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RC Circuit Discharge: Time Constant and Charge Decay - Prof. Michael J. Moloney, Assignments of Physics

An explanation of how a charged capacitor discharges through a resistor, introducing the concept of time constant (rc). It includes exercises to calculate the time it takes for a capacitor to discharge a certain percentage of its charge or reach a specific voltage.

Typology: Assignments

Pre 2010

Uploaded on 08/18/2009

koofers-user-uwh
koofers-user-uwh 🇺🇸

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Download RC Circuit Discharge: Time Constant and Charge Decay - Prof. Michael J. Moloney and more Assignments Physics in PDF only on Docsity! RC Worksheet MJM February 4, 2005 Name_____________________ Box _____ C = q/V (See text, p. 985) We start with a charged capacitor I C having a charge q on it. The initial charge is Qo . + + + + + We will close a switch at t=0 and begin to discharge C q + the capacitor through the resistor. - - - - - - - R The charge q on the capacitor will be decreasing, so - dq/dt will be negative. We want the current i in the circuit to be positive, so we need i = -dq/dt . The loop equation during discharge is q/C - iR = 0, or q/C - R dq/dt = 0. We could multiply this by C and get q + RC dq/dt = 0. The second term must have units of charge, so RC must be a time! And indeed 1 ohm-farad = 1second. RC is called the 'time constant'. Separating the equation gives dq/q = -dt/(RC). this integrates to Qo q dq/q = -0 t dt/(RC) => ln(q/Qo) = -t/(RC) => q = Qo exp(-t/(RC)) Exercise: Write the boxed equation in terms of the capacitor voltage V. Exercise: Calculate how much of the charge (or voltage) is left after we have waited a time equal to 3 RC. Exercise: Ditto except that we wait 6 time constants . Exercise: Given a R = 2.5  and C = 21000  F, how long must we wait until this capacitor has only 5% of its original charge left? (more on the back) 1
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