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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
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 time-dependent 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).
(c) Consider Eq. 32-28 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. 32-28 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. 31-29 yields (for all values of n) ⎛
5π π ⎞
+ ⎟ = (3.91× 10−3 A) ⎜ −
i = I sin ⎜ 2nπ −
⎜ 2 ⎟ = −3.38 × 10 A,
or | i | = 3.38 ×10−2 A. ...
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