16 HMK #4 Chp 26a

16 HMK #4 Chp 26a - Ch. 26 Add. Probs. PHYSICS 2B Spring...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Ch. 26 Add. Probs. PHYSICS 2B Spring 2011 Homework Solutions – Chapter 26 Additional Problems Author: Evan Grohs - egrohs@physics.ucsd.edu Ch. 26 Problems: 1, 2, 3, 5, 6, 8, 11, 17, 55, 59, 62, 70, 72 Page 1/6
Background image of page 1

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

View Full DocumentRight Arrow Icon
Ch. 26 Add. Probs. page 2/6 D IRECT -C URRENT C IRCUITS 26.1. I DENTIFY : The newly-formed wire is a combination of series and parallel resistors. S ET U P : Each of the three linear segments has resistance R /3. The circle is two R /6 resistors in parallel. E XECUTE : The resistance of the circle is R /12 since it consists of two R /6 resistors in parallel. The equivalent resistance is two R /3 resistors in series with an R /12 resistor, giving R equiv = R /3 + R /3 + R /12 = 3 R /4. E VALUATE : The equivalent resistance of the original wire has been reduced because the circle’s resistance is less than it was as a linear wire. 26.2. I DENTIFY : It may appear that the meter measures X directly. But note that X is in parallel with three other resistors, so the meter measures the equivalent parallel resistance between ab . S ET U P : We use the formula for resistors in parallel. E XECUTE : 1/(2.00 ) = 1/ X + 1/(15.0 ) + 1/(5.0 ) + 1/(10.0 ), so X = 7.5 . E VALUATE : X is greater than the equivalent parallel resistance of 2.00 . 26.3. I DENTIFY : The emf of the battery remains constant, but changing the resistance across it changes its power output. S ET U P : The power output across a resistor is P V 2 R . E XECUTE : With just 1 R , 2 1 1 V P R and 11 (36.0 )(25.0 ) 30.0 V VP R   is the battery voltage. With 2 R added, tot 40.0 R  . 22 tot (30.0 V) 22.5 W 40.0 V P R . E VALUATE : The two resistors in series dissipate electrical energy at a smaller rate than 1 R alone. 26.5. I DENTIFY : The equivalent resistance will vary for the different connections because the series-parallel combinations vary, and hence the current will vary. S ET U P : First calculate the equivalent resistance using the series-parallel formulas, then use Ohm’s law ( V = RI ) to find the current. E XECUTE : (a) 1/ R = 1/(15.0 ) + 1/(30.0 ) gives R = 10.0 . I = V / R = (35.0 V)/(10.0 ) = 3.50 A. (b) 1/ R = 1/(10.0 ) + 1/(35.0 ) gives R = 7.78 . I = (35.0 V)/(7.78 ) = 4.50 A 26
Background image of page 2
Ch. 26 Add. Probs. page 3/6 (c) 1/ R = 1/(20.0 ) + 1/(25.0 ) gives R = 11.11 , so I = (35.0 V)/(11.11 ) = 3.15 A. (d) From part (b), the resistance of the triangle alone is 7.78 . Adding the 3.00- internal resistance of the battery gives an equivalent resistance for the circuit of 10.78 . Therefore the current is I = (35.0 V)/(10.78 ) = 3.25 A E VALUATE : It makes a big difference how the triangle is connected to the battery.
Background image of page 3

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

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 12/18/2011 for the course PHYS 2B 2b taught by Professor Hirsch during the Spring '10 term at UCSD.

Page1 / 6

16 HMK #4 Chp 26a - Ch. 26 Add. Probs. PHYSICS 2B Spring...

This preview shows document pages 1 - 4. Sign up to view the full document.

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