0
$\begingroup$

Can a inductor and capacitor series combination control current in a resistance-less wire, connected with finite dc voltage battery, from reaching infinity?

Why it should

  1. In the intial stage at $t=0$ the current wants to rise to infinity, since $R=0$, but the change in current is restricted by the inductor. So current rises very slowly.

  2. Meanwhile the capacitor is being charged with whatever current flowing in the circuit in initial stages thus voltage at its plates keep rising.

  3. After some time, when inductor is no more able to restrict the rate of change of current it is restricted by the voltage developed at plates of capacitor.

Why it shouldn't

  1. The capacitor and inductor combination can atmost restrict the rate of current i.e it may take very long time for current to reach infinity but surely it eventually will.

What will happen?

Improve this question
$\endgroup$
6
  • $\begingroup$ On its own, a circuit with nothing but a capacitor will not reach infinite current as time approaches inifnity: it will reach zero. Can you explain why adding an inductor might change that? $\endgroup$
    – Cort Ammon
    Commented Apr 27 at 3:03
  • $\begingroup$ Since in R-C circuit with dc voltage (V) applied across it, the current is given as$ i_t= i_o e^{-t/RC}$. So in a pure capacitive circuit, the intial(at t=0) current $i_t = i_o$ where $i_o$ is V/R , since R=0,$ i_o $= infinite, so capacitor go boom. The role of inductor is to stop capacitor from going boom in initial stages. This is what i think. $\endgroup$
    – SHINU_MADE
    Commented Apr 27 at 3:16
  • $\begingroup$ If the inductor matters in the initial stages, would the circuit not act more like a capacitor in the long run? $\endgroup$
    – Cort Ammon
    Commented Apr 27 at 3:18
  • $\begingroup$ Yes indeed...... I understand your point i need to think more now. $\endgroup$
    – SHINU_MADE
    Commented Apr 27 at 3:20
  • $\begingroup$ If it woudl help, I can write an answer using the differential equations, but I do feel the intuition about the circuit is more valuable. $\endgroup$
    – Cort Ammon
    Commented Apr 27 at 3:28

2 Answers 2

Reset to default
2
$\begingroup$

Consider the voltage across the capacitor. It is necessarily less than or equal to your DC input voltage because that input voltage is divided between the capacitor and inductor. Okay, maybe that step is a bit of a leap. It may not be true if the current ever flows in the opposite direction of the input voltage. I think it should be intuitive that this is the case. A formal proof would involve taking advantage of the fact that all elements in the circuit are linear, and in a linear circuit one cannot find variables that vary at any frequency that isn't present. DC only has one frequency ($\omega=0$). At AC frequencies, a LC circuit is called a "tank" circuit, and does indeed show gains. But at DC, it is hopefully obvious that the current never switches direction.

Because of this, we can be confident that the current at any point in time is always less than the current that would have gone through a capacitor on its own. A capacitor on its own would receive all of the voltage of the input. This capacitor, with an inductor in series, always receives less voltage.

In the steady-state case, as t approaches infinity, the current in a capacitor with a constant voltage approaches 0 (not infinity). The current in this circuit must be less than or equal to the current in the capacitor-only circuit, which means it also must approach 0.

This only works in DC. AC behavior of LC circuits is different.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Could you look at the question i asked at physics.stackexchange.com/q/812050/385600 if this gets clear i'll understand everything eventually. $\endgroup$
    – SHINU_MADE
    Commented Apr 27 at 8:01
  • $\begingroup$ I put an answer there. Also, that answer pointed out a failure in my analysis here. While I am correct in that the voltage must be below that of a capacitor on its own, in my head I replaced the inductor with a variable resistor, and considered its behavior as that resistance went to 0. That's the risks associated with being too familiar with these circuits -- I took a leap that was tricky. $\endgroup$
    – Cort Ammon
    Commented Apr 27 at 21:53
2
$\begingroup$

Well, instead of just hypothesising what will happen we could easily understand the situation mathematically.

Let at any time $t$ there be $q$ charge on the capacitor(Capacitance=$C$) and current $i$ be flowing through the circuit. Let inductance of inductor be $L$.

LC are in series so sum of potential drop across the the combination should be same as emf($E$) of the cell. [By KVL]

$$\frac qC+L\frac{di}{dt}=E\implies\frac{d^2q}{dt^2}=\frac EL-\frac q{LC}$$

This is a double differential equation with general solution as:

$$q-CE=A\sin\left(\frac{1}{\sqrt{LC}}t+\phi\right)$$

From this it is quite understood that the charge in the circuit will oscillate with an angular frequency of $\frac 1{\sqrt{LC}}$ and so will the current.

By using the using the initial conditions $q(0)=0,i(0)=0$

$$q(t) = C E \left[ 1 - \cos \left( \frac 1{\sqrt{LC}} t \right)\right]$$

Improve this answer
$\endgroup$
4
  • $\begingroup$ Your solution for $q (t)$ satisfies the differential equation $$\frac{d^2 q}{d t^2} + \frac{q}{LC} = 0$$ $\endgroup$
    – jim
    Commented Apr 27 at 19:17
  • 1
    $\begingroup$ @jim I had added a $CE$ in the solution but someone editted that out I will correct it $\endgroup$
    – Yash Shrivastava
    Commented Apr 28 at 6:52
  • $\begingroup$ it would be useful to point out that using the initial conditions $ q(0) = 0, i(0)= 0$ gives $$ q(t) = C E ( 1 - \cos ( \omega_0 t )) $$ $\endgroup$
    – jim
    Commented May 8 at 16:46
  • $\begingroup$ @jim it was not really necessary to understand the behaviour of charge on the capacitor so just saved myself some work but since you have pointed that out I did add it too $\endgroup$
    – Yash Shrivastava
    Commented May 8 at 16:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.