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This chapter is from the book

Ohm's Law

There is an extremely important formula that relates the flow of electrons through a resistor. It is called Ohm's Law (after George Simon Ohm):

Equation 3-1

03equ01.gif


  • where V is the voltage force (in volts)

  • i is the current (in amps), and

  • R is the resistance (in ohms, or Ω)

Other variations on Ohm's Law are:

039equ01.gif


Given any two values, the third can be calculated from Ohm's Law. For example, if V = 5 volts and R = 1K (1,000 Ω), then i will be 5 mA (.005 amps).

This is perhaps the most important formula in the book! If there is only one thing in this book you need to commit to memory, this is it. Absolutely anyone and everyone in our industry needs to be able to write down and work with Ohm's Law without having to think about it.

There are a couple of simple observations we can make by looking at Figure 3-1. For example, the flow of water (and the current) must be the same at every point in the pipe (or wire or trace). If this were not true, where would the electrons or the water go? You can't pump water out of a pump if you don't have water coming into the pump. And you can't force electrons out of a battery if they are not also flowing in.

Consider the potential impact if this were not true. If we could force electrons out of the battery without returning them in the other side, then somewhere in the circuit there would be a big buildup of electrons and somewhere else there would be a big shortage. These two areas would then have large negative and positive charges, respectively. Then the two areas would really be hard to handle, attracting or repelling lots of other things around them. We know intuitively that this doesn't happen. So, current flows in a loop, and when it does so it must be the same everywhere in that loop.

We will see later that current might divide into branches, and we will discuss the laws relating to that at that time. But current in any branch is the same at every point in the branch, and ultimately returns to its source.

A second observation relates to the forces around the loop. Consider the voltage at the battery in Figure 3-1 (or the pressure at the pump). It is the same as the voltage across the resistor (or the pressure across the crimp). If we were to go around the loop and add up the voltages (or pressures) they would sum to zero. For example, if we went clockwise around the loop the voltage across the resistor would be +V and the voltage across the battery would be measured as –V (in a clockwise rotation we would see the negative terminal of the battery first). This is a simple illustration, but nevertheless it is a fundamental truth that voltages around even a more complex loop must always sum to zero. Always. This is known as Kirchhoff's (Gustav Robert Kirchhoff) 2nd Law (see Chapter 5).

Let's now look at two resistors (crimps) in series (Figure 3-2). If one resistor (crimp) is very large, very little will flow, regardless of how small the other resistor is. For there to be a large flow, both resistors (crimps) need to be relatively small. We will see later that the effect of two resistors (crimps) in series is additive; that is, the equivalent resistance, Req, is given by Equation 3-2.

03fig02.gifFigure 3-2. Hydraulic analog of two resistors connected in series.

Equation 3-2

03equ02.gif


Of particular note, if R2, for example, is very much larger than R1, then the equation for equivalent series resistance reduces to Equation 3-3.

Equation 3-3

03equ03.gif


If we put two resistors (crimps) in parallel, as in Figure 3-3, the situation is a little different. First, notice that the current out of the battery must equal the current into the battery, as we discussed before, just as the water flowing out of the pump must equal the water flowing into the pump. But the current (water) divides into two branches. Since we can't create current (or water) from nothing, the sum of the current (water) in the two branches must equal the current (water) through the battery (pump).

03fig03.gifFigure 3-3. Hydraulic analog of two resistors connected in parallel.

In these figures, there are two nodes: one at the top where the two branches break off from the main line, and the other at the bottom. The current (water) into a node must equal the current (water) out of the node. If this were not true, where would the electrons (or water) come from or go?

Also, since there can only be a single voltage (pressure) at a node, then the voltage across R1 must be the same as the voltage across R2.

Now consider what happens if R1 and R2 are significantly different, say R1 is very much larger than R2. In that case, most of the current will flow through R2. In fact, no matter how large R1 is, the effective resistance of their combination will never be greater than R2. The equivalent resistance of their combination will be smaller than R2 as R1 reduces in size. It is reasonably intuitive to see that if R1 = R2, then the current through R1 will equal the current through R2 (since the voltage across them is the same, and in fact equals V in this illustration.) If i1 = i2, then the equivalent (single) resistor that looks like the parallel pair of resistors R1 and R2 (where R1 = R2) would be R/2. The formula for the equivalent resistance, Req, of two parallel resistors is given by Equation 3-4:

Equation 3-4

03equ04.gif


This is called the parallel combination of two resistors. We will see later that a similar formula will apply to some combinations (not necessarily parallel ones) of inductance, capacitance, reactance, and impedance.

Of particular note, if R2 is very much larger than R1, then the equation for the equivalent parallel resistance reduces to Equation 3-5 (compare that with Equation 3-2).

Equation 3-5

03equ05.gif


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