- #1
terryds
- 392
- 13
Homework Statement
Please help
Homework Equations
I = V/R
I_capacitor = I e^(-t/RC)
The Attempt at a Solution
I only know the equation V=IR and the capacitor current function. Please help
If you are studying RC circuits you must have already covered other circuit analysis methods, laws, and formulas.terryds said:I only know the equation V=IR and the capacitor current function. Please help
gneill said:If you are studying RC circuits you must have already covered other circuit analysis methods, laws, and formulas.
You should be able to make some attempt either by writing the differential equation for the circuit and solving, or by determining initial and final conditions and applying the known response functions that govern RC circuits in general.
I note that your problem doesn't specify a value for the supply voltage. Is this intentional?
No, show an attempt first. That's the rule here.terryds said:The voltage source is 5V.
Actually I just want to know how to get the function, not about the value.
Please guide me :\
gneill said:No, show an attempt first. That's the rule here.
Please look at my reply above. I forgot to quote you beforegneill said:I see that you've chosen to write the differential equations. Okay, what you've done so far is fine. You''ve got three unknowns though (i1, i2, and i3) so you'll need three equations. You've got two so far.
It is true that the potential across the capacitor will increase over time, but this is not what "rise" or "drop" terms refer to in this context. They refer to the change in potential seen when you sum the potential changes around the circuit according to KVL.terryds said:Before I continue, why is there voltage drop across the capacitor? Isn't it supposed to be voltage rise across the capacitor?
gneill said:It is true that the potential across the capacitor will increase over time, but this is not what "rise" or "drop" terms refer to in this context. They refer to the change in potential seen when you sum the potential changes around the circuit according to KVL.
If you do a "KVL walk" around the loop in a clockwise direction (the direction of the current flow) there's a potential rise across the battery, then potential drops across the resistor and capacitor.
The value of the voltage source doesn't matter. It's the polarity of the change in potential when you "walk over" the component that matters. If you did your KVL walk in the counterclockwise direction (against the current flow in this case) then there would be potential rises on the the capacitor and resistor and a potential drop across the voltage source.terryds said:So, the voltage drop refers to the voltage of something - voltage source?
Is it correct?
I mean, even if potential across capacitor is increasing, but since the value of it is still less than the voltage source, so we can say that it is voltage drop, right?
gneill said:The value of the voltage source doesn't matter. It's the polarity of the change in potential when you "walk over" the component that matters. If you did your KVL walk in the counterclockwise direction (against the current flow in this case) then there would be potential rises on the the capacitor and resistor and a potential drop across the voltage source.
gneill said:@terryds, See my edit to my last post.
Your differential equation looks fine. You can simplify that first term a bit more; You should see a familiar expression involving resistors when you do.
It's the initial current. Go back to the circuit and find the initial current value for ##i_2##.terryds said:It becomes
(R1*R2/(R1+R2)) di2/dt - i2/C1 = 0
Let (R1*R2)/(R1+R2) be R_parallel
And then, I get
##i_2 = k e^{\frac{t}{C_1 R}}##
Is it correct?
But, how to know the value of k?
When t=0, this simplifies to ##i_2 = k##terryds said:And then, I get
##i_2 = k e^{\frac{t}{C_1 R}}##
Is it correct?
But, how to know the value of k?
gneill said:It's the initial current. Go back to the circuit and find the initial current value for ##i_2##.
NascentOxygen said:When t=0, this simplifies to ##i_2 = k##
So k will be the initial value of ##i_2## before current has had a chance to add charge to the capacitor plates.
gneill said:No. At time = 0+, what does an uncharged capacitor "look like" in a circuit? What is it equivalent to?
According to your figure (and equations) in post 12, ##i_2## is the capacitor current. I noticed at the time that you had changed your current definitions from your initial post (where i2 went through the 100 Ohm resistor) , but since you wrote correct equations with this new definition I didn't bother to mention the change.terryds said:It is short circuit.
So i2 initial is zero, since all the current will rather go through the capacitor than r2 right?
But, it's very funny to say that k is 0
gneill said:According to your figure (and equations) in post 12, ##i_2## is the capacitor current. I noticed at the time that you had changed your current definitions from your initial post (where i2 went through the 100 Ohm resistor) , but since you wrote correct equations with this new definition I didn't bother to mention the change.
Yes.terryds said:i2 initial is V/R1, right?
So,
##I_{2i} = \frac{V}{R_1}e^{\frac{t}{C_1 R}}##
Is it correct now?
An RC circuit is a type of electrical circuit that contains a resistor (R) and a capacitor (C) connected in series or parallel. The R represents the resistance of the circuit, while the C represents the capacitance of the circuit.
In an RC circuit, current flows from the positive terminal of the voltage source through the resistor and into the capacitor. The capacitor then charges up until the voltage across it is equal to the voltage of the source. Once this happens, no more current will flow through the circuit.
The current in an RC circuit is not constant, but instead varies with time. Initially, the current is high as the capacitor charges up, but it decreases as the capacitor becomes fully charged. Eventually, the current will reach zero when the capacitor is fully charged and no more current can flow through the circuit.
The resistance and capacitance values in an RC circuit affect the rate at which the capacitor charges up and the time it takes for the current to reach zero. A higher resistance will result in a longer charging time and a slower decrease in current, while a higher capacitance will result in a shorter charging time and a faster decrease in current.
The time constant, denoted by τ (tau), is a measure of how quickly a capacitor charges up in an RC circuit. It is calculated by multiplying the resistance (R) by the capacitance (C), such as τ = RC. This value represents the time it takes for the current to reach 63.2% of its maximum value in the circuit.