Current in a RC circuit: Difference between revisions
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Mathematically: | Mathematically: | ||
Looking at the equation for current in an RC circuit, <math>I = ({\frac{V}{R}})e^{\frac{-t}{RC}} </math> , and with the knowledge that <math> t \approx 0 </math>, we can see that the term <math> e^{\frac{-t}{RC}} \approx 1 </math> . Therefore, the current in the circuit is the same as it would be as if the capacitor wasn't there. <math> I = frac{V}{R} </math> . | Looking at the equation for current in an RC circuit, <math>I = ({\frac{V}{R}})e^{\frac{-t}{RC}} </math> , and with the knowledge that <math> t \approx 0 </math>, we can see that the term <math> e^{\frac{-t}{RC}} \approx 1 </math> . Therefore, the current in the circuit is the same as it would be as if the capacitor wasn't there. <math> I = {\frac{V}{R}} </math> . | ||
===Middling=== | ===Middling=== | ||
Revision as of 12:48, 2 December 2015
Short Description of Topic
The Main Idea
Now that we have an understanding of steady state current, we can begin to examine the current in a RC circuit.
The current in a RC circuit differs from the current in a simple circuit because the capacitor acquires and releases charge; this varies the current.
This a graphical representation of the changing current and voltage on a capacitor with respect to time.
A Mathematical Model
The graph presented in the previous section is representative of a exponential equation. The current across a capacitor falls of like [math]\displaystyle{ I = ({\frac{V}{R}})e^{\frac{-t}{RC}} }[/math] where V is the voltage driving the current, R is the resistance of the circuit, t is time, and C is the capacitance of the capacitor.
As can be seen, as t approaches infinity, I approaches 0. In plain english, if the circuit is closed for a "very long time" the current in the circuit will approach zero.
A Computational Model
To aid in visualization, I will provide a "way of thinking".
When the switch is closed, the circuit is complete and the current begins to run. Remember current is just mobile electrons. The excess of electrons on one plate (along with the the consequential deficit of electrons on the other plate) provide a surface charge that repels the incoming electrons. Decreasing the flow of electrons decreases the current at the rate provided in the previous section.
Examples
Be sure to show all steps in your solution and include diagrams whenever possible
Simple
What is the current in the following circuit immediately after the switch is closed?
V_s = 5 V
R = 2 Ohms
C = 1 Farad
SOLUTION
We can solve this problem mathematically or analytically.
Mathematically:
Looking at the equation for current in an RC circuit, [math]\displaystyle{ I = ({\frac{V}{R}})e^{\frac{-t}{RC}} }[/math] , and with the knowledge that [math]\displaystyle{ t \approx 0 }[/math], we can see that the term [math]\displaystyle{ e^{\frac{-t}{RC}} \approx 1 }[/math] . Therefore, the current in the circuit is the same as it would be as if the capacitor wasn't there. [math]\displaystyle{ I = {\frac{V}{R}} }[/math] .
Middling
Difficult
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