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# ch27 - 1059 Chapter 27 1(a Let i be the current in the...

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Unformatted text preview: 1059 Chapter 27 1. (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 : 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 = 2 1 i R (0.50 A) 2 (4.0 ) = 1.0 W, (c) and for R 2 , P 2 = 2 2 i R (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 = 1 i (0.50 A)(12 V) = 6.0 W (e) and for 2 , P 2 = 2 i (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. 2. The current in the circuit is i = (200 V – 50 V)/(3.0 + 2.0 ) = 30 A. So from V Q + 200 V – (2.0 ) i = V P , we get V Q = 100 V + (2.0 )(30 A) –200 V = –40 V. 3. (a) The potential difference is V = + ir = 12 V + (50 A)(0.040 ) = 14 V. (b) P = i 2 r = (50 A) 2 (0.040 ) = 1.0×10 2 W. CHAPTER 27 1060 (c) P' = iV = (50 A)(12 V) = 6.0×10 2 W. (d) In this case V = – ir = 12 V – (50 A)(0.040 ) = 10 V. (e) P r = i 2 r =(50 A) 2 (0.040 ) = 1.0 2 W. 4. (a) The loop rule leads to a voltage-drop across resistor 3 equal to 5.0 V (since the total drop along the upper branch must be 12 V). The current there is consequently i = (5.0 V)/(200 ) = 25 mA. Then the resistance of resistor 1 must be (2.0 V)/ i = 80 . (b) Resistor 2 has the same voltage-drop as resistor 3; its resistance is 200 . 5. 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. 6. (a) The cost is (100 W · 8.0 h/2.0 W · h) (\$0.85) = \$3.4 2 . (b) The cost is (100 W · 8.0 h/10 3 W · h) (\$0.06) = \$0.048 = 4.8 cents. 7. (a) The energy transferred is U Pt t r R 2 2 2 0 2 0 60 50 80 ( . ( . min) ( ....
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ch27 - 1059 Chapter 27 1(a Let i be the current in the...

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