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RC Circuit Analysis: Determining Time Constants, Capacitor Charging, and Filtering - Prof., Exams of Chemistry

Solutions to various problems related to rc circuits, including determining the time required for a capacitor to charge to a certain voltage, analyzing a circuit with multiple branches, and using rc circuits as high pass and low pass filters. It includes calculations and explanations of concepts such as time constants, impedance, and reactance.

Typology: Exams

Pre 2010

Uploaded on 08/19/2009

koofers-user-b2n
koofers-user-b2n 🇺🇸

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Download RC Circuit Analysis: Determining Time Constants, Capacitor Charging, and Filtering - Prof. and more Exams Chemistry in PDF only on Docsity! Key CHM 341 Electronics 2 EASY 1. Determine the time required to charge a capacitor in an RC circuit to 90% of full capacity given the following values of resistance and capacitance: R = 100 ohms, C = 1000 pF This is like a first order kinetics problem: the charge dissipation rate is proportional to the R C stored charge. We should expect exponential charge (or discharge). The applicable eqn. is: vcapacitor = Vi*(1- e-t/RC) which can be “rearranged” (if you would like) e-t/RC = -(vcap - Vi)/ Vi -t/RC = ln((E i - ecap)/E i) btw, the RC term is known as the time constant, τ, for the RC circuit: τ = RC = (100 Ω)(1,000x10-12 F) τ = 1.00x10-7 sec = 0.100 µsec We want to know when the voltage in the capacitor rises to 90% (or 0.90) of the applied voltage, Ei. So, plug in 0.90 for ecap and 1 for E and substitute in the known value for τ (i.e. RC): t = 2.3x10-7 sec or 0.23 µsec Note: it will always take 2.303*τ to charge a cap to 90% of the applied voltage. It will also take 2.303*τ to discharge a cap to 10% of the initial voltage. In one τ unit of time we achieve 37% charging (or discharging). You should be comfortable with this standard ln or exp math. A BIT HARDER 2. A 100 mF capacitor is connected in series with a 10.0 Ω resistor. This combination is connected in parallel with a 25.0 Ω resistor. Both branches are then connected in parallel to a 4.50 V battery that can be switched on and off. The capacitor starts off fully discharged. (a) What is the time constant in both branches when the switch is closed? (b) What is the maximum charge that the capacitor can attain after the switch is closed? (c) When will the voltage drop across the 10.0 Ω resistor be equal to 1.50 V after the switch is closed? (d) If the switched is opened, what will be the value of the new time constant and in which direction will the current flow through the 10.0 Ω resistor? (e) If the switch is opened after the capacitor is fully charged, how long will it take for there to be only one electron on the capacitor? (f) If the switch is opened after the capacitor is fully charged, how long will it take for the voltage across the 25.0 Ω resistor to drop to 1.50 V? (answer on next page) 1
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