Ohm's Law

WIP -- Claimed by Max Trussell

Ohm's law is very famous equation discovered by Georg Ohm describing the proportional relationship between voltage and current through some conductor. Most commonly this equation is seen in the form of

, with I representing current in amperes, V representing electric potential in volts, and R the resistance in ohms.

The Main Idea

Stripped down to its most basic, Ohm's Law exists so that either the electric potential, current, or total resistance of some conductor may be found when two out of the three are known quantities. This is possible because of the simple, linear relationship between the three.

A simple Ohmic circuit displaying I, V, and R as relevant to Ohm's Law.

A Mathematical Model

While most often represented as [math]\displaystyle{ {I = \frac{|\Delta V|}{R}} }[/math], Ohm's Law may also be represented as [math]\displaystyle{ V = IR }[/math] or [math]\displaystyle{ {R = \frac{V}{I}} }[/math]. Noteworthy is the fact that Ohm's Law depends upon Ohmic resistance and near ideal conductors to be accurate. Fortunately, most simple circuits without capacitance or inductance fit this criteria as the wires offer minuscule resistance when compared to the various resistors in the circuit. Additionally at any given point in time, Ohm's Law applies to both alternating current and direct current. Also worth noting is that V does not necessarily represent the potential difference across a single source of electric potential (e.g. a battery) but rather the absolute value of the potential difference across an entire circuit.

A Computational Model

There are numerous programs to be found that simulate circuits via Ohm's Law, with many allowing for analysis beyond the standard scope of Ohm's Law. One such example is Phet's Circuit Construction Kit, a free, lightweight tool to create and analyze simple DC current circuits. Here is a video briefly describing its installation and use!

Examples

Simple

The most basic example of Ohm's Law in action involves a closed circuit with a single power source and a single Ohmic resistor.

For this example the power source will be 12 volts and the resistor will be 4 ohms. To solve for the current, simply plug in the given values to get [math]\displaystyle{ {\frac{12V}{4\Omega} = 3A} }[/math], meaning that the current is 3 amperes.

Middling

A slightly more complex example involves a closed circuit with a single power source and two Ohmic resistors in parallel.

In this case, R₁ is given to be 2 ohms, R₂ is given to be 4 ohms, and I is given to be 3 amperes. The first step is to remember that the R in the Ohm's Law equation represents total resistance, not individual resistors. In this case (Ohmic resistors in parallel), [math]\displaystyle{ {\frac{1}{R_{Net}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}} }[/math], plug in R₁ and R₂ to find [math]\displaystyle{ R_{Net} = \frac{4}{3} \Omega }[/math].

Now using the form of Ohm's Law [math]\displaystyle{ V = IR }[/math], plug in known values to get [math]\displaystyle{ (3A)(\frac{4}{3}\Omega) = 4V }[/math], so the electric potential across this circuit has been shown to be 4 volts.

Difficult

A more challenging application of Ohm's Law involves a closed circuit with two identical power sources connected in series to three resistors, two of which are in parallel.

Step 1: Find the Total Resistance

The total resistance for this circuit is going to equal the resistance of R₁ plus the resistance of R₂ and R₃ in parallel. Using the equation from the previous problem, [math]\displaystyle{ {\frac{1}{R_{P}} = \frac{1}{R_{2}} + \frac{1}{R_{3}}} }[/math], plug in R₂ and R₃ to get [math]\displaystyle{ R_{P} = 2.4\Omega }[/math]

Then for total resistance, R_P must be added to R₁ to get [math]\displaystyle{ 2.4\Omega + 2\Omega = 4.4\Omega }[/math]

Step 2: Find the Total Voltage

The total voltage for this circuit in this case is simply equal to the voltages of each power source added together, meaning [math]\displaystyle{ V_{T} = 2(4V) = 8V }[/math]

Step 3: Plug Into Appropriate Version of Ohm's Law

Now just plug the total resistance and voltage into [math]\displaystyle{ I = \frac{V}{R} }[/math] to get a current of 1.82 amperes

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