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Electrical engineering
Course: Electrical engineering > Unit 2
Lesson 4: Natural and forced response- Capacitor i-v equations
- A capacitor integrates current
- Capacitor i-v equation in action
- Inductor equations
- Inductor kickback (1 of 2)
- Inductor kickback (2 of 2)
- Inductor i-v equation in action
- RC natural response - intuition
- RC natural response - derivation
- RC natural response - example
- RC natural response
- RC step response - intuition
- RC step response setup (1 of 3)
- RC step response solve (2 of 3)
- RC step response example (3 of 3)
- RC step response
- RL natural response
- Sketching exponentials
- Sketching exponentials - examples
- LC natural response intuition 1
- LC natural response intuition 2
- LC natural response derivation 1
- LC natural response derivation 2
- LC natural response derivation 3
- LC natural response derivation 4
- LC natural response example
- LC natural response
- LC natural response - derivation
- RLC natural response - intuition
- RLC natural response - derivation
- RLC natural response - variations
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RC natural response - example
An example of the exponential natural response of an RC circuit, with real component values. Created by Willy McAllister.
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according to the relation i(t) = v(t)/ R .. when t equals zero,
the value of i(t) will be 2V/1000 so will be 2 mA . why have you made it 1.8 ?(11 votes)
You are correct. The peak of the current at time=0 should be 2 mA, not 1.8.(6 votes)
In the graph for current, why aren't the values for the current negative if, in a capacitor, i = C dv/dt? Since the voltage is decreasing with time, then i should be less than zero, correct?(4 votes)- Jamil - You make a good point. I forgot to copy the current arrow from the previous video onto the schematic in this video. The current arrow points to the left, into the top of the resistor. This means i is defined to flow up out of the capacitor (the opposite of the sign convention for passive components). This introduces a minus sign in the capacitor i-v equation. The capacitor equation is i = -C dv/dt, (with a minus sign). And that's why the current plot shows a positive spike. Thanks for pointing this out.(4 votes)
So the capacitor will never get empty!(3 votes)- If the math is a perfect model of the RC, then you are right, the exponential never reaches 0V unless you wait infinity time. However, it's important to remember the math is a model. It models two ideal components, R and C. The model does not take into account the real world of making R's and C's. It doesn't account for how these components are made of atoms, or how the atoms are vibrating due to the temperature. So in reality, the model has limitations. In the real world, the charge on the capacitor gets so low you can't tell it apart from normal electron activity at room temperature. After 5 or 10 time constants, for all practical purposes the charge on the capacitor is zero.(4 votes)
This equation implies that the capacitor never discharged completely nor the current falls to zero... Is this phenomenon real?(2 votes)- Good question. If you look at the mathematical equation you know that the exponential term never actually goes to zero. The thing to remember is that these equations are models of the real-world circuits we build. And those models are really useful but they are incomplete, and don't capture absolutely everything about a real capacitor and real resistor. Like for example the model knows nothing about temperature and vibrating atoms and Brownian motion. So from a practical engineering point of view, if time is allowed to go out for 5 or 10 times the value of RC, we recognize that the RC response is pretty much over. The voltage gets very very small and "sinks into the noise," becoming indistinguishable from 0 volts.(4 votes)
In the current-graph, the peak at zero is not exactly on the y-axis (current), but a little bit to the right.
Is this intentional? I.e., does this mean that current starts flowing a little later than the voltage starts dropping?(1 vote)- The current plot has a slight slope away from zero right after t=0. That's an artifact of the Excel plot I used to prepare this illustration. The "nearly vertical" line connects two successive time points.
In the ideal case the curve should follow exactly what the equation says, with an infinitely abrupt vertical rise at t(0+) (the moment in time just after the switch closes).(1 vote)
Video transcript
- [Voiceover] We just
derived what the current is and the voltage. These are both the natural
response of the RC. Now what I did is I went
ahead and I plotted out this using a computer, just using Excel, to plot out what these
two expressions look like, and let me show you that. So let's do a real quick example here with real values just to
see how this equation works. We'll say that this is 1,000 ohms. That's our resistor. And we'll say C is one microfarad. And what we wanna work out is R times C equals 10 to the third times 10 to the minus six and that equals 10 to
the minus three seconds or one millisecond. That's the product of R and C. I forgot V-naught. Let's say we put two
volts on this capacitor to start with, like that. And now we can say V of
t equals V-naught two volts times e to the minus t
over one millisecond. And that's our natural response
for this particular circuit. Now let me show you what that looks like. This is V of t on this side. Equals two e to the minus
t over one millisecond. And you can see it starts at two volts and then sags down as we predicted, and that's an exponential curve. And then over here on this side, i, as we said before, starts out at zero in the capacitor, the current
in the capacitor is zero, and as soon as we throw open that switch, the charge charges over
through the resistor and this is the equation here, i of t equals two volts over 1,000 e to the minus t over RC or t over one millisecond. So this is what we call
the natural response of an RC circuit, and you'll run into this in almost every circuit you ever build.