Homework 05-solutions - sanne (as42476) Homework 05 Yao...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: sanne (as42476) Homework 05 Yao (59110) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points The current in a wire decreases with time according to the relationship I = (2 . 88 mA) e a t where a = 0 . 13328 s 1 . Determine the total charge that passes through the wire from t = 0 to the time the current has diminished to zero. Correct answer: 0 . 0216086 C. Explanation: I = dq dt q = integraldisplay t t =0 I dt = integraldisplay t =0 (0 . 00288 A) e . 13328 s 1 t dt = (0 . 00288 A ) e . 13328 s 1 t- . 13328 s 1 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = . 0216086 C . 002 10.0 points If 4 . 1 10 21 electrons pass through a 50 re- sistor in 6 min, what is the potential difference across the resistor? Correct answer: 91 . 225 V. Explanation: Let : n = 4 . 1 10 21 , t = 6 min , R = 50 , and q e = 1 . 602 10 19 C . The current is defined as I = Q t . Since each electron carries a q e , we have I = n q e t = ( 4 . 1 10 21 ) (1 . 602 10 19 C) 6 min parenleftbigg 1 min 60 s parenrightbigg = 1 . 8245 A . Then from Ohms Law, V = I R = (1 . 8245 A) (50 ) = 91 . 225 V . 003 10.0 points A conductor with cross-sectional area 9 cm 2 carries a current of 12 A. If the concentration of free electrons in the conductor is 7 10 28 electrons / m 3 , what is the drift velocity of the electrons? Correct answer: 0 . 00118886 mm / s. Explanation: Let : I = 12 A , n = 7 10 28 electrons / m 3 , q e = 1 . 60218 10 19 C , and A = 9 cm 2 = 0 . 0009 m 2 . The current in a conductor is given by I = n q v d A , where n is the number of charge carriers per unit volume, q is the charge per carrier, v d is the drift velocity of the carriers and A is the cross section of the conduction. Solving for v d , we have v d = I n q A = 12 A 7 10 28 electrons / m 3 1 1 . 60218 10 19 C 1 . 0009 m 2 1000 mm 1 m = . 00118886 mm / s . sanne (as42476) Homework 05 Yao (59110) 2 004 (part 1 of 3) 10.0 points Consider two cylindrical conductors made of the same ohmic material. Conductor 1 has a radius r 1 and length 1 while conductor 2 has a radius r 2 and length 2 . Denote: The currents of the two conductors as I 1 and I 2 , the potential differences between the two ends of the conductors as V 1 and V 2 , and the electric fields within the conductors as E 1 and E 2 . V 1 vector E 1 I 1 1 r 1 b V 2 vector E 2 I 2 2 r 2 b If 2 = 1 , r 2 = 2 r 1 , 2 = 3 1 and V 2 = V 1 , find the ratio R 2 R 1 of the resistances....
View Full Document

Page1 / 10

Homework 05-solutions - sanne (as42476) Homework 05 Yao...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online