Is this true that the present in the circuit consisting of an inductor, a capacitor and also a resistor(all in series) is zero too at the moment simply after cshedding the switch? I tried creating a differential equation and also resolving it yet it is composed of attributes of Q and also it derivative and its double derivative, which I cannot solve.

What and how have the right to I conclude about the existing in this circuit just after switch is closed. (Consider that the switch was opened up for a lengthy time prior to closing)

You are watching: What is the battery current immediately after the switch closes?

homework-and-exercises electric-circuits electrical-resistance capacitance inductance

Share

Cite

Improve this question

Follow

edited May 22 "16 at 18:11

Qmechanic♦

154k2828 gold badges372372 silver badges18201820 bronze badges

asked May 18 "16 at 14:34

user1825567user1825567

29311 gold badge33 silver badges88 bronze badges

$endgroup$

4

Add a comment |

## 4 Answers 4

Active Oldest Votes

2

See more: What Does Money Doesn T Grow On Trees Meaning & Examples, Money Doesn'T Grow On Trees Origin

$egingroup$

So, we have actually series LCR circuit. $V$ is a consistent voltage resource. $L$, $C$, and also $R$ represents the inductance, capacitance and also resistance in the circuit respectively. A existing $I$ flows via the circuit.

Now, the current with each component is the exact same. So, the potential difference between each component added up together gives the emf $V$. Hence the differential equation becomes:

$$LfracdIdt+fracQC+IR=V$$

wright here $Q$ is the charge on the capacitor and also is concerned the present by $I=displaystylefracdQdt$. This suggests we have only one unwell-known in the equation if we rearea all $I$ in regards to $Q$:

$$Lfracd^2Qdt^2+RfracdQdt+fracQC=V$$

which is a 2nd order differential equation. Differentiating aget w.r.t $t$ and rewriting in terms of $I$, we acquire

$$Lfracd^2Idt^2+RfracdIdt+fracIC=fracdVdt$$

Since we have actually a constant dc voltage source, $displaystylefracdVdt=0$. Hence

$$Lfracd^2Idt^2+RfracdIdt+fracIC=0$$

Dividing throughout by $L$, we have

$$fracd^2Idt^2+fracRLfracdIdt+fracILC=0$$ or

$$fracd^2Idt^2+2alphafracdIdt+omega_0^2 I=0$$

where $displaystylealpha=fracR2L$ and $displaystyleomega_0=frac1sqrtLC$

This is an ODE via constant coefficients. The characteristic equation of this differential equation is offered by:

$$s^2+2alpha s+omega_0^2=0$$

The roots of this equation in $s$ are:

$s_1=-alpha +sqrtalpha^2-omega^2$ and also $s_2=-alpha -sqrtalpha^2-omega^2$

The general solution is given by:

$$I(t)=A_1e^s_1t+A_2e^s_2t$$.

Now, at $t=0$, let the existing be zero. On switching on the present, then the existing rises to a maximum worth tremendously. Otherwise, it takes a finite time for the current to have actually a constant worth in the circuit . The current will not instantly rises to a maximum worth. This is as a result of the visibility of inductance and also capacitance in the circuit. This is why we say, unprefer in the resistive circuit, in an LCR circuit, the present will be zero, simply prompt after the switch is closed.