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Unformatted text preview: 32. (a) The circuit consists of one generator across one capacitor; therefore, εm = VC.
Consequently, the current amplitude is
I= εm
XC = ω Cε m = (377 rad / s)(4.15 × 10−6 F)(25.0 V) = 3.91 × 10−2 A . (b) When the current is at a maximum, the charge on the capacitor is changing at its
largest rate. This happens not when it is fully charged (±qmax), but rather as it passes
through the (momentary) states of being uncharged (q = 0). Since q = CV, then the
voltage across the capacitor (and at the generator, by the loop rule) is zero when the
current is at a maximum. Stated more precisely, the timedependent emf ε(t) and current
i(t) have a φ = –90° phase relation, implying ε(t) = 0 when i(t) = I. The fact that φ = –90°
= –π/2 rad is used in part (c).
1
(c) Consider Eq. 3228 with ε = − 2 ε m . In order to satisfy this equation, we require
sin(ωdt) = –1/2. Now we note that the problem states that ε is increasing in magnitude,
which (since it is already negative) means that it is becoming more negative. Thus,
differentiating Eq. 3228 with respect to time (and demanding the result be negative) we
must also require cos(ωdt) < 0. These conditions imply that ωt must equal (2nπ – 5π/6) [n
= integer]. Consequently, Eq. 3129 yields (for all values of n) ⎛
5π π ⎞
3⎞
⎛
−2
+ ⎟ = (3.91× 10−3 A) ⎜ −
i = I sin ⎜ 2nπ −
⎜ 2 ⎟ = −3.38 × 10 A,
⎟
6 2⎠
⎝
⎝
⎠
or  i  = 3.38 ×10−2 A. ...
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Charge, Current

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