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I've recently learned a little about R-L series circuit.

Consider a simple L-R circuit, as shown in the given image- -

(https://i.stack.imgur.com/ovfbQ.jpg)

An inductor L is connected in series with a resistor R which itself is connected with a battery. At time t=0, the switch is closed (before that time there was no current in the circuit and the switch was opened).

Now, of course I've come across the differential equation used to find the current in the circuit as a function of time and solved it, but no matter how hard I try, i'm not able to 'THINK' about the circuit in an intuitive way (or qualitative way). I'll tell what problems I'm facing -

Firstly, let's imagine that there was no inductor in the circuit (rest of circuit was exactly same), as soon as the switch is turned ON, the current would have risen from zero to a finite value (V/R, where V is the EMF of battery and R is the resistance) instantaneously.

Now there's an inductor present, then of course at time t=0, there will be a sudden surge for the current to rise, so di/dt will be positive, the inductor will do its usual job to generate an opposing EMF. So initially, the current in the circuit will be HAVE TO BE LESS than V/R, because the EMF of the battery is being 'opposed' by that of inductor.

I fail to understand why the current in the circuit, just AFTER the switch is turned ON is zero, I mean the inductor will generate an opposing EMF but WHY this opposing EMF is equal to V in magnitude, it could be less than V, giving me a current LESS than V/R, but GREATER than zero.

My second problem is to get a 'feel' of this circuit, by 'feel', I mean to understand the circuit in a 'qualitative' manner.

Let's assume(of course I need a reason for this.)that initially the current in circuit is zero, now I have to predict what will happen to current at the next instant. For that I need to know what will happen to the EMF of the inductor at next instant (whether it will increase or decrease, remember I'm just analyzing the circuit in a qualitative way), but the EMF of the inductor is controlled by the time derivative of current, for that I need to know..... what will happen to the current at the next instant.

See? I'm struck in a circular path, again returning to my same question. This confuses me. (I hope you all can understand my problem.)

Can someone explain the behaviour of current as a function of time in a qualitative manner?

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    $\begingroup$ Can Google the link between LCR circuits and their mechanical counterpart. $\endgroup$ – sbp Sep 26 '18 at 20:04
  • $\begingroup$ To support what sbp was saying a bit; the wikipedia page on mechanical-electrical analogies is a pretty good introduction to the topic. As someone more mechanically inclined than electrically inclined, noticing all the connections here was really helpful in understanding more complicated circuits, and just showing very interesting analogies between seemingly unrelated systems. $\endgroup$ – JMac Sep 26 '18 at 20:31
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I fail to understand why the current in the circuit, just AFTER the switch is turned ON is zero

Since the voltage across an (ideal) inductor is proportional to the time derivative of the current through

$$v_L = L \frac{di_L}{dt}$$

the inductor current must be continuous for finite $v_L$.

Since the inductor current is zero just before the switch is thrown, the inductor current must be zero just after the switch is thrown, i.e., the inductor current must be continuous across the time the switch is thrown.

but WHY this opposing EMF is equal to V in magnitude, it could be less than V, giving me a current LESS than V/R, but GREATER than zero.

If it were not the case that $v_L = V$ just after the switch is thrown, there would be a non-zero voltage across the resistor which implies, by Ohm's law, a non-zero current through the resistor.

But the resistor is in series with the inductor and so they must have identical current through.

Since we've already established that the inductor current is zero just after the switch is thrown, the resistor current is also zero then which requires that the voltage across the resistor is zero just after the switch is thrown.

The only inductor voltage compatible with zero volts across the resistor is $v_L = V$.

In summary (assuming the switch is thrown at t = 0):

(1) Continuity of inductor current requires $i_L(0+) = 0$

(2) Series connection implies $i_R(0+) = 0$

(3) Ohm's law requires $v_R(0+) = 0$

(4) KVL implies $v_L(0+) = V$


See? I'm struck in a circular path, again returning to my same question. Can someone explain the behaviour of current as a function of time in a qualitative manner?

It's not circular if you try it this way:

$i_L(t + dt) = i_L(t) + \frac{di_L(t)}{dt}dt = i_L(t) + \frac{v_L(t)}{L}dt$

We've already established that $i_L(0+) = 0,\,v_L(0+) = V$ thus (dropping the $+$ for simplicity)

$$i_L(0 + dt) = i_L(0) + \frac{V}{L}dt = \frac{V}{L}dt$$

Now, by KVL, the voltage across the inductor must satisfy

$$v_L(t) = V - v_R(t) = V - Ri_L(t)$$

Thus,

$$v_L(0 + dt) = V - Ri_L(0 + dt) = V\left(1 - \frac{R}{L}dt\right)$$

Qualitatively, since there is an initial positive voltage across the inductor, the inductor current must initially be increasing at a rate proportional to that voltage.

But, a moment later, the inductor current has increased which, due to the series resistor, decreases the voltage across the inductor which decreases the rate at which the inductor current changes.

A moment later, the inductor current has increased but not as much as during the previous moment. Once again, the increased current decreases the voltage across the inductor which again decreases the rate at which the inductor current changes.

It may or may not occur to you at this point that, from all of this, you can induce that

$$v_L(0 + \Delta t) = V\left[1 - \frac{R}{L}\Delta t + \frac{1}{2}\left(\frac{R}{L}\Delta t\right)^2 - \cdots \right] = Ve^{-\frac{R}{L}\Delta t}$$

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    $\begingroup$ Exactly answers my question.. I now understand that although in the initial moment di/dt is positive but the rate of increase of di/dt is negative, in other words current increases faster intially but as time progresses, current increases less and less faster due to the decrement of the potential across inductor (as a resistor is attached to it in series). THANKS! For explaining it in such a nice manner $\endgroup$ – Shivansh J Sep 27 '18 at 14:15
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There are good analogies for each circuit component using water and pipes.

A resistor is like a thin piece of pipe where higher pressure difference corresponds linearly to the flow rate through.

A capacitor is like a tank with a rubber membrane obstructing the flow of water. No net water can pass through, but water can be stored on either side by creating a pressure difference and making the membrane stretch. Any sudden change in current will not immediately change the pressure, because the membrane will stretch to compensate.

An inductor is like a heavy water mill. When there is a pressure difference, it's heavy so it takes a while to get moving. It basically has inertia. But once that mill is turning it doesn't want to stop, so even if you turn off the source it will "overshoot" and keep pumping water in the direction it was previously going. Any sudden change in pressure will not immediately change the current, because the mill is heavy and has inertia.

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The voltage applied to the circuit is divided between the incuctor and the resistor.

The voltage on the resistor, per Ohm's law, is equal $IR$, so, since the initial current is zero, the initial voltage drop on the resistor is zero and, therefore all battery voltage is applied to the incductor.

As the current through the inductor gradually increases, the voltage drop on the resistor proportionally increases, so the voltage applied to the inductor decreases accordingly.

Why is the initial current is zero and why does it take time for the current to grow?

It helps to observe that a current has a magnetic field associated with it and the magnetic field has energy stored in it. Any time we deal with energy, we know that some work has to be done to transfer it and some time is needed to perform that work.

In application to electric circuits, the work is done by a battery, speeding up electrons in a wire. The initial speed of electrons is zero, so the initial current also should be zero, even if there is no inductor in the circuit, since any wire will have some inductance and a current flowing through any wire will have some magnetic field associated with it, which would require some work and some time to build up.

Due to a physical configuration of an inductor (basically, a coil), the concentration and energy of the magnetic field in it is much greater than that of a straight wire of the same length, carrying the same current.

Because of that, given the same applied voltage, the electrons in an inductor would not accelerate as fast as electrons in a regular wire and it would take more work and more time get the electrons to the same speed or get the current to the same magnitude.

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