1. An 8-V battery is connected to an RC circuit (R = 1 ? and C = 9 ?F). Initiall
ID: 2236341 • Letter: 1
Question
1. An 8-V battery is connected to an RC circuit (R = 1 ? and C = 9 ?F). Initially, the capacitor is uncharged. What is the final charge on the capacitor (in ?C)? 2. A 5-V battery is connected to an RC circuit (R = 13 ? and C = 3.0 ?F). If the capacitor is initially uncharged, what is the current in the circuit immediately after the connection is made? 3. The quantity t1/2=? ln 2 is called the half-life of an exponential decay, where ?=RC is the time constant in an RC circuit. The current in a discharging RC circuit drops by half whenever t increases by t1/2. For a circuit with R=2 k? and C=4 ?F, if the current is 6 mA at t=5 ms, at what time (in ms) will the current be 3 mA?Explanation / Answer
1. The RC circuit will contain current until the capacitor's voltage equals the battery's voltage. At this point, the charge on the capacitor is
(Q = CV = (9 mu F)(8V) = 72mu C)
2. Immediately after the circuit is closed, there is no charge on the capacitor, so it has no voltage. Therefore, the voltage of the resistor must balance out the voltage of the battery, and
(i_{initial} = V/R = (5V)/(13Omega) = 0.38A)
3. Whenever a time passes that equals the half life, the current drops by half. Since the current goes from 6mA to 3mA (which is a halving of the current), the time that passes is one half life:
(t_{1/2} = au ln(2) = (RC)ln2 = (2*10^3Omega)(4*10^{-6} F)ln(2) = 5.55*10^{-3}s )
So, from the initial time, 5.55ms passes. The final time is therefore 10.55ms.
Hope this helps
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