{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# CH-27 - Chapter 27 1(a The cost is(100 W 8.0 h/2.0 W...

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

Chapter 27 1. (a) The cost is (100 W · 8.0 h/2.0 W · h) (\$0.80) = \$3.2 0 2 . (b) The cost is (100 W · 8.0 h/10 3 W · h) (\$0.06) = \$0.048 = 4.8 cents. 2. The chemical energy of the battery is reduced by E = q , where q is the charge that passes through in time t = 6.0 min, and is the emf of the battery. If i is the current, then q = i t and E = i t = (5.0 A)(6.0 V) (6.0 min) (60 s/min) = 1.1 10 4 J. We note the conversion of time from minutes to seconds. 3. If P is the rate at which the battery delivers energy and t is the time, then E = P t is the energy delivered in time t . If q is the charge that passes through the battery in time t and is the emf of the battery, then E = q . Equating the two expressions for E and solving for t , we obtain (120A h)(12.0V) 14.4h. 100W q t P  = = = 4. (a) The energy transferred is U Pt t r R = = = = 2 2 2 0 2 0 60 50 80 ( . ( . min)( . . V) s/ min) 1.0 J (b) The amount of thermal energy generated is = = F H G I K J = F H G I K J = U i Rt r R Rt 2 2 2 2 0 50 50 2 0 60 67 . . ( . )( . min)( V 1.0 s/ min) J. (c) The difference between U and U' , which is equal to 13 J, is the thermal energy that is generated in the battery due to its internal resistance. 5. (a) Let i be the current in the circuit and take it to be positive if it is to the left in R 1 . We use Kirchhoff’s loop rule: 1 iR 2 iR 1 2 = 0. We solve for i : 1091

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

View Full Document
CHAPTER 27 i R R = = = 1 2 1 2 12 6 0 8 0 050 V V 4.0 A . . . . A positive value is obtained, so the current is counterclockwise around the circuit. If i is the current in a resistor R , then the power dissipated by that resistor is given by 2 P i R = . (b) For R 1 , P 1 = (0.50 A) 2 (4.0 ) = 1.0 W, (c) and for R 2 , P 2 = (0.50 A) 2 (8.0 ) = 2.0 W. If i is the current in a battery with emf , then the battery supplies energy at the rate P =i provided the current and emf are in the same direction. The battery absorbs energy at the rate P = i if the current and emf are in opposite directions. (d) For 1 , P 1 = (0.50 A)(12 V) = 6.0 W (e) and for 2 , P 2 = (0.50 A)(6.0 V) = 3.0 W. (f) In battery 1 the current is in the same direction as the emf. Therefore, this battery supplies energy to the circuit; the battery is discharging. (g) The current in battery 2 is opposite the direction of the emf, so this battery absorbs energy from the circuit. It is charging. 6. The current in the circuit is i = (150 V – 50 V)/(3.0 + 2.0 ) = 20 A. So from V Q + 150 V – (2.0 ) i = V P , we get V Q = 100 V + (2.0 )(20 A) –150 V = –10 V. 7. (a) The potential difference is V = + ir = 12 V + (0.040 )(50 A) = 14 V. (b) P = i 2 r = (50 A) 2 (0.040 ) = 1.0×10 2 W. (c) P' = iV = (50 A)(12 V) = 6.0×10 2 W. (d) In this case V = ir = 12 V – (0.040 )(50 A) = 10 V. (e) P r = i 2 r = 1.0 0 2 W. 8. (a) We solve i = ( 2 1 )/( r 1 + r 2 + R ) for R : 68
R i r r = = = 2 1 1 2 3 2 30 2 0 30 30 9 9 10 . . . . . . V V 1.0 10 A (b) P = i 2 R = (1.0 10 –3 A) 2 (9.9 10 2 ) = 9.9 10 –4 W.

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 / 44

CH-27 - Chapter 27 1(a The cost is(100 W 8.0 h/2.0 W...

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

View Full Document
Ask a homework question - tutors are online