810_Physics ProblemsTechnical Physics

810_Physics ProblemsTechnical Physics - 150 Direct Current...

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

View Full Document Right Arrow Icon
150 Direct Current Circuits (b) The hot pot is designed to carry current I V == = P 500 417 W 120 V A .. It has resistance R V I = 120 28 8 V 4.17 A 28.8 240 V 144 FIG. P28.56(b) In terms of current, since 5 . A 0.833 A = , we can place five light bulbs in parallel and the hot pot in series with their combination. The current in the generator is then A and it delivers power = IV 4 17 240 1 000 A V W af . P28.57 The current in the simple loop circuit will be I Rr = + ε . (a) VI r R ter =− = + and V ter as R →∞ . (b) I = + and I r as R 0. (c) + IR R 22 2 d dR R = + + + = 2 0 2 3 2 2 εε Then 2 RRr =+ and = . FIG. P28.57 P28.58 The potential difference across the capacitor
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