Ohm's Law Formulas

Why This Matters

In Electrical Circuits Systems I, you're being tested on your ability to analyze circuit behavior and solve for unknown quantities—not just plug numbers into equations. Ohm's Law and its power-related extensions form the foundation of every circuit analysis problem you'll encounter. Whether you're calculating current through a resistor, determining power dissipation, or selecting components that won't overheat, these formulas are your essential toolkit.

Here's the key insight: these aren't ten separate formulas to memorize. They're algebraic rearrangements of two core relationships—Ohm's Law ($V = IR$) and the power equation ($P = VI$). Understanding when to use each form matters more than memorizing them all. Don't just know the formulas—know which variables you have, which you need, and which equation gets you there fastest.

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The Core Relationship: Ohm's Law

Ohm's Law describes the fundamental relationship between voltage, current, and resistance in a conductor. It states that current through a conductor is directly proportional to voltage and inversely proportional to resistance.

$V = IR$ (Voltage Form)

• Voltage (V) is the electrical pressure or potential difference that drives charge through a circuit—think of it as the "push"
• Direct proportionality—doubling either current or resistance doubles the voltage drop across a component
• Circuit analysis essential—use this form when you know current and resistance and need to find voltage drops in series circuits

$I = \frac{V}{R}$ (Current Form)

• Current (I) represents charge flow rate in amperes—this form solves for how much charge moves per second
• Inverse relationship with resistance—as $R$ increases, $I$ decreases for constant voltage, explaining why resistors limit current
• Most common application—use when analyzing a circuit with a known voltage source and you need to determine current draw

$R = \frac{V}{I}$ (Resistance Form)

• Resistance (R) quantifies opposition to current flow in ohms ($\Omega$)—this form lets you measure or verify component values
• Experimental determination—apply a known voltage, measure current, calculate resistance to characterize unknown components
• Troubleshooting tool—comparing calculated resistance to expected values helps identify faulty components

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Primary Power Relationships

Power measures the rate of energy transfer in a circuit. These three forms of the power equation let you calculate wattage using whichever two variables you know.

$P = VI$ (Basic Power Form)

• Power (P) is measured in watts (W) and represents energy consumed or delivered per second
• Universal application—works for any circuit element, whether resistive, capacitive, or inductive
• Component rating verification—multiply operating voltage by current draw to ensure components won't exceed their wattage limits

$P = I^2R$ (Current-Resistance Form)

• Squared current term—power scales with current squared, so doubling current quadruples power dissipation
• Heat generation focus—this form directly shows resistive losses; critical for wire sizing and thermal management
• Use when voltage is unknown—ideal for analyzing current-carrying conductors where you know the current and resistance

$P = \frac{V^2}{R}$ (Voltage-Resistance Form)

• Squared voltage term—power scales with voltage squared, making this form essential for high-voltage system analysis
• Load power calculation—when you know the supply voltage and load resistance, this gives power directly
• Efficiency analysis—helps determine power delivered to a load versus power lost in distribution

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Derived Formulas: Solving for Secondary Variables

These formulas are algebraic rearrangements that let you solve for voltage, current, or resistance when power is one of your known quantities. They're derived by substituting Ohm's Law into the power equation.

$V = \sqrt{PR}$ (Voltage from Power and Resistance)

• Square root relationship—voltage increases with the square root of power, not linearly
• Power system design—determines what voltage is needed to deliver a specific power to a known resistance
• Derived from $P = \frac{V^2}{R}$ by solving for $V$—recognize the algebraic origin to reduce memorization

$I = \sqrt{\frac{P}{R}}$ (Current from Power and Resistance)

• Current determination—calculates the current flowing when you know power consumption and circuit resistance
• Safety verification—ensures current stays within conductor ampacity ratings for a given power load
• Derived from $P = I^2R$ by solving for $I$—understanding this derivation helps you reconstruct the formula if forgotten

$R = \frac{V^2}{P}$ (Resistance from Voltage and Power)

• Component selection—determines what resistance value will dissipate a specific power at a given voltage
• Heater and load design—essential for sizing resistive heating elements to achieve target power output
• Rearrangement of $P = \frac{V^2}{R}$—multiply both sides by $R$, then divide by $P$

$R = \frac{P}{I^2}$ (Resistance from Power and Current)

• Alternative resistance calculation—useful when current and power are measured but voltage isn't directly accessible
• Current-limited applications—helps design resistance values for circuits with fixed current sources
• Rearrangement of $P = I^2R$—simply divide both sides by $I^2$

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Quick Reference Table

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Self-Check Questions

1. You know a resistor's power rating and its resistance value. Which two formulas could you use to find the maximum safe voltage and current?

2. Why does doubling the current through a resistor quadruple the power dissipation, while doubling the voltage across it also quadruples power? Which formula reveals each relationship?

3. Compare $P = I^2R$ and $P = \frac{V^2}{R}$: In a series circuit where current is constant, which formula would you use to compare power dissipation across different resistors? Why?

4. A technician measures 12V across a component and 2A through it. Using only Ohm's Law and the basic power formula, calculate resistance and power—then verify your power answer using one of the derived formulas.

5. You're designing a circuit and need to select a resistor that will dissipate exactly 5W when connected to a 10V source. Which formula gives you the required resistance directly, and what value do you get?