{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

GPII - Homework 4, Circuits - solutions

# GPII - Homework 4, Circuits - solutions - honea(cth632...

This preview shows pages 1–2. Sign up to view the full content.

honea (cth632) – H4Circuits – 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(part1of4)10.0points A 47 Ω lamp and a 59 Ω lamp are connected in series and placed across a potential of 47 V. What is the equivalent resistance of the circuit? Correct answer: 106 Ω. Explanation: Let : R 1 = 47 Ω , R 2 = 59 Ω , and V = 47 V . The equivalent resistance of the circuit is R = R 1 + R 2 = 47 Ω + 59 Ω = 106 Ω . 002(part2of4)10.0points What is the current in the circuit? Correct answer: 0 . 443396 A. Explanation: I = V R = 47 V 106 Ω = 0 . 443396 A . 003(part3of4)10.0points What is the voltage drop across the first lamp? Correct answer: 20 . 8396 V. Explanation: This is a series circuit, so the voltage drop across the first lamp is V 1 = IR 1 = (0 . 443396 A) (47 Ω) = 20 . 8396 V . 004(part4of4)10.0points What is the power dissipated in the first lamp? Correct answer: 9 . 24021 W. Explanation: P 1 = IV 1 = (0 . 443396 A) (20 . 8396 V) = 9 . 24021 W 005 10.0points Resistances of 2 . 4 Ω, 3 . 6 Ω, and 5 . 1 Ω and a 27 V battery are all in series. Find the potential difference across the first (2 . 4 Ω) resistor. Correct answer: 5 . 83784 V. Explanation: Let : R 1 = 2 . 4 Ω , R 2 = 3 . 6 Ω , R 3 = 5 . 1 Ω , and V = 27 V .

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 5

GPII - Homework 4, Circuits - solutions - honea(cth632...

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

View Full Document
Ask a homework question - tutors are online