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Formulas and Concepts

At this stage we're ready to start analyzing some simple circuits. So far all we are dealing with are sources of emf (batteries) and resisters, placed in series and parallel arrangements. The two rules which all you to solve any combination of this sort are Kirchoff's Rules:
  1. Junction: The total change in potential around any closed loop in a circuit must be zero. This says that the potential always has one value at every point.
  2. Loop: The total current entering a junction is equal to the total current leaving. This says that no charge is building up at a junction.

The procedure to solve a circuit is sort of like drawing a free body diagram. In each branch you put an arrow labeling the current in that branch. The direction of the arrow is arbitrary! But, when you solve for the currents, a negative sign indicates that you chose the wrong direction: i.e. the current is flowing opposite the direction you chose. The is just like when you draw a force arrow in one direction and its value comes out negative.

Now, when writing down the Junction Rule, we start at one point, and write down the voltage change each element contributes in order until we've returned to the starting point. Note that the direction we traverse the loop need not be in the direction of the current! To do this, you need to know what voltage change each element introduces. We start with resisters.

R
If you go through a resister in the direction of the current then the voltage decreases by $IR$.

R$+$
If you go through a resister opposite the direction of the current then the voltage increases by $IR$
And next we have the batteries.
$\mathscr{E}-$
If you go through an EMF from the $+$ plate to the plate, regardless of the current direction, then the voltage decreases by $\mathscr{E}$.

$\mathscr{E}+$
If you go through an EMF from the plate to the $+$ plate, regardless of the current direction, then the voltage increases by $\mathscr{E}$
Note that resisters depend on the current direction, but batteries do not!
\includegraphics[scale=.5]{R-.eps} \includegraphics[scale=.5]{R+.eps} \includegraphics[scale=.55]{E-.eps} \includegraphics[scale=.55]{E+.eps}
$V=-IR$ $V=+IR$ $V=-\mathscr{E}$ $V=+\mathscr{E}$

Next, one subtlety concerns the power in a circuit, specifically the battery. Generally, a real battery has two parts: an source of EMF and an internal resistance. Both parts are internal to the battery and thus not part of the external circuit! So we have the following quantities:

  1. Power generated by the EMF: This is the product $I\mathscr{E}$ of the current in the battery and the EMF in the battery.
  2. Power consumed by the battery: This is the product $I^2R_{int}$ of the current in the battery squared and the internal resistance of the battery.
  3. Power delivered by the battery to the circuit: This is the difference $I\mathscr{E}-I^2R_{int}$ of the total power generated by the EMF and power eaten by the resistance.
Make careful note of the distinction between these three, and also that they can be asked in different ways, not just the way mentioned here. Finally, this means that the total power consumed by the external circuit must be equal to the total power delivered to it, which is the third item on this list.


next up previous
Next: Solution Up: TDEC 115 - Week Previous: TDEC 115 - Week
Daniel Cross 2006-09-18