# Influence of R and L in RL circuit

1. Sep 15, 2015

### cgiustini

Hi,

I have a qualitative question about RL circuits. I'm trying to summarize how R and L each affect the transient response time L/R from the perspective of energy transfers. I'm pretty confident in my description of the influence of L. I'm not very confident in my description of R in the case of a charging RL circuit, when an EMF is connected to an inductor and resistor in series. I would appreciate if someone could review my thinking.

Description of L's influence:
Inductance L represents magnetic flux through the inductor per unit of current flowing the inductor. Higher L means that more magnetic flux is stored (therefore more magnetic energy) is stored in the inductor per unit of current. Qualitatively, it makes sense for higher inductance to increase the current rise time in a charging RL circuit since the inductor has to drain more energy from the source EMF to achieve a given current. Similarly, when a charged inductor is connected to a resistor, a higher inductance means that more energy is stored per unit of current: therefore, higher inductance means that it will take more time for that energy to dissipate into the resistor.

Description of R's influence:
In a charging RL circuit, the resistance R limits the maximum current that can be set in the inductor. As a result, higher R reduces the total energy that needs to be transferred to the inductor to achieve steady state current. In this case, is this a correct explanation for why the transient response time is inversely proportional to R?
In a discharging RL circuit, increasing R increases the resistor power, ie the rate at which the resistor is dissipating energy from the inductor. Therefore higher R results in a quicker transfer of magnetic energy into thermal energy.

Do these descriptions make sense?

Thanks,
Carlo

2. Sep 15, 2015

### Staff: Mentor

You need to expand your thinking to include rates of change. V=L di/dt; voltage aross the inductor is inductance times rate of change of current. That is the important effect of L in a RL circuit, and it applies to both DC and AC circuits.

So I challenge you to begin thinking about rates of change. That should be your first and most basic step in understanding, not increase/decrease.

3. Sep 15, 2015

### vanhees71

The math is more telling than pure words. So let's write down the equation for an LR-circuit.
$$L \frac{\mathrm{d} i}{\mathrm{d} t}+R i=U(t),$$
where $U(t)$ is the external EMF.

To make things simpler, we can also consider the homogeneous equation with $U=0$. You can think of it to switch off a constant EMF at time $t$, thereby closing the circuit. Then you have to solve the equation for $U=0$ with a given initial current $i(0)=i_0$. The solution is simply
$$i(t)=i_0 \exp \left (-\frac{R}{L} t \right)=i_0 \exp \left (-\frac{t}{\tau} \right),$$
where $\tau=L/R$ is the relaxation time.

The interpretation is easy, when you remember where the terms of the equation come from. $L \dot{i}$ comes from Faraday's Law. The change in current implies a change of the magnetic field within the, which induces an electric field which counteracts the cause of the change, i.e., it tries to keep the current up. Thus the larger $L$, the longer it takes to switch off the current.

Now there's also the resistance, which dissipates of energy by converting it to heat within the wire. The power (dissipating energy per time) is $R i^2$. Thus the larger $R$ the more heat is dissipated per unit time, which makes the relaxation time short, that's why $R$ is in the denominator of the relaxation time.

We can also show that energy is conserved. At $t=0$ there is a certain amount of energy stored in the magnetic field within the coil. During the decay of the current the total amount of energy dissipated in the resistance is
$$E=\int_0^{\infty} \mathrm{d} t R i^2(t)=R i_0^2 \int_0^{\infty} \mathrm{d} t \exp \left (-\frac{2 t}{\tau} \right )=\frac{R \tau}{2} i_0^2=\frac{L}{2} i_0^2.$$
Thus, the energy initially stored in the magnetic field within the coil is $E=L i_0^2/2$.

4. Sep 15, 2015

### cgiustini

Thanks for the replies. I appreciate it. From Faraday's Law, I understand mathematically that increasing L reduces the rate of change of current through an inductor. What I'm trying to understand is the qualitative behavior behind this. vanhees71, your response states that a changing current changes the magnetic field in the inductor that induces an electric field that counteracts the current change (which is a verbal description of Faraday's Law). Faraday's law shows that the induced electric field is also proportional to inductance, meaning that larger L causes the counteracting electric field to be higher. In the case of slowing current (inductor discharge), the electric field pushes the charge through the inductor in the direction of the original current. In the case of increase current, the electric field pushes back on the charge trying to flow through. The question becomes why does higher L cause a higher electric field? In the case of a simple solenoid, is the answer simply that larger L --> larger area --> larger changes in magnetic flux when the current running through the solenoid changes --> larger electric field around the solenoid wire, which is simply a restatement of Faraday's Law.

My reasoning in the paragraph above seems circular because I simply use Faraday's law to explain Faraday's law. I struggle with this kind of situation a lot, where I understand the math and can do all the problems, but I seem to think that I don't qualitatively understand the physics. At the end of the day, I understand that higher L increases the current rise/fall time in a circuit because Faraday's law dictates that the change in magnetic flux due to current variation in the inductor is proportional to the induced counteracting electric field. If L is a measure of magnetic flux per unit current, higher L (induced by larger inductor area) will therefore produce a stronger counteracting electric field to more effectively slow the current change rate. Beyond this, is there anything else to attempt to understand aside from why Faraday's Law is what it is? Does it even make sense to understand the "why" behind Faraday's Law as opposed to just taking it for granted? I would appreciate some insights.

Carlo

5. Sep 16, 2015

### vanhees71

I think the reasoning is valid. To illustrate it a bit better, think of a solenoidal coil with its axis in $z$-direction of a Cartesian coordinate system and a stationary current flowing through it. Then you have $\vec{B} \simeq B \vec{e}_z$, $B \propto i$. If you now switch off the battery but still keep the circuit closed in the first moment the current will stay as it is but through the resistor it decreases a bit. So you have $\partial_t \vec{B}=\dot{B} \vec{e}_z$ with $\dot{B}<0$. The electric field is given by Faraday's Law
$$\vec{\nabla} \times \vec{E}=-\frac{1}{c} \partial_t \vec{B}=-\frac{1}{c} \dot{B} \vec{e}_z.$$
From the curl in cylinder coordinates you have
$$\vec{\nabla} \times \vec{E}=\vec{e}_z \left (\frac{1}{\rho} \partial_{\rho} (\rho E_{\varphi})-\partial_{\varphi} E_{\rho} \right )+\cdots$$
Due to rotational symmetry the second term should vanish. Also $B$ within the coil is quite homogeneous and thus it doesn't depend on $\rho$ and thus also $E_{\varphi}$ shouldn't depend on $\rho$ much. So we have $\vec{\nabla} \times \vec{E} \simeq \vec{e}_z E_{\varphi}/\rho$. Since now $\dot{B}<0$ we have $E_{\varphi}>0$, which means that the induced electric field tends to keep the current going.

This is, of course, in accordance with Lenz's Rule, which says that the law of induction acts such that the magnetic field tries to prevent its change with time.