# RL Circuit

A resistor connected in series with an inductor is called RL series circuit. The R stands for resistance and L means inductance, so the term RL is used in short. The RL circuit is also called LR circuit; all that's required is an inductor and a resistor connected in series so you can say either RL or LR series circuit.

Here we deal with the relationship of rate of change of current to the inductance and how the current changes with time in an RL circuit. We know that an inductor maintains the steady current by opposing any changes in the current.

You know from self inductance (equation of voltage) that the rate of change of current is inversely proportional to the inductance and directly proportional to the voltage across inductor terminals. The greater the rate of change of current, greater the voltage across the inductor terminals, and if the less the rate of change of current, less the voltage across the inductor terminals. If you do not know what's been going on here review self inductance article.

And the greater the inductance, less the rate of change of current which means if the inductance is greater, the current changes more slowly. If the current is suddenly interrupted, a large voltage appears across the inductor terminals which depends on the inductance and the rate of change of current. If an inductor with large inductance is used and the current is interrupted, that leads to very high voltage across the inductor terminals and that's dangerous!

That's the reason why the arc appears across the switch contacts when the steady current in a circuit with inductor is interrupted by opening the switch. The rate of change of current will be very high and the back emf across the inductor terminals will also be very high and that leads to arc across the switch contacts.

## Current Growth in RL Series Circuit

An electric circuit is shown in Figure 1 where an inductor of inductance $L$ and a resistor of resistance $R$ are connected in series. We saw in self inductance that, even if there is a nonelectrostatic electric field within the inductor, the potential difference across the inductor is real and we can apply Kirchhoff's laws.

If $i$ is the current at an instant in the circuit, after applying Kirchhoff's laws, we get (we have moved clockwise in the upper loop):

$\mathcal{E} - L\frac{di}{dt} - iR = 0 \tag{1}\label{1}$

You know the potential difference across an inductor is $L\,di/dt$. Now we solve this equation to determine the expression of current as a function of time. We first rearrange Equation \eqref{1} into the form (separating variables $i$ on the left and $t$ on the right):

$\frac{di}{i - \mathcal{E}/R} = -\frac{R}{L}dt$

To solve this equation, let $x = i-\mathcal{E}/R$, and therefore, $dx = di$, so the left side will be $dx/x$. Integrate from $x_i$ to $x_f$ on the left hand side where $x_i$ is the value of $x$ at $t = 0$. And integrate the right hand side from $0$ to $t$. Note that $x_i = -\mathcal{E}/R$ at $t = 0$ because $i = 0$. The final solution of the current as a function of time is

$i = \frac{\mathcal{E}}{R}(1-e^{-Rt/L}) \tag{2}\label{2}$

The value $L/R$ is a constant called time constant of the RL circuit denoted by $\tau$. Note that this value has SI unit of second (s), that is $\Omega \cdot s / \Omega = s$. And it is the time interval required for the current required to reach the value of $1 - e^{-1} = 0.632 = 63.2\%$ of its final value. This value is useful to know how quickly the current reaches its final value for a given RL circuit configuration.

You can understand from the ratio $L/R$, that the greater the inductance, more slowly the current rises. In $2\tau$, the current reaches $1-e^{-2}$ of its final value and in $3\tau$, it reaches $1-e^{-3}$ of its final value an so on. In terms of time constant, $\tau$, the above equation can be rewritten as,

$i = \frac{\mathcal{E}}{R}(1-e^{-t/\tau}) \tag{3}\label{3}$

Now we have the equation of current as a function of time, which tells us that the current increases exponentially from zero at $t=0$ to its final equilibrium value at $t = \infty$. When the time is infinity, the current reaches its maximum equilibrium value $\mathcal{E}/R$, in other words the current approaches the maximum equilibrium value asymptotically as shown in Figure 2 below.

This situation is illustrated by Figure 1 above in which the switch $S_1$ is closed and $S_2$ is open. In this case the charge flows through the inductor and resistor and it's rate of flow (current) gradually rises exponentially. Let's examine the situation of rate of change of current ($di/dt$) as a function of time. To do that let's first differentiate Equation \eqref{3} with respect to time and we get,

$\frac{di}{dt} = \frac{\mathcal{E}}{L}e^{-t/\tau} \tag{4} \label{4}$

Now the above equation tells us that how the rate of change of current, that is $di/dt$ changes with time. At first when $t=0$, $di/dt = \mathcal{E}/R$. It means, at first (when $t = 0$) it is maximum. And when $t = \infty$, $di/dt = 0$. It means, as the time approaches infinity, the rate of change of current approaches zero. Note that the rate of change of current approaches zero asymptotically as illustrated in the Figure 3 below.

If the inductance is zero, the Equation \eqref{3} suggests that, $i=\mathcal{E}/R$, in this case there is no time dependence of current, and the rate of change of current is zero. It is the same as there is no inductor and the current does not change exponentially to the final value but it is instantly maximum. You know now that a simple coil made from a straight piece of wire can behave quiet differently in an electric circuit.

If you multiply Equation \eqref{1} by $i$ and rearrange, you can write,

$\mathcal{E}i = Li\frac{di}{dt} - i^2R$

The rate at which the source delivers energy (power input) to the circuit is $\mathcal{E}i$, and the rate at which the energy is delivered (power input) to resister is $i^2R$ and the rate at which the energy is delivered (power input) to the inductor is $L\,i\,di/dt$. The energy is only delivered to the inductor if the current is increasing. Note that we don't say the energy is dissipated as it can mean the energy is disappeared violating the law of conservation of energy.

## Current Decay in RL Series circuit

This situation is illustrated by Figure 4 where the switch $S_1$ is open, it means there is no battery and now the inductor acts as the source of emf. What happens is that, in the absence of battery, there is no current due to the battery. But as the switch $S_1$ is opened, the current maintained by the battery is zero, that leads to the collapse of magnetic field of the inductor. As the magnetic field starts to decrease, the flux is changed in the inductor that leads to induced emf in the inductor.

The Lenz's law tells us that the induced emf is in a direction to oppose the change in flux. That means, the inductor has polarity opposite to that of when the current was growing, that is the current has the same direction (in the sense of flow of positive charge) and the inductor allows to maintain steady current. In this case there is no battery and the inductor is acting as the source of emf. Where does this energy come from? The inductor stores energy in the magnetic field, and as the magnetic field starts to collapse, the energy stored in the magnetic field is released.

Do not confuse with the energy stored in the magnetic field. We saw in case of capacitors, that the energy was stored in the electric field between the plates of the capacitor. The energy that comes from magnetic field has something to do with the flux change. As the current that maintains the magnetic field no longer exists (due to the presence of battery), that magnetic field starts to decrease and there is the change in flux and hence the energy. Since the collapsing magnetic field always causes the change in flux, the energy is indeed stored in the magnetic field.

Similarly we apply Kirchhoff's law in the lower loop in Figure 4 (or put $\mathcal{E} = 0$ in Equation \eqref{1}), and we can write,

$iR + L\frac{di}{dt} = 0 \tag{5} \label{5}$

The solution to the above differential equation (note that electromagnetism is mathematically demanding, so you must know how to solve the above equation), is

$i = \frac{\mathcal{E}}{R}e^{-t/\tau} = I_0e^{-t/\tau} \tag{6} \label{6}$

The above Equation \eqref{5} tells us that the current approaches zero exponentially. Initially the current is maximum at $t= 0$, that is $I_0$, and when $t$ approaches infinity, the current approaches zero asymptotically as shown in the graph of current as a function of time below. We explained similar situations in RC circuit (resistor and capacitor in series).

Again time constant $\tau = L/R$ in Equation \eqref{6} is the measure of how quickly the current decays. In time $\tau$, the current decreases to $e^{-1} = 37\%$ of its initial value $I_0$. In time $2\tau$, the current decreases to $e^{-2}$ of its initial value and so on.

If you multiply the Equation \eqref{5} by $i$ and rearrange,

$L\frac{di}{dt} = -i^2R \tag{7} \label{7}$

In this case the inductor acts as the source of energy and the rate at which the energy is being delivered to the resistor is negative as the current is decreasing. The negative sign in above equation means that the energy is decreasing in the inductor and in the same rate the energy is being transferred to the resistor.