ch31-p032

# ch31-p032 - 32. (a) The circuit consists of one generator...

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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 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). 1 (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⎞ ⎛ −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.

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