{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

mat-sln-asn-hwk19-spr02 - Solution for Homework 19 RL...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
Solution for Homework 19 RL Circuits Solution to Homework Problem 19.1(Resistance Given Time Constant) Problem: A RL circuit has a time constant, τ , of 40s and an inductance, L, of 3H what is the resistance of the circiut? Select One of the Following: (a) 0 (b-Answer) 0 . 075Ω (c) 10Ω (d) 26Ω (e) 42Ω Solution The time constant for a RL circuit is defined by, τ = L R solve for R, R = L τ sustitute, R = 3 H 40 s evaluate, R = . 075Ω Total Points for Problem: 3 Points Solution to Homework Problem 19.2(Long/Short RL Problem) Problem: A coil of resistance 12Ω and self-inductance 2mH is connected across an ideal 12V battery. (a)What is the initial current? Select One of the Following: (a-Answer) 0A (b) 1A (c) 144A (d) 0 . 024A (e) 6000A (b)What is the final current? Select One of the Following: 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
(a) 0A (b-Answer) 1A (c) 144A (d) 0 . 024A (e) 6000A Solution to Part(a) When the circuit is first connected, the inductor allows zero initial current flow, I (0) = 0 , since an inductor resists changes in current and does not allow instantaneous changes in current. Grading Key: Part (a) 2 Points Solution to Part(b) After a long time, I is not changing, so Δ V L = dI dt = 0 , and substituting into the loop equation, Δ V battery - IR = 0 , gives I = Δ V battery R = 1 A. Grading Key: Part (b) 2 Points Total Points for Problem: 4 Points Solution to Homework Problem 19.3(Magnetic Field of Two WIres) Problem: Two infinite straight wires carry current I = 2A as shown below. One wire is parallel to the y axis and one wire is parallel to the z axis. The current directions are as drawn. The distance from each wire to point P is 3cm . Calculate the magnetic field at point P . Be sure to report the magnetic field as a vector. Select One of the Following: (a) 2 . 4 × 10 - 5 Tˆy + 2 . 4 × 10 - 5 z (b) 1 . 3 × 10 - 5 Tˆx - 1 . 3 × 10 - 5 Tˆy (c) 2 . 4 × 10 - 5 Tˆy - 2 . 4 × 10 - 5 z (d-Answer) 1 . 3 × 10 - 5 Tˆy - 1 . 3 × 10 - 5 z (e) 0 2
Background image of page 2
I d d x y I P Wire 1 Wire 2 Solution (a) The field of an infinite wire: The field of on infinite wire is given by B = μ 0 I 2 πR R is the same for both wire one and wire two. The direction can be read off using the right-hand rule. (b) Compute the field of wire one: vector B 1 = μ 0 I 2 πR = (4 π × 10 - 7 Tm A )(2A) 2 π (0 . 03m) = 1 . 3 × 10 - 5 Tˆy (Correct Sig Figs) (c) Compute the field of wire two : vector B 2 = μ 0 I 2 πR = (4 π × 10 - 7 Tm A )(2A) 2 π (0 . 03m) = - 1 . 3 × 10 - 5 z (Correct Sig Figs) (d) Compute the total field: The total field is the vector sum of the two vector B tot = vector B 1 + vector B 2 = 1 . 3 × 10 - 5 Tˆy - 1 . 3 × 10 - 5 z Total Points for Problem: 3 Points Solution to Homework Problem 19.4(Velocity of Alpha Particle) 3
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Problem: An alpha particle is a Helium nucleus and has twice the charge, q = 2 q p , and four times the mass, m = 4 m p , of a proton (the mass of a proton is m p = 1 . 67 × 10 - 27 kg and the charge of a proton is q p = +1 . 6 × 10 - 19 C ).
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}