794_Physics ProblemsTechnical Physics

794_Physics ProblemsTechnical Physics - 134 Direct Current...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
134 Direct Current Circuits P28.4 (a) Here ε =+ IR r af , so I Rr = + = + = 12 6 00800 248 . . . V 5.00 A ΩΩ bg . Then, ∆Ω VI R == = 2 48 5 00 12 4 .. . A V a fa f . (b) Let I 1 and I 2 be the currents flowing through the battery and the headlights, respectively. Then, II 12 35 0 . A , and −−= Ir Ir 0 so + = 22 35 0 0 080 0 5 00 12 6 . . V b g giving I 2 193 = . A . Thus, V 2 500 965 . V . FIG. P28.4 Section 28.2 Resistors in Series and Parallel P28.5 VIR R 11 1 200 A and VIR R R = + 21 2 1 160 300 Therefore, 2 00 1 60 3 00 ... RR or R 1 12 0 = . P28.6 (a) R p = + = 1 1700 1100 412 . RRRR s =++= + + = 123 400 412 900 171 . (b) R = 34 0 17 1 V = I I = 199 . A for 4 00 , 9 00 resistors. Applying R = , 818 . V Ω= 700 = I a f so I = 117 . A for 7 00 resistor 100 = I so I = 0818 . A for 10 0 resistor. FIG. P28.6 P28.7 For the bulb in use as intended,
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.
Ask a homework question - tutors are online