection
3 pages

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Lecture 20 : In Class Review for Exam 2
2 pages

Lecture 19 : Electromagnetic Spectrum
3 pages

Lecture 18 : Electromagnetic Waves
5 pages

Lecture 17 : Self Inductance, Transformers, and Circuits Containing Inductors
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Lecture 14 : Electromagnetic Induction
4 pages

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Lecture 12 : Currents through Magnetic Fields
3 pages

Lecture 10 : Kirchhoff's Rules and RC Circuits
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Lecture 9 : Resistors and Capacitors.
4 pages

Lecture 8 : Electric Currents, Ohm's Law, and
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Lecture 7 : Capacitors and Dielectrics
3 pages

Lecture 6 : Electric Potential Energy and Electric Potential
4 pages

Lecture 5 : Electric Fields and Forces (18.9,19.1)
3 pages

Lecture 4 : Electric Field Lines
3 pages

2 pages

Lecture 1 : Charge and Charge Transfer
2 pages

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Unformatted text preview:
Lecture 11 Outline of Last Lecture II. Kirchhoff’s Rules a. Junction rule: the total current going into a junction equals the total current out of the junction i. ΣIin = ΣIout ii. Derived from the law of conservation of charge b. Loop rule: the sum of changes in potential around a closed circuit loop is zero i. ΣΔVclosed loop = 0 c. Problem: Find the values for current through the closed circuit below given: ε1 = 10V, ε2 = 14V, R1 = 6Ω, R2 = 4Ω, R3 = 2Ω. i. First, draw currents through this circuit. Direction does not matter at this point in the problem. When you get values for the current, if a value is negative then current flows in the opposite direction as predicted. The magnitude will still be the same. 1. Since ΣIin = ΣIout, for junction A, I1 + I2 = I3. ii. Second, mark each side of every element in the circuit with higher (+) or lower () potential. Remember: current flows from high potential to low potential. Physics 012 1st Edition
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