German physicist Georg Simon Ohm published his remarkable findings in 1827. This was the relationship between voltage, current, and resistance. We refer to this as Ohm’s Law.
What Exactly is Ohm’s Law?
Ohm’s Law states that the current flowing through a conductor is directly proportional to the voltage across the conductor and inversely proportional to the resistance of the conductor.
The Mathematical Expression
We can express this law using three equivalent equations\[V = I \times R\]\[I = \frac{V}{R}\]\[R = \frac{V}{I}\]Where,
R = Resistance (we measure this in Ohms, Ω)
V = Voltage (we measure this in Volts, V)
I = Current (we measure this in Amperes, A)
Ohm’s Law Triangle
This simple diagram helps we remember all three formulas instantly.
We can use the triangle this way. At first, we cover the variable we want to find. Then remaining two variables show the formula. When they sit side by side, multiply them. Again, when one sits above the other, divide the top by the bottom.
Examples
- To find V: Cover V, we see I × R, so V = I × R
- For I: Cover I, we see V/R, so I = V / R
- And for R: Cover R, we see V/I, so R = V / I
Practical Examples
Problem: Suppose we have connected a 12V battery to a 6Ω resistor. How much current flows through the circuit?
Solution: Here, V = 12 V, R = 6 Ω. We know: \[I = \frac{V}{R} = \frac{12V}{6\Omega} = 2A \] Answer: 2 Amperes of current flows through the circuit.
Problem: A circuit draws 3A of current through a 15Ω resistance. What voltage appears across the resistor?
Solution: Here, I = 3 A, R = 15 Ω \[V = I \times R = 3A \times 15\Omega = 45 V\]Answer: We’ll measure 45 Volts across the resistor.
Problem: A device operates at 120V and draws 5A of current. What resistance does it have?
Solution: Here, V = 120 V, I = 5 A\[R = \frac{V}{I} = \frac{120V}{5A} = 24\Omega\]Answer: The device presents a resistance of 24 Ohms.
Problem: We want to connect an LED to a 9V battery. The LED requires 20mA (0.02A) and drops 2V across it. What series resistor do we need?
Solution: Here, voltage across resistor V = 9 V – 2 V = 7 V\[R = \frac{V}{I} = \frac{7V}{0.02A} = 350\Omega\]Answer: Choose a 350 Ω resistor (or pick the nearest standard value like 330 Ω or 390 Ω).
Combining Power with Ohm’s Law
We can combine Ohm’s Law with the power equation to create powerful problem-solving tools. Obviously, the Basic Power Equation \(P = V \times I\). We measure P (power) in Watts (W). When we substitute Ohm’s Law into the power equation, we get,\[P = V \times I (our basic power formula)\]\[P = I^2 × R \]\[(use this when we know current and resistance)\]\[P = \frac{V^2}{R} \]\[(use this when we know voltage and resistance)\]
Calculating Power Dissipation
Problem: A 100Ω resistor carries 0.5A of current. How much power does it dissipate?
Solution: Here, R = 100 Ω, I = 0.5 A, \[P = I^2 \times R\]\[ = (0.5A)^2 \times 100\Omega\]\[ = 0.25 \times 100 = 25W\]Answer: The resistor dissipates 25 Watts of power as heat.
Limitations of Ohm’s Law
While incredibly useful, we’ll find Ohm’s Law has limitations. Some materials don’t follow Ohm’s Law linearly. Diodes and transistors (semiconductor devices), light bulbs (their resistance changes with temperature), thermistors (their resistance varies with temperature), arc lamps, and gas discharge tubes.
In AC circuits containing inductors or capacitors, we must replace simple resistance with impedance, which includes both resistance and reactance. While the principles stay similar, our calculations become more complex.
At microwave frequencies and beyond, other electromagnetic effects become significant. Simple Ohm’s Law calculations won’t suffice.
Most materials change their resistance with temperature. For precise calculations, we must consider temperature coefficients.