P28_058 - to their next task(such as handling the sensitive electronic equipment(c We solve V = V e − t/RC for R with the new values V = 1400 V

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58. (a) In the process described in the problem, no charge is gained or lost. Thus, q = constant. Hence, q = C 1 V 1 = C 2 V 2 = V 2 = V 1 C 1 C 2 = (200) µ 150 10 = 3000 V . (b) Eq. 28-36, with τ = RC , describes not only the discharging of q but also of V .Thu s , V = V 0 e t/τ = t = RC ln µ V 0 V = ( 300 × 10 9 )( 10 × 10 12 F ) ln µ 3000 100 which yields
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Unformatted text preview: to their next task (such as handling the sensitive electronic equipment). (c) We solve V = V e − t/RC for R with the new values V = 1400 V and t = 0 . 30 s. Thus, R = t C ln( V /V ) = . 30 s (10 × 10 − 12 F) ln(1400 / 100) = 1 . 1 × 10 10 Ω ....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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