Current in inductor just after switch is closed

I have read that the current in an inductor resistor circuit just after closing the switch is zero. This is derived by finding the differential equation in terms of current by using KVL.

Is this true that the current in the circuit consisting of an inductor, a capacitor and a resistor(all in series) is zero too at the time just after closing the switch? I tried writing a differential equation and solving it but it consists of functions of Q and it derivative and its double derivative, which I cannot solve.

What and how can I conclude about the current in this circuit just after switch is closed. (Consider that the switch was opened for a long time before closing)

• Could you add the equation you derived to the question? It might be helpful. – dahemar May 18 '16 at 15:02
• Can you think about it in terms of the phsyics in Faraday's Law instead of brute force solving the de? – Declan May 18 '16 at 15:38
• @David Herrero Marti. I am using mobile and I find it difficult to write such equation by formatting, so I have written it in a paper and posted the pic if it. I hope you all don't mind. – user1825567 May 19 '16 at 12:11
• @user1825567, if the capacitor is initially discharged, the current immediately after the switch is closed will be zero. Then the current will increase and the capacitor will charge. As the capacitor becomes fully charged, the current will drop back to zero. However, if the capacitor starts off fully charged, no current will flow at any time after the switch is closed. Therefore, the answer will depend on your starting conditions. – David White May 22 '16 at 18:21

So, we have series LCR circuit. $V$ is a constant voltage source. $L$, $C$, and $R$ represents the inductance, capacitance and resistance in the circuit respectively. A current $I$ flows through the circuit.

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

$$L\frac{dI}{dt}+\frac{Q}{C}+IR=V$$

where $Q$ is the charge on the capacitor and is related to the current by $I=\displaystyle{\frac{dQ}{dt}}$. This means we have only one unknown in the equation if we replace all $I$ in terms of $Q$:

$$L\frac{d^2Q}{dt^2}+R\frac{dQ}{dt}+\frac{Q}{C}=V$$

which is a second order differential equation. Differentiating again w.r.t $t$ and rewriting in terms of $I$, we get

$$L\frac{d^2I}{dt^2}+R\frac{dI}{dt}+\frac{I}{C}=\frac{dV}{dt}$$

Since we have a constant dc voltage source, $\displaystyle{\frac{dV}{dt}=0}$. Hence

$$L\frac{d^2I}{dt^2}+R\frac{dI}{dt}+\frac{I}{C}=0$$

Dividing throughout by $L$, we have

$$\frac{d^2I}{dt^2}+\frac{R}{L}\frac{dI}{dt}+\frac{I}{LC}=0$$ or

$$\frac{d^2I}{dt^2}+2\alpha\frac{dI}{dt}+\omega_0^2 I=0$$

where $\displaystyle{\alpha=\frac{R}{2L}}$ and $\displaystyle{\omega_0=\frac{1}{\sqrt{LC}}}$

This is an ODE with constant coefficients. The characteristic equation of this differential equation is given by:

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

The roots of this equation in $s$ are:

$s_1=-\alpha +\sqrt{\alpha^2-\omega^2}$ and $s_2=-\alpha -\sqrt{\alpha^2-\omega^2}$

The general solution is given by:

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

Now, at $t=0$, let the current be zero. On switching on the current, then the current rises to a maximum value exponentially. Otherwise, it takes a finite time for the current to have a constant value in the circuit . The current will not instantly rises to a maximum value. This is due to the presence of inductance and capacitance in the circuit. This is why we say, unlike in the resistive circuit, in an LCR circuit, the current will be zero, just immediate after the switch is closed.

• And can we say that the charge on the capacitor remains unchanged just after the switch is closed? – user1825567 May 19 '16 at 15:20
• No. The charge should also rise exponentially – UKH May 20 '16 at 1:16
• the time taken for charging can be calculated from the time constant – UKH May 20 '16 at 1:17

From Kirchhoff's second law, the sum of all the voltages around a loop is equal to zero. That is, the sum of the voltages across the three elements of your circuit, R, L and C, must be equal to the time varying voltage from the source:

$$V_R+V_L+V_C = V(t)$$

As $V_R=RI$, $V_L=L\frac{dI}{dt}$ and $V_C=\frac{Q}{C}$, we get your equation, which is correct:

$$LI'(t) + RI(t) + \frac{1}{C}Q(t)=V(t)$$

By substituting $I=\frac{dQ}{dt}$ and differentiating, we arrive to the second order ODE:

$$LI′′(t) + RI′(t) + \frac{1}{C}I(t)= V'(t)$$

And assuming that your voltage source does not varies with time, $V'(t)=0$.

You can find a step by step solution to the ODE in this link (or in any standard textbook). As you see, there are three different possibilities for the general solution, depending on the variables of your specific circuit, so start substituting your $R$, $L$, $C$ values on the $\alpha$, $\omega_0$, etc. expressions (those are coefficients which are used to simplify the calculations), and find out which one you need. Then, you can try substituting different initial conditions (i. e., the values of the currents in the circuit at the onset) on your general solution, and exploring how will the system behave for each of them. This way, you should be able to find a response to your question.

Setting up the differential equation

$$L \frac {dI}{dt} + RI + \frac {Q}{C} = V$$

will not necessarily answer your question, "What and how can I conclude about the current in this circuit just after switch is closed."

If you look at the methods of solving the differential equation somewhere on the way to the solution initial conditions are needed, one of which is often the condition you have asked about - the initial current is zero.

There are a number of simplistic ways of considering what might happen:

1. Before the switch is closed the average velocity/momentum of the mobile electrons in the circuit is zero.
The switch is closed and almost instantaneously there is a net electric field set up in the circuit.
That electric field exerts a finite force on the mobile electrons.
The finite force on the mobile electrons produces a finite acceleration of the mobile electrons.
An electric current is the net movement of mobile electrons in a circuit.
The mobile electrons cannot start moving instantaneously so this is the reason for the initial current in the circuit being zero.

2. The energy stored in an inductor is $\frac 1 2 L I^2$.
An instantaneous finite value of current would require infinite power being delivered to the inductor.

So the current when the switch is closed is zero and the rate of change of current at that time assuming that the capacitor is initially uncharged is $\frac V L$

An interesting example which follows on from this is of the charging of a capacitor $C$ by a battery of voltage $V$ through a series resistor $R$.
The differential equation for this arrangement is

$$RI + \frac {Q}{C} = V$$

Assuming that the capacitor has no charge on it when the switch is closed the initial condition used is that the current is $\frac VR$ - an instantaneous change from zero current.
When the experiment is done it appears that this is exactly what happens.

In truth the current increases from zero to approximately $\frac VR$ in a time which is very much shorter than the time constant of the circuit $CR$.
The reason is that there is inductance in the circuit as it is a loop of wire but of a very small value but significant value just after the switch is closed.

The component that ensures the current is zero just after the switch is closed is the inductor. Inductors do not like changes in current, since a change in current means the magnetic field linking the inductor is changing and this generates a back emf that opposes the change.

If you replace the inductor with a piece of wire you would have an RC circuit and the current flowing after the switch is closed would be $E/R$. Capacitor's inhibit voltage changes, so that just after the switch is closed the voltage across the capacitor is zero.