# Trouble with Kirchoffs Law and EMF

So I'm having some trouble using Kirchhoff's second law . For example suppose I have a battery that supplies a constant emf $$\epsilon_0$$ and is connected to a resistance R and inductance L. I don't know in wich side of the equation to put my electromotive force generated by the inductance L , can anyone give me some intuition please ? Thanks.

Also when im applying Ohm's Law , $$V=IR$$ to discover the current induced by an eletromotive force do i apply it to $$-\epsilon$$ because this is going to be my potencial difference or do i apply it to $$\epsilon$$?

• Even for a circuit as simple as this, it will hep to draw a circuit and label all the components so we can visulaize it and discuss it using the same names for all the components. – The Photon Jun 10 at 16:15
• Ur right , ive edited. – Pedro Santos Jun 10 at 16:18
• Can you 1. Crop the image so that the schematic can be read clearly. 2. Annotate the schematic with what direction you chose to call positive current and positive voltage for each component? – The Photon Jun 10 at 16:20
• yeah sorry i have no idea how to do that – Pedro Santos Jun 10 at 16:21
• There is a "Snipping Tool" that lets you cut one section of your screen. Click on the Window icon on bottom left and start typing Snipping Tool. A shortcut should pop up. Click the shortcut to run the program. – Bill N Jun 10 at 18:34

The text above the circuit diagram has been cut off so I'm not sure what is being asked.

To apply Kirchhoff's voltage law, pick a direction for the loop current and then assign polarities for the inductor and resistor as + to - in the direction of current flow. Since the only current source is the battery, and conventional current is that of positive charge out of the battery, the obvious choice of the direction of current is clockwise. Per Kirchhoff's voltage law (KVL) the algebraic sum of the voltages around a loop equals zero. That means your loop equation going clockwise will be

$$+ε_{o}-V_{L}-IR=0$$

To continue your analysis you need to know the initial conditions of the circuit. This was not described in your picture, but from the graph of current versus time it is obvious that the current at time $$t=0$$ is zero. Set $$I=0$$ for time $$t=0$$ in your loop equation and that will tell you what the voltage across the inductor and resistor is immediately after they are connected to the battery, that is at $$t=0$$.

As time continues as shown in the graph, the current builds in the circuit but levels off until it becomes constant at $$t=$$ infinity. An ideal inductor has zero voltage across it (looks like a short circuit) when current is constant, because the voltage across an inductor is given by

$$V_{L}=L\frac{di(t)}{dt}$$

And if current is constant, $$\frac{di(t)}{dt}=0$$

If the voltage across the inductor is zero at $$t=∞$$, then from KVL all the battery voltage must be across the resistor.

I leave it to you as an assignment to determine the current in the circuit as a function of time.

Hope this helps.

Given that the solution starts with the statement

The total emf in this circuit is that provided by the battery plus that resulting from the self-inductance.

it would seem that you need to equate the sum of all the potential differences on one side of the equation, $$IR$$ in your example, to all the emfs on the other side of the Kirchhoff's voltage law equation $$\mathcal E_0$$ and $$- L \frac {dI}{dt}$$ giving $$IR = \mathcal E_0 - L\frac{dI}{dt}$$

This is exactly the saying that there is a potential difference of $$L\frac{dI}{dt}$$ across the inductor which then gives $$IR + L\frac{dI}{dt}= \mathcal E_0$$

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A lot of people use the passive sign convention eg the MITx course 6002.1x Circuits and Electronics 1: Basic Circuit Analysis which has an excellent textbook linked to it.

In the case of your circuit it is useful to first label the nodes $$A,\,B$$ and $$C$$ and a current $$I$$.

For the resistor and the inductor label the terminal where the current enters the circuit element $$+$$ and the terminal where the current leaves the circuit element $$-$$.
Label the voltages across the circuit elements $$V_{\rm R}$$ and $$V_{\rm L}$$.
The label $$V_{\rm R}$$ together with the $$+$$ and $$-$$ label is interpreted as follows:

• the potential of the terminal labelled $$+$$ is $$V_{\rm R}$$ is higher in potential than the potential of the terminal labelled $$-$$.

Now start at node $$A$$ and walk around the circuit in a clockwise direction as follows:

• Node $$A$$ to node $$B$$ is an increase in potential of $$\mathcal E_0$$
• Node $$B$$ to node $$C$$ is a decrease in potential of $$L \frac{dI}{dt}$$
• Node $$C$$ to node $$A$$ is a decrease in potential of $$RI$$

Applying Kirchhoff's voltage law gives

$$\mathcal E_0 - L\frac{dI}{dt}- IR = 0$$

The important thing is to make sure you know which convention you are using and stick rigidly to it.