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Series circuits - from the point of view of the current

Think back to the tanks connected by a pipe with a restriction in it described in Chapter 2.1. All of the water flowing from one tank to the other must flow through the pipe, and therefore, through the restriction. The flow of the water is, in fact, determined by the resistance of the restriction (we're assuming that the pipe does not impede the flow of water... just the restriction...)

What would happen if we put a second restriction on the same pipe? The water flowing though the pipe from tank to tank now ``sees'' a single, bigger resistance, and therefore flows more slowly through the entire pipe.

The same is true of an electrical circuit. If we connect two resistors, end to end, and connect them to a battery as is shown in the diagram below, the current must flow from the positive battery terminal through the fist resistor, through the second resistor, and into the negative terminal of the battery. Therefore the current leaving the positive terminal ``sees'' one big resistance equivalent to the added resistances of the two resistors.

Figure 2.9: Two resistors and a battery, all connected in series.
\includegraphics[width=1.5in]{02electronics/graphics/series_resistors}

What we are looking at here is called an example of resistors connected in series. What this essentially means is that there are a series of resistors that the current must flow through in order to make its way through the entire circuit.

So, how much current will flow through this system? That depends on the two resistances. If we have a 9 V battery, a 1.5 kohm resistor and a 500 $\Omega $ resistor, then the total resistance is 2 kohms. From there we can just use Ohm's law to figure out the total current running through the system.


$\displaystyle V$ $\textstyle =$ $\displaystyle I R$ (3.50)
$\displaystyle \textrm{therefore}$     (3.51)
$\displaystyle I$ $\textstyle =$ $\displaystyle \frac{V}{R}$ (3.52)
  $\textstyle =$ $\displaystyle \frac{9 \textrm{ V}}{1500 \Omega + 500 \Omega}$ (3.53)
  $\textstyle =$ $\displaystyle \frac{9 \textrm{ V}}{2000 \Omega}$ (3.54)
  $\textstyle =$ $\displaystyle 0.0045 \textrm{ A}$ (3.55)
  $\textstyle =$ $\displaystyle 4.5 \textrm{ mA}$ (3.56)

Remember that this is not only the current flowing through the entire system, it's also therefore the current running through each of the two resistors. This piece of information allows us to go on to calculate the amount of voltage drop across each resistor.


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Geoff Martin 2006-10-15

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