Subsection 4.1.8 Ohm’s Law
Ohm’s law describes the relationship between current, voltage, and resistance in an electric circuit. According to Ohm’s Law, the current flowing through a circuit is directly proportional to the voltage across it and inversely proportional to the resistance of the circuit. Mathematically, this relationship is expressed as
\begin{equation}
I = V/R\tag{4.1.6}
\end{equation}
where:
\(I\) is the the current in amperes
\(V\) is the voltage in volts, and
\(R\) is the resistance in ohms.
In other words, an increase in voltage leads to a corresponding increase in current if the resistance remains constant, while an increase in resistance results in a decrease in current for a given voltage.
It’s important to note that Ohm’s Law is most applicable to linear resistors under constant conditions. In AC circuits with reactive components like capacitors and inductors, Ohm’s Law needs to be extended to include the concept of impedance, which takes into account both resistance and reactance.
Use this digram for the three example problems below.
Example 4.1.12. Find Current.
Given \(E\) = 24 V and \(R\) = 100 \(\Omega\text{,}\) determine the current \(I\text{.}\)
Answer.
Solution.
\begin{equation*}
I = \frac{E}{R} = \frac{24 \text{ V}}{100\ \Omega} = 0.24 \text{ A} = 240 \text{ mA}\text{.}
\end{equation*}
Example 4.1.13. Find Resistance.
Given \(E\) = 6.2 kV and \(I\) = 233 A, determine resistance \(R\text{.}\)
Answer.
Solution.
Before applying
(4.1.6) the given voltage must be converted from kilovolts to volts. (Recall that1 kV = 1000 V)
\begin{equation*}
I = \frac{E}{R} \implies R = \frac{E}{I} = \frac{6200 \text{ V}}{233 \text{ A} } = 26.6\ \Omega\text{.}
\end{equation*}
Example 4.1.14. Find Voltage.
Given \(I\) = 5 mA and \(R\) = 50 \(\Omega\text{,}\) determine voltage \(E\text{.}\)
Answer.
Solution.
\begin{equation*}
I = \frac{E}{R} \implies E = IR = (0.005 \text{ A})(50\ \Omega) = 0.25 \text{ V}\text{.}
\end{equation*}
