Ohm's Law

We have examined voltage (\(U\)), current (\(I\)), and resistance (\(R\)) separately. Ohm's Law relates these three quantities in the case of "simple" conductors, known as ohmic conductors.

For an ohmic conductor (for example, an ideal resistor), the voltage \(U\) across its terminals is proportional to the current \(I\) flowing through it:

• as the current increases, the voltage increases in the same proportion.

This relationship is written as:

• \(U\): voltage in volts \((\mathrm{V})\)
• \(R\): resistance in ohms \((\mathrm{\Omega})\)
• \(I\): current in amperes \((\mathrm{A})\)

It can be viewed as a "rule of three" between \(U\), \(R\), and \(I\).

Interpretation of Ohm's Law

For a given resistance (\(R\) fixed):

• if we increase the voltage \(U\), the current \(I\) increases;
• if we decrease the voltage, the current decreases.

For a given voltage (\(U\) fixed):

• if we increase the resistance \(R\), the current \(I\) decreases;
• if we decrease the resistance, the current increases.

This reflects the intuitive idea:

• the greater the resistance, the harder it is for the current to flow; the greater the voltage, the more it "pushes" the current through the resistance.

Simple Examples

1. Calculating the voltage

   A \(10\ \mathrm{\Omega}\) resistor is carrying a current of \(2\ \mathrm{A}\). The voltage across it is:

   \(\(U = R \times I = 10\ \mathrm{\Omega} \times 2\ \mathrm{A} = 20\ \mathrm{V}\)\) 2. Calculating the current

   A \(100\ \mathrm{\Omega}\) resistor is connected to a \(50\ \mathrm{V}\) voltage source. The current is:

   \(\(I = \frac{U}{R} = \frac{50\ \mathrm{V}}{100\ \mathrm{\Omega}} = 0{,}5\ \mathrm{A}\)\) 3. Calculating resistance

   A resistor carries a current of \(0.2\ \mathrm{A}\) under a voltage of \(12\ \mathrm{V}\). Its value is:

   \[R = \frac{U}{I} = \frac{12\ \mathrm{V}}{0.2\ \mathrm{A}} = 60\ \mathrm{\Omega}\]