{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Midterm2Solution

Midterm2Solution - PHYSICS 9C MIDTERM 2 I certify by my...

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

PHYSICS 9C MIDTERM 2 March 10, 2008 I certify by my signature below that I will abide by the UC Davis Code of Academic Conduct. This includes not copying from anyone else’s exam not letting any other student copy from my exam not discussing this exam with any student who has not yet taken it, nor providing any information, written or oral, that might get to a student who has not yet taken it. Anyone suspected of cheating will be automatically reported to Student Judicial Affairs. LAST NAME: FIRST NAME: STUDENT ID: (P R I N T) (P R I N T) (LAST 4 DIGITS) Signature: Problem 1. (2 points) In the sketch, all bulbs have equal resistances. The brightness of bulbs B and C added together, compared with the brightness of bulb A , is A) four times as large B) twice as large. C) the same. D) half as large. E) impossible to tell Solution: The current inside bulb A is i A =V/R, while the current inside bulb B or C is i B =i C =V/(2R). The power of bulb A is P A =i A 2 *R=V 2 /R while the power of bulbs B+C is P B +P C =i B 2 R+i C 2 R=V 2 /(4R)+V 2 /(4R)=V 2 /(2R). Therefore the brightness of bulbs B and C added together is half as large as the brightness of bulb A.

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

View Full Document