# Ohm’s law

- the mathematical formulation of Ohm’s law and the basic inverse relation that it encodes
- how a water pipe analogy can help in understanding the meaning of Ohm’s law.

## Ohm’s law

Ohm’s law states the if \(\normalsize{V}\) is the voltage (measured in volts) across a resistor \(\normalsize{R}\) (measured in ohms) which draws a current \(\normalsize{I}\) (measured in amperes), then \[\Large{V=IR}.\] A resistor is an object that uses electrical power and converts it to something else, perhaps heat or light. A toaster is an example. The electricity that runs through a toaster is powered by a voltage differential, supplied by the electrical power outlet. The greater the voltage, the more current \(\normalsize{I}\) runs through the toaster. So for a fixed resistor \(\normalsize{R}\), Ohm’s law establishes a linear proportionality between voltage and current.^{Toaster Filaments Nick carson at en.wikipedia CC BY 3.0, via Wikimedia Commons}However we can look at the law in other ways. If we regard the voltage \(\normalsize{V}\) as fixed, then the resistance and current are inversely proportional, since their product is constant and equal to the fixed voltage. If we increase the resistance, then the current decreases, while if we decrease the resistance, then the current increases. This is the situation with a circuit run by a battery, or with the electricity in our house, where the voltage supplied is constant (\(\normalsize{110-120}\) volts in most of the Americas, \(\normalsize{220-230}\) volts in Europe, Australia and most countries in Asia). However strictly speaking in this case the voltage is alternating in direction. In the limiting case where the resistance becomes zero, for example if you replace the toaster with a wire, then an indefinitely large current flows. Then a

*short*erupts, often with disastrous consequences, particularly if you don’t have a fuse connected to break the circuit in such an emergency situation.

## Some examples

If we connect a lamp to a circuit run by a 6 volt battery, and draw a current of 3 amps, then the resistance \(\normalsize R\) is given by \[\Large R=\frac {V}{I}=\frac 63= 2\;\text{ohms}.\] Now if we connect that same lamp to a 10 volt battery, then the current \(\normalsize I\) will be \[\Large I=\frac {V}{R}=\frac {10}2= 5\;\text{amps}.\] If we want to make the light brighter we need to increase the current, let’s say to 8 amps then we need to increase our voltage to \[\Large V=IR=8\times2=16\;\text{volts}.\]Q1(E): An electrical device is connected to 120 volts. Find the current if the resistance is 480 ohms.Q2(E): Suppose that we have a battery of some constant voltage lighting a small lamp, and an ammeter reads 40 mA, where mA stands for milliamps, which are one thousandths of an amp. If the current drops to 20 mA, what has happened to the resistance?

## How does a resistor resist?

*resisted*, less gets through. Some materials have low resistance, such as copper wire, which allows electrons to flow through it without much trouble at all. Other materials, such as wood, have high resistance, stopping electrical flow almost immediately. In practice we have things like lamps and toasters that generate light or heat from the electrons, slowing them down, but still letting them through.

*directly proportional*to the length \(\normalsize L\): double the length of the wire tube and its resistance doubles. But \(\normalsize R\) is also

*inversely proportional*to the cross-sectional area \(\normalsize A\): double the area and the resistance halves.

Q3.(E): A tubular resistor is in the shape of a wire. If we triple its length and halve its diameter, what happens to its resistance?

## A hydraulic analogy

^{Bernoullis Law Derivation Diagram by MannyMax CC-BY-SA-3.0, via Wikimedia Commons}

## Answers

\[\Large{I=\frac{V}{R}=\frac{120}{480}= 0.25\;\text{amps}}.\]A1.By Ohm’s law we can find the current byA2.When the voltage is constant, the relation between the current and the resistance is inverse. Therefore, if the current halves, the resistance doubles.A3.Tripling the length of the resistor increases its resistance by a factor of 3, while halving its diameter multiplies the cross-sectional area by 1/4. Overall the resistance changes by a factor of \(\frac{3}{1/4}=12\).

#### Maths for Humans: Linear, Quadratic & Inverse Relations

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