16 HMK #4 Chp 26a

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

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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

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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
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.

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## 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.

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

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