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quiz_solutions08

Fundamentals of Physics, (Chapters 21- 44) (Volume 2)

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PY212 Solutions Quiz 8 This problem can be solved in two ways: assuming R is small (valid assumption) or doing it exactly. We shall first show here the approximation. We do not have an equation for the current as a function of time, but we do have the charge; Eq. 31-25: ( ) ( ) φ ω + = t Qe t q L Rt cos 2 / with /H 2 . 0 2 Ω = L R A and rad/s 1 . 129 1 ) 2 / ( 2 2 2 2 = + = + = ω ω ω A LC L R . The current can be found by take the derivative with respect to time: ( ) ( ) ( ) + + = = φ ω ω φ ω t e t e L R Q dt dq t i L Rt L Rt sin cos 2 2 / 2 / The problem states that initially the current is zero, so we have a situation very similar 1 to Fig. 31-1 with (a) representing the situation at t = 0. According to that figure the charge on the capacitor q ( t ) is at a maximum for t = 0. Hence, from ( ) φ cos 0 Q q = we conclude that φ = 0, and therefore C 10 5 . 0 ) 0 ( 6 × = = q Q . In Fig. 31-1 we see that the current is at a maximum after ¼ of the period. Hence, the current is maximum at ω π 2 4 1 4 1 4
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