Initial current in an RLC circuit

  • #1
mymodded
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TL;DR Summary
in RLC circuit with capacitor fully charged initially, ##q(t) = q_{max} e^{-Rt/2L} \cos(\omega t)## , if we want ##i(0) = \frac{d q}{dt}|_{t=0}## we don't get 0 as expected, whats wrong?
suppose that we have an RLC circuit where initially, the capacitor is fully charged, the charge in the capacitor is given by ##q(t) = q_{max} e^{-Rt/2L} \cos(\omega t)##, if we want to find the current, we would differentiate the charge, so $$\Large i(t) = \frac{d q}{dt} = \frac{d}{dt} (q_{max} e^{-Rt/2L} \cos(\omega t)) = q_{max} (-\frac{R}{2L}e^{-Rt/2L} \cos(\omega t) -\omega e^{-Rt/2L}\sin(\omega t))$$

we know that the current at t = 0 has to be equal to 0A (since the capacitor is fully charged), but if we plug in t = 0 in i(t) we don't get 0A, so what's wrong?
 
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  • #2
I think we need a circuit diagram.
RLC, series or parallel?
Fully charged to what?
 
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  • #3
Baluncore said:
I think we need a circuit diagram.
RLC, series or parallel?
Fully charged to what?
Sorry forgot to mention, it's a simple RLC circuit where they are all in series. I'm not exactly sure what you mean by fully charged to what. We fully charge a capacitor using a battery then disconnect and connect the capacitor to an inductor and a resistor
 
  • #5
mymodded said:
we know that the current at t = 0 has to be equal to 0A (since the capacitor is fully charged)
The capacitor has an initial charge. Explain why you think that makes the initial current zero?
 
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  • #6
Mister T said:
The capacitor has an initial charge. Explain why you think that makes the initial current zero?
The RLC circuit has two initial conditions (states) for the transient response, which are best represented (IMO) as capacitor voltage and inductor current. He doesn't have to "think" the inductor current is zero, he's telling us the initial inductor current is zero. You/he could choose whatever value you like.

I suspect much of his confusion is applying these ICs to the general transient solution. But, then he said the Q(t) solution was "given", so I guess that's one IC. I haven't actually reviewed the math, sometimes it's easier to give the answer than troubleshoot.

If the OP wants a better (more exact) response, he'll need to ask the question with a bit more rigor.
 
  • #7
I mean, the simple resolution is that the given charge function just does not comply with the initial conditions of the circuit.
 
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  • #8
Mister T said:
The capacitor has an initial charge. Explain why you think that makes the initial current zero?
Because when the capacitor starts discharging, the inductor opposes the change in current.

Edit: also that's the maximum charge on the capacitor
 
  • #9
mymodded said:
Because when the capacitor starts discharging
That's a current.
mymodded said:
the inductor opposes the change in current.
That just means that the change in current is slower than it otherwise would be. It doesn't mean the flow of current is halted, if it did the capacitor wouldn't be discharging.
 
  • #10
mymodded said:
Edit: also that's the maximum charge on the capacitor
This is an erroneous argument. The capacitor (to the linear approximation) will carry a charge that is proportional to its capacitance. There is no max charge except under a given circuit voltage in something like an RC circuit.

mymodded said:
Because when the capacitor starts discharging, the inductor opposes the change in current.
This is correct, but caution is advised. You need to understand the actual differential equation governing the current to make the argument proper.
 
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  • #11
Mister T said:
That just means that the change in current is slower than it otherwise would be. It doesn't mean the flow of current is halted, if it did the capacitor wouldn't be discharging.
This is also wrong. The derivative can be zero even when the second derivative is not, this is what will be going on here. The full differential equation governing the circuit (based on one Kirchhoff loop) is
$$
\frac{q}{C} + iR + Li' = 0 \quad \Longrightarrow \quad
\frac{q}{CL} + \frac{R}{L} q' + q'' = 0
$$
This differential equation needs two initial conditions. One on ##q(0)## and one on ##q'(0)##. If ##t = 0## is the moment the circuit is closed, then at that moment ##q'(0) = 0##.

Edit: The differential equation is that of a damped harmonic oscillator with the damping given by the resistance. The exact functional behaviour will depend on whether the system is under-, over-, or critically damped.
 
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  • #12
mymodded said:
Because when the capacitor starts discharging, the inductor opposes the change in current.
Inductance opposes a change in current, as you said. But that doesn't address the question of what the current is initially, i.e. what current level is it trying to change from? That is an initial condition that can't be calculated, it must be defined in advance for transient response problems. This is exactly analogous to the capacitor voltage, which could start at any given value.

This is more clear in the integral form of the inductor equation ##i(t)=\frac{1}{L} \int_0^t v(\tau) \, d\tau + i(0)##

PS: The circuit behavior can be solved from these equations, which explicitly include the ICs:
$$i(t)=\frac{1}{L} \int_0^t v_l(\tau) \, d\tau + i(0)$$
$$v_c(t)=\frac{1}{C} \int_0^t i(\tau) \, d\tau + v_c(0)$$
$$ v_r(t)=Ri(t)$$
$$v_c(t)+v_l(t)+v_r(t)=0$$
 
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  • #13
Orodruin said:
This is also wrong. The derivative can be zero even when the second derivative is not, this is what will be going on here.
Right, so the current is zero at that instant, but it's beginning to increase. I wasn't thinking correctly about the current being halted.
 
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  • #14
Voltage source is initially 1 V, then goes to 0 V at t=0.

RLC_plot.png

RLC_schematic.png
 
  • #15
DaveE said:
That is an initial condition that can't be calculated, it must be defined in advance for transient response problems.
I think this is it. My main confusion came when I was trying to solve a question in my textbook (and sorry if I'm not allowed to reply with a problem) that said


"An LC circuit consists of a capacitor, C = 2.50 μF, and an inductor, L = 4.00 mH. The capacitor is fully charged using a battery and then connected to the inductor. An oscilloscope is used to measure the frequency of the oscillations in the circuit. Next, the circuit is opened, and a resistor, R, is inserted in series with the inductor and the capacitor. The capacitor is again fully charged using the same battery and then connected to the circuit. The angular frequency of the damped oscillations in the RLC circuit is found to be 20.0% less than the angular frequency of the oscillations in the LC circuit.
a) Determine the resistance of the resistor.
b) How long after the capacitor is reconnected in the circuit will the amplitude of the damped current through the circuit be 50.0% of the initial amplitude?
c) How many complete damped oscillations will have occurred in that time?"



(question b, a and c are easy to solve), from question a, the resistance is found to be 48 ohms, and we find that it is an under damped situation. I believe that the main confusion is that when my textbook first derived the formula for the charge, they gave it to us directly as (for underdamped) ## q(t) = q_{max} e^{-Rt/2L} \cos(\omega t)## without showing us how we solved the differential equation (they did give us the differential equation, but I think because the method of solving it is similar to the one in a weakly damped mechanical oscillator, which we weren't taught for some reason, it wasn't shown), and I know that the solution is actually in terms of some constants A1 and A2 and ##q_{max}## and ##\omega## are related to them. And maybe differentiating the new q(t) function (the one that they gave us) made some complications or something, not sure. Which is why when we set t = 0, it didn't give us 0A (even though it should, according to what happened in the question)

Also the way that my textbook solved question b is really weird, they said "Even though the current is actually a damped oscillation, the magnitude of the oscillation is describe simply by the exponential decay, similar to that of the charge. Therefore, the current is at half of its maximum when the exponential is at a half" and procceeded to solve for t in the equation $$1/2 I_{max} = i_{max} e^{-Rt/2L}$$ and I really don't know how they are allowed to do that, if it's a damped oscillation, how are they allowed to drop the cosine and sine?

nvm they asked for when the AMPLITUDE becomes half, not when the current becomes half, my bad.
 
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  • #16
mymodded said:
and I know that the solution is actually in terms of some constants A1 and A2 and qmax and ω are related to them
This is incorrect. While the amplitude ##q_{\rm max}## is related to the constants, ##\omega## is directly expressed in terms of C/L/R. The other constant should be a phase shift ##\phi## in ##\omega t \to \omega t + \phi##.
 
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  • #17
Orodruin said:
This is incorrect. While the amplitude ##q_{\rm max}## is related to the constants, ##\omega## is directly expressed in terms of C/L/R. The other constant should be a phase shift ##\phi## in ##\omega t \to \omega t + \phi##.
Yes yes my bad
 
  • #18
Orodruin said:
This is an erroneous argument. The capacitor (to the linear approximation) will carry a charge that is proportional to its capacitance. There is no max charge except under a given circuit voltage in something like an RC circuit.
Agreed. The analysis really has to be tackled formally and with maths (of course). The initial conditions are whatever is stated (and they must be stated). IMO, there's no point in just chatting about the situation, except when it comes to describing what the 'graph' tells you.

The "max charge" can only be determined at the end of the calculation.
 
  • #19
mymodded said:
The capacitor is fully charged using a battery
Note that this is different from simply saying "fully charged". Saying it's fully charged using a battery simply means it was connected to the battery for a very long time, long enough for the voltage across the capacitor to match the voltage across the battery.
 
  • #20
sophiecentaur said:
The "max charge" can only be determined at the end of the calculation.
It's simply equal to ##CV## where ##V## is the voltage across the battery that was used to charge the capacitor before it was connected to the inductor.
 
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  • #21
Mister T said:
It's simply equal to ##CV## where ##V## is the voltage across the battery that was used to charge the capacitor before it was connected to the inductor.
Only if the system starts with no current. i.e. appropriate initial conditions That may have been assumed.
 
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  • #22
mymodded said:
... we know that the current at t = 0 has to be equal to 0A (since the capacitor is fully charged), ...
We know the current was zero prior to the connection, since there was no circuit before the connection at t=0, so no current could flow.

At the instant the charged capacitor is connected, the capacitor voltage appears across the inductor. The current then begins to rise, as di/dt = V/L .
 
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  • #23
Oh yes. No current initially defines the max voltage
 
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