\(\def \u#1{\,\mathrm{#1}}\) \(\def \abs#1{\left|#1\right|}\) \(\def \ast{*}\) \(\def \deg{^{\circ}}\) \(\def \ten#1{\times 10^{#1}}\) \(\def \redcancel#1{{\color{red}\cancel{#1}}}\) \(\def \BLUE#1{{\color{blue} #1}}\) \(\def \RED#1{{\color{red} #1}}\) \(\def \PURPLE#1{{\color{purple} #1}}\) \(\def \th#1,#2{#1,\!#2}\) \(\def \lshift#1#2{\underset{\Leftarrow\atop{#2}}#1}}\) \(\def \rshift#1#2{\underset{\Rightarrow\atop{#2}}#1}}\) \(\def \dotspot{{\color{lightgray}{\circ}}}\)
Chapter 16: Circuits
3.

Resistors

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If you connect an object to the terminals of a battery, so that there is a potential difference across it, then chances are at least a small amount of current will flow through that object. The higher the potential difference, typically, the higher the current will be. In many objects, the ratio between these two, $\Delta V/I$, is constant, and is called the resistance of the object:

$$R={\Delta V\over I}$$
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Resistance is represented with the symbol $R$, and it is measured in Ohms (Ω) or in Volts/Amperes. Devices which are specifically used for their resistance are called resistors. The symbol of a resistor is a zig-zag line, as shown here.

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When the equation above is written as $$\Delta V = IR$$ then it is generally known as Ohm's Law, and it tells the potential drop across a resistor, given the current through it and its resistance. Ohm's Law should always be applied to single resistors in a single circuit, not to a circuit as a whole.

While 1Ω resistors are not uncommon, in electronics it is common to see resistors in the kΩ or MΩ range, with corresponding currents in the mA or µA range. It may be useful to remember that metric prefixes are just stand-ins for numbers, and can be worked on mathematically. For instance, suppose a 5mA current runs through an 8kΩ resistor: then the potential drop across that resistor is $\Delta V = (5\u{mA})(8\u{kΩ}) = 40\u{(mk)}\u{V}$. But since ${\rm m}=10^{-3}$ and ${\rm k}=10^{+3}$, ${\rm (mk)}=1$ and the two cancel each other out, and we get $\Delta V=40\u{V}$. If you remember that ${\rm m}\times{\rm k}\hbox{ (milli $\times$ kilo)}= 1$ and $\mu \times {\rm M}\hbox{ (micro$\times$mega)}=1$, it can save time when the resistances get large.