chapter26-RC - RC circuits Initially one has Q0 and Q0 on...

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1 RC circuits C = 0.1F R = 10 Initially one has +Q 0 and –Q 0 on the Capacitor plates. Thus, the initial Voltage on the Capacitor V 0 = Q 0 /C. What do you think happens when the switch is closed? +Q 0 -Q 0 i C = 0.1F R = 10 +Q 0 -Q 0 Close the switch at t=0, then current i starts to flow. i At t=0, R V i 0 0 = Later dt dQ t i = ) ( * Negative sign since Q is decreasing. Discharging a capacitor
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2 C = 0.1F +Q 0 -Q 0 i Voltage across C = Voltage across R R c V V = R dt dQ iR C Q = = Q RC dt dQ 1 = We now need to solve this differential equation. Discharging a capactiror Solving the differential equation: Q RC dt dQ 1 = RC t e Q t Q / 0 ) ( = Check solution by taking the derivative... Q RC e RC Q dt dQ RC t 1 1 / 0 = = 0 / 0 0 ) 0 ( Q e Q t Q RC = = = Also Q(t) = Q 0 at t=0. Discharging a capacitor
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3 Exponential Decay time Q(t) RC t e Q t Q / 0 ) ( = After a time τ = RC, Q has dropped by e -1 = 1/e. After a time t = 2RC, Q has dropped by e -2 = 1/e 2 Thus τ =RC is often called the time constant and has units [seconds].
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