6 7 V-8 1 2 3 V-9 1 2 3 100 J 4 5 V-10 1 0.36 MJ = 0.1 , 10 hrs we use 1, cost: 5¢ E1+ V-12 1 Loop 1: I1r1 + I1R1 = I2R2 E2 Loop 2: I2R2 + I3R3 + I3r2 = 2 Point A: I1 + I2 = I3 B: I3 = I1 + I2 In-Text Fill-In Answers Page Fill-In # Fill-In Answer V-13 1 I. 6V + I1(1) + I1(8) = I2(4) 6 = 4I2 9I1 – II. I2(4) + I3(7) + I3(1) = 10V 5 = 2I2 + 4I3 III. I1 + I2 = I3 2 I1 = A, I2 = 0.99 A, I3 = 0.76 A –0.23 3 a. = (I1)(R1) = (0.23 A)(8) = 1.8 V V1 4 b. P = (I1)(E1) = (0.23 A)(6 V) = 1.4 W 5 c. #2 = (E2) (I3r2) = (10 V) (0.76 A)(1) = 9.3 V VBatt. – – V-14 1 d. Conversion: 10 min x 60 s/min = 600 s E2 = (P)(t) = (I2)2(R2)(t) = (0.99 A)2( 4 )( 600 s ) = 2352 J ~ 2.4 kJ 2 d. E3 = P)(t) = (I3)2(R3)(t) = (0.76 A)2(7)( 600 s ) = 2426 J ~ 2.4 kJ V-18 1 Series,
6 7 V-8 1 2 3 V-9 1 2 3 100 J 4 5 V-10 1 0.36 MJ = 0.1 , 10 hrs we use 1, cost: 5¢ V-12 1 Loop 1: E1+ I1r1 + I1R1 = I2R2 Loop 2: I2R2 + I3R3 + I3r2 = E2 2 Point A: I1 + I2 = I3 B: I3 = I1 + I2 In-Text Fill-In Answers Page Fill-In # Fill-In Answer V-13 1 I. 6V + I1(1) + I1(8) = I2(4) 6 = 4I2 – 9I1 II. I2(4) + I3(7) + I3(1) = 10V 5 = 2I2 + 4I3 III. I1 + I2 = I3 2 I1 = –0.23 A, I2 = 0.99 A, I3 = 0.76 A 3 a. V1 = (I1)(R1) = (0.23 A)(8) = 1.8 V 4 b. P = (I1)(E1) = (0.23 A)(6 V) = 1.4 W 5 c. VBatt. #2 = (E2) – (I3r2) = (10 V) – (0.76 A)(1) = 9.3 V V-14 1 d. Conversion: 10 min x 60 s/min = 600 s E2 = (P)(t) = (I2)2(R2)(t) = (0.99 A)2( 4 )( 600 s ) = 2352 J ~ 2.4 kJ 2 d. E3 = P)(t) = (I3)2(R3)(t) = (0.76 A)2(7)( 600 s ) = 2426 J ~ 2.4 kJ V-18 1 Series,
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