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mat-sln-asn-hwk18-spr02

# mat-sln-asn-hwk18-spr02 - Solution for Homework 18...

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Unformatted text preview: Solution for Homework 18 Inductance Solution to Homework Problem 18.1(What Makes Second Speaker Work?) Problem: In lab, the second speaker we built was made with a fixed magnet and a coil attached to a vibrating board. What physical law in most important in describing the operation of this speaker? Select One of the Following: (a) Faraday’s Law. (b) Ampere’s Law (c-Answer) The Lorentz Force (d) Biot-Savart Law (e) Coulomb’s Law Solution Lorentz Force Total Points for Problem: 3 Points Solution to Homework Problem 18.2(Flux, Energy from Inductance) Problem: I’ve been surfing the internet for inductors since my kid’s electronics kit didn’t have one. The Series 4922 Power Inductor from Delevan has a maximum current rating of . 050A for the . 022H inductor. (a)At maximum current, what is the magnetic flux through the inductor? Select One of the Following: (a) 25Tm 2 (b) 5 . 3Tm 2 (c) 4 . 4 × 10- 1 Tm 2 (d-Answer) 1 . 1 × 10- 3 Tm 2 (e) 3 . 1 × 10- 5 Tm 2 (b)How much energy is stored in the inductor? Select One of the Following: (a) 1 . 1 × 10- 1 J (b) 9 . 1 × 10- 2 J (c) 6 . 8 × 10- 2 J (d) 3 . 1 × 10- 3 J (e-Answer) 2 . 75 × 10- 5 J Solution to Part (a) 1 Find flux with inductance: φ m = LI = (0 . 022H)(0 . 05A) = 1 . 1 × 10- 3 Tm 2 Grading Key: Part (a) 2 Points Solution to Part (b) Compute energy from inductance: One of the formulas for the energy of a magnetic field is U m = 1 2 LI 2 So U m = 1 2 (0 . 022H)(0 . 05A) 2 = 2 . 75 × 10- 5 J Grading Key: Part (b) 2 Points Total Points for Problem: 4 Points Solution to Homework Problem 18.3(Potential Difference Across the Primary Coil of a Transformer Given the Potential Difference of the Secondary and Numbers of Turns) Problem: A given transformer is made up of a primary coil with 200 turns and a secondary coil with 50 turns. If the secondary coil has a potential difference across it of V , what is the potential difference across the primary coil? Select One of the Following: (a) The potential difference across the primary coil is 1 16 V . (b) The potential difference across the primary coil is 1 4 V . (c-Answer) The potential difference across the primary coil is 4 V . (d) The potential difference across the primary coil is 16 V . Solution If the secondary coil of a transformer has N s windings and a potential difference of V s and the primary coil of the transformer has N p windings and a potential difference V p , then V p = V s N p N s . The potential difference V p across the primary coil of the transformer in the problem is then equal to V p = V 200 turns 50 turns = 4 V ....
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mat-sln-asn-hwk18-spr02 - Solution for Homework 18...

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