{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

GPII - Homework 3, Electric Current - solutions

GPII - Homework 3, Electric Current - solutions -...

Info iconThis preview shows pages 1–2. 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 Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: honea (cth632) – H3ElectricCurrrent – avram – (1956) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points In a particular television picture tube, the measured beam current is 94 . 9 μ A. How many electrons strike the tube screen every 26 s? Correct answer: 1 . 5402 × 10 16 electrons. Explanation: Let : I = 94 . 9 μ A = 9 . 49 × 10 − 5 C / s and Δ t = 26 s . The current is I = q Δ t = n q e Δ t . Thus n = I Δ t q e = (9 . 49 × 10 − 5 C / s) (26 s) 1 . 602 × 10 − 19 C / electron = 1 . 5402 × 10 16 electrons . 002 10.0 points A wire is made of a material with a resistiv- ity of 1 . 63303 × 10 − 8 Ω · m. It has length 6 . 72317 m and diameter 0 . 79857 mm. What is the resistance of the wire? Correct answer: 0 . 219206 Ω. Explanation: Let : ρ = 1 . 63303 × 10 − 8 Ω · m , ℓ = 6 . 72317 m , and r = 0 . 399285 mm = 0 . 000399285 m . By definition, the resistance of the wire is R = ρ ℓ A = ρ ℓ π r 2 = ( 1 . 63303 × 10 − 8 Ω · m ) (6 . 72317 m) π (0 . 000399285 m) 2 = . 219206 Ω . 003 (part 1 of 2) 10.0 points The potential difference in a simple circuit is 15 V and the resistance is 20 Ω....
View Full Document

{[ snackBarMessage ]}

Page1 / 4

GPII - Homework 3, Electric Current - solutions -...

This preview shows document pages 1 - 2. Sign up to view the full document.

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