{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

903_Physics ProblemsTechnical Physics

# 903_Physics ProblemsTechnical Physics - Chapter 32 P32.12...

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

Chapter 32 243 P32.12 L N I NBA I NA I NI R N A R B = = = Φ µ π µ π 0 0 2 2 2 FIG. P32.12 P32.13 ε ε = = − 0 e L dI dt kt dI L e dt kt = − ε 0 If we require I 0 as t → ∞ , the solution is I kL e dq dt kt = = ε 0 Q Idt kL e dt k L kt = = = − z z ε ε 0 0 0 2 Q k L = ε 0 2 . Section 32.2 RL Circuits P32.14 I R e Rt L = ε 1 e j : 0 900 1 3 00 2 50 . . . ε ε R R e R = s H a f exp . . . . ln . . F H G I K J = = = R R 3 00 2 50 0 100 2 50 10 0 1 92 s H H 3.00 s a f P32.15 (a) At time t , I t e R t a f e j = ε τ 1 where τ = = L R 0 200 . s . After a long time, I e R R max = = −∞ ε ε 1 e j . At I t I a f = 0 500 . max 0 500 1 0 200 . . a f e j ε ε R e R t = s 1 0.5 0 0 0.2 0.4 0.6 t (s) I (A) I max FIG. P32.15 so 0 500 1 0 200 . . =
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}