Voltage and Current for 0.2-F Capacitor

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    Capacitor Voltage
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The discussion focuses on determining the expression for current i(t) through a 0.2-F capacitor given the voltage equation v(t)=4(1-e^(-10t)) for t>0. The constant e represents the base of the natural logarithm, which is crucial for taking the derivative. The current can be calculated using the formula i(t)=c (dv(t))/dt, where c is the capacitance. As time increases, the voltage approaches a maximum, indicating the capacitor is charging, and the current will decrease to zero as it becomes fully charged. Understanding the behavior of the voltage function is essential for accurately analyzing the current over time.
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Homework Statement


Voltage across 0.2-F cap is given by

v(t)=4(1-Є^(-10t)) for t>0


determine expression for the current i(t)

Homework Equations



i(t)=c (dv(t))/dt

The Attempt at a Solution



I'm really not sure were to begin with this problem what is the Є symbol and how do you take the derivative of it?
 
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pmontone, you're in luck. That's not a Є symbol, but rather the constant e (= 2.718), the basis for the exponential function.
 
pmontone said:

Homework Statement


Voltage across 0.2-F cap is given by

v(t)=4(1-Є^(-10t)) for t>0


determine expression for the current i(t)

Homework Equations



i(t)=c (dv(t))/dt

The Attempt at a Solution



I'm really not sure were to begin with this problem what is the Є symbol and how do you take the derivative of it?

First, look at your given equation v(t) and try to understand its meaning. What happens to v(t) as t increases? Is this equation representative of a "charging" or "discharging" capacitor? Now, based on this answer, how do you expect the current across the capacitor to behave over time? Use your equation and then analyze the result for physical accuracy. Remember that a fully-charged capacitor acts as an open circuit (no current).
 

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