Understanding a very simply RL circuit

[zenterix]

Homework Statement

My question is about the relatively simple RL circuit shown below.

Relevant Equations

$\mathcal{E}=\oint\vec{E}\cdot d\vec{l}=-L\dot{I}=-\Phi'$

As far as I can tell we have

$$\mathcal{E}_L=\oint \vec{E}\cdot d\vec{l}=IR-\mathcal{E}=-L\dot{I}=-\dot{\Phi}\tag{1}$$

This differential equation can be written

$$\dot{I}+\frac{R}{L}I=\frac{\mathcal{E}}{L}\tag{2}$$

which is easily solved

$$I(t)=\frac{\mathcal{E}}{R}\left ( 1-e^{-\frac{Rt}{L}}\right )\tag{3}$$

My question is about power and work.

The power generated by the back emf in the inductor is

$$P_L=\mathcal{E}_L I=-LI\dot{I}\tag{4}$$

My first question is about the negative sign here. I don't think it should be here. Is power defined as the absolute value of emf times current? Let's assume that this is the case.

If we start with zero current, to obtain a current $I(t)$ the total work that is

$$W=\int P_Ldt=\int_0^t (LI\dot{I})dt=\frac{LI(t)^2}{2}\tag{5}$$

My second question is about the power and work associated with the voltage source.

$$P=\mathcal{E} I=(IR-\mathcal{E}_L)I$$

$$W=\int I^2 Rdt - \int\mathcal{E}_LIdt$$

$$=R\frac{\mathcal{E}^2}{R^2}\int_0^t \left (1-e^{-\frac{Rt}{L}}\right )^2 dt +\frac{LI(t)^2}{2}\tag{6}$$

Does this calculation make sense?

My final questions are about how to interpret everything.

From (3) it seems that $\mathcal{E}$ only affects the final value of the current. The speed with which such a current is achieved is affected by the self-inductance of the inductor and the resistance in the circuit.

If the self-inductance $L$ is relatively low then there is less back emf.

If $R$ is relatively high then

 

[DaveE]

The short answer about polarities is that they ALL have to be defined in schematics and such. Then given those definitions you write equations that reflect how the real world works.

Yes, you can define things so that power is negative. It's a real thing, for example, in electrical distribution networks. At my house the power flow to the city* changes polarity based on how much sun is hitting the solar panels.

Power is defined as voltage times current, in this EE context, with appropriate polarity definitions, and (maybe) "-" signs in your equations if needed to make it work like the real world.

In practice most all EEs subscribe to the "passive sign convention" so that the voltage polarity of any component in the circuit is positive where the current enters and negative where the current leaves. Then, of course, current flow is backwards from the direction electrons go (thanks Ben Franklin!). But, this convection can't always be followed, with things like transistors or networks. So everyone still has to define them explicitly. Some times this is implicit in your equations, since we know the underlying physics. With your schematic, if you write $E = L \dot{I}$, I know $E$ must be larger on the left side because of your current direction definition, that's how inductors work.

*note the polarity definition here, the opposite of power flow from the city.

PS: Using the passive sign convention, when the power is positive, it is flowing into the device (a "load", as opposed to a "source"). For example making heat in a resistor, increasing the energy in a magnetic or electric field, charging a battery, etc. It's a bit more complex in EM waves, but there is still a polarity (the poynting vector, for example) which is based on defining the terms used in equations to make sense IRL.