# Ohm's Law, Kirchhoff's Loop Rule and Power

> AP Physics 2 · Electric Circuits
> Source: https://www.owlsprep.com/study/ap-physics-2-u4-ohm-s-law-kirchhoff-s/

This module covers Ohm's Law for ohmic/non-ohmic materials, Kirchhoff's Loop Rule from conservation of energy, and power calculation for DC circuit components, aligned to AP Physics 2 CED requirements.

**Prerequisites:** Basic definitions of current, voltage, and resistance; Conservation of energy for closed systems; Difference between series and parallel circuit connections

## Learning objectives

- Apply Ohm's Law to ohmic and non-ohmic circuit components
- Use Kirchhoff's Loop Rule to analyze single-loop DC circuits
- Calculate electric power for any DC circuit component
- Follow correct sign conventions for loop analysis

## Ohm's Law

**Ohm's Law** — An empirical relationship stating that potential difference across a component equals the product of current through the component and its resistance. Ohmic materials have constant resistance at fixed temperature; non-ohmic materials have variable resistance that changes with voltage or current.

*Notation:* V = IR

*Example:* Metallic conductors at constant temperature are ohmic; diodes and incandescent filaments are non-ohmic.

On a V-I graph, ohmic materials produce a straight line passing through the origin, with slope equal to constant resistance. While $R = V/I$ can calculate instantaneous resistance for non-ohmic materials at a single operating point, this does not make the component ohmic.

**Worked example:** A student measures current through an unknown component at three different applied voltages: $V=2.0\ \text{V}, I=0.50\ \text{A}$; $V=4.0\ \text{V}, I=0.90\ \text{A}$; $V=6.0\ \text{V}, I=1.20\ \text{A}$. (a) Calculate the resistance of the component at each voltage, (b) determine if the component is ohmic.

1. Rearrange Ohm's Law to solve for resistance:
2. $$R = \frac{V}{I}$$
3. Calculate resistance for each measurement:
4. - At 2.0 V: $R = \frac{2.0}{0.50} = 4.0\ \Omega$
5. - At 4.0 V: $R = \frac{4.0}{0.90} \approx 4.4\ \Omega$
6. - At 6.0 V: $R = \frac{6.0}{1.20} = 5.0\ \Omega$
7. By definition, an ohmic component has constant resistance across all applied voltages (within experimental error). The resistance here increases systematically with voltage, so the component is non-ohmic.

> **Exam tip:** When asked to identify an ohmic material from a V-I graph, remember the requirement is a straight line passing through the origin, not just any straight line. A non-zero intercept means the component is not ohmic.

## Kirchhoff's Loop Rule

Kirchhoff's Loop Rule is a direct application of conservation of energy to closed circuits. It states that the net change in electric potential around any closed loop in a circuit is zero:

Physically, this means the total energy gained from energy sources (like batteries) equals the total energy lost to resistive components when moving a charge around the full loop. Correct application depends entirely on following a consistent sign convention:

1. Choose any traversal direction (clockwise or counterclockwise; direction does not change the final result if you are consistent).
2. For batteries: moving from negative to positive terminal gives $
Delta V = +\varepsilon$ (potential gain); moving from positive to negative gives $
Delta V = -\varepsilon$ (potential drop).
3. For resistors: moving in the same direction as conventional current gives $
Delta V = -IR$ (potential drop); moving opposite current gives $
Delta V = +IR$ (potential gain).

**Worked example:** A 9.0 V battery with negligible internal resistance is connected in series with a 2.0 Ω resistor and a 4.0 Ω resistor. Use Kirchhoff's Loop Rule to find the current in the circuit.

1. Assume conventional current flows clockwise out of the battery's positive terminal, and traverse the loop clockwise starting at the battery's negative terminal.
2. Sum potential changes per the sign convention:
3. - Moving negative to positive through the battery: $
Delta V = +9.0\ \text{V}$
4. - Moving through 2.0 Ω in the direction of current: $
Delta V = -2.0I$
5. - Moving through 4.0 Ω in the direction of current: $
Delta V = -4.0I$
6. Set the sum equal to zero per the loop rule:
7. $$9.0 - 2.0I - 4.0I = 0$$
8. Solve for $I$:
9. $$9.0 = 6.0I \implies I = 1.5\ \text{A}$$
10. The positive sign confirms the assumed direction of current is correct.

> **Exam tip:** If your final current comes out negative, do not change the magnitude. The negative sign only indicates your assumed direction was wrong; AP Physics 2 accepts negative current values as correct if reasoning is accurate.

## Electric Power in Circuits

Electric power is the rate of energy conversion in a circuit component. The general formula applies to any component, whether it supplies power (like a battery) or dissipates power (like a resistor):

Positive $P$ means the component absorbs or dissipates power; negative $P$ means the component supplies power to the circuit. For ohmic components, substitute Ohm's Law to get two alternative useful forms:

$P=I^2 R$ is most convenient for series circuits (all components share the same current), while $P=V^2/R$ is easiest for parallel circuits (all components share the same voltage). Always use the potential difference across the specific component you are analyzing, not the total battery emf, unless the component is connected directly across the battery.

**Worked example:** A 12.0 V car battery with internal resistance connected in series to a 5.5 Ω headlight bulb supplies 2.0 A of current to the bulb. (a) Calculate the actual power dissipated by the bulb, (b) find the voltage drop across the battery's internal resistance.

1. We know current and resistance for the bulb, so use $P=I^2 R$:
2. $$P = (2.0\ \text{A})^2 (5.5\ \Omega) = 22\ \text{W}$$
3. Find the voltage drop across the bulb with Ohm's Law to confirm:
4. $$V = IR = (2.0\ \text{A})(5.5\ \Omega) = 11.0\ \text{V}$$
5. Confirm power with $P=VI$: $P = (11.0\ \text{V})(2.0\ \text{A}) = 22\ \text{W}$, which matches the first calculation.
6. Total battery emf equals the sum of voltage drops across the internal resistance and the bulb, so:
7. $$\Delta V_{\text{internal}} = \varepsilon - V_{\text{bulb}} = 12.0\ \text{V} - 11.0\ \text{V} = 1.0\ \text{V}$$

> **Exam tip:** Always use the potential difference across the specific component you are calculating power for, not the total emf of the battery, unless the component is connected directly across the battery with zero internal resistance.

## AP-Style Worked Practice Questions

**Check your understanding**

1. Two ohmic resistors, $R_1 = 2.0\ \Omega$ and $R_2 = 6.0\ \Omega$, are connected in parallel across a $6.0\ \text{V}$ battery with negligible internal resistance. What is the total power dissipated by the two resistors?

   - 4.5 W
   - 12.0 W
   - 24.0 W
   - 36.0 W

   *Why:* For parallel resistors, voltage across each equals the battery voltage. Total power is $P_1 + P_2 = \frac{(6.0)^2}{2.0} + \frac{(6.0)^2}{6.0} = 18 + 6 = 24.0\ \text{W}$.

**Worked example:** A single-loop circuit has two opposing batteries (positive terminals facing each other) in series with two resistors. $
varepsilon_1 = 10\ \text{V}$, $
varepsilon_2 = 4\ \text{V}$, $R_1 = 2\ \Omega$, $R_2 = 1\ \Omega$, all batteries have negligible internal resistance. (a) Use Kirchhoff's Loop Rule to find the magnitude and direction of current, (b) calculate power dissipated by each resistor, (c) identify which battery is being charged and find the power it absorbs.

1. (a) Assume current flows clockwise, traverse the loop clockwise, sum potential changes:
2. $$\varepsilon_1 - \varepsilon_2 - IR_1 - IR_2 = 0$$
3. Substitute values and solve:
4. $$10 - 4 - I(2 + 1) = 0 \implies 6 = 3I \implies I = 2\ \text{A}$$
5. The positive result confirms current is 2 A, flowing clockwise.
6. (b) Power dissipated by resistors is $P=I^2R$:
7. $P_1 = (2)^2(2) = 8\ \text{W}$, $P_2 = (2)^2(1) = 4\ \text{W}$
8. (c) Current flows into the positive terminal of $
varepsilon_2$, so $
varepsilon_2$ is being charged. Power absorbed is:
9. $$P = \varepsilon_2 I = (4\ \text{V})(2\ \text{A}) = 8\ \text{W}$$

**Worked example:** A portable phone charger is rated for 5.0 V output at maximum 2.4 A current. The charger is 85% efficient, meaning 85% of input power from the wall is delivered to the phone, with the rest dissipated as heat. Find the maximum output power, total input power drawn from the wall, and wasted heat power.

1. Maximum output power to the phone is $P_{\text{out}} = VI$:
2. $$P_{\text{out}} = (5.0\ \text{V})(2.4\ \text{A}) = 12\ \text{W}$$
3. Efficiency $
eta = P_{\text{out}} / P_{\text{in}}$, rearrange for input power:
4. $$P_{\text{in}} = \frac{P_{\text{out}}}{\eta} = \frac{12\ \text{W}}{0.85} \approx 14.1\ \text{W}$$
5. Wasted heat power is the difference between input and output power:
6. $$P_{\text{heat}} = P_{\text{in}} - P_{\text{out}} = 14.1\ \text{W} - 12\ \text{W} = 2.1\ \text{W}$$

## Common pitfalls

- **Wrong:** Using $P = V^2/R$ to compare power of series-connected bulbs, assuming $V$ across each bulb equals the source voltage.
  - Why it fails: Students forget $V$ in power formulas refers to the component's own potential difference, not the source voltage. Series components split the source voltage.
  - Correct: For series-connected components, use $P=I^2 R$ for power calculations, since all series components share the same current.
- **Wrong:** Always assigning a negative potential change to resistors when applying the loop rule, regardless of traversal direction.
  - Why it fails: Students associate resistors with voltage drops and forget the sign depends on traversal direction relative to current direction.
  - Correct: Write the sign of every potential change explicitly based on your traversal and current directions before solving the loop equation.
- **Wrong:** Calling a component ohmic because it satisfies $V=IR$ at a single operating point.
  - Why it fails: Students confuse the general formula $R=V/I$ (valid for any component) with Ohm's Law, which requires constant resistance across all voltages.
  - Correct: Confirm resistance is constant across a range of voltages before labeling a component ohmic; on a V-I graph, check for a straight line through the origin.
- **Wrong:** For multi-battery circuits, setting the sum of voltage drops equal to the largest battery's emf instead of setting total sum to zero.
  - Why it fails: Students memorize a simplified rule for single-battery circuits and generalize it incorrectly.
  - Correct: Always set the sum of all potential changes around any closed loop equal to zero, regardless of the number of batteries.
- **Wrong:** Assuming all metallic conductors are non-ohmic because incandescent filament resistance increases with current.
  - Why it fails: Students generalize the behavior of temperature-dependent filaments to all metallic conductors.
  - Correct: Ohmic materials are defined as having constant resistance when temperature is held constant. Resistance change from temperature variation does not make a material non-ohmic by definition.

## Cheatsheet

| Category | Formula | Notes |
| --- | --- | --- |
| Ohm's Law | $V = IR$ | Works for any single component for instantaneous resistance; requires constant resistance for ohmic materials |
| Ohmic Material | $R = \text{constant}$ | Resistance must be constant across a range of voltages at fixed temperature, not just one point |
| Kirchhoff's Loop Rule | $\sum \Delta V = 0$ | Applies to all closed DC loops; derived from conservation of energy |
| Battery Sign (Loop Rule) | $\Delta V = +\varepsilon$ (neg → pos), $-\varepsilon$ (pos → neg) | Sign depends on traversal direction, not just the battery |
| Resistor Sign (Loop Rule) | $\Delta V = -IR$ (same as current), $+IR$ (opposite current) | Uses conventional current direction |
| General Power | $P = VI$ | Works for any component, supplying or absorbing power |
| Power (Ohmic, Series) | $P = I^2 R$ | Best for series circuits with identical current for all components |
| Power (Ohmic, Parallel) | $P = \frac{V^2}{R}$ | Best for parallel circuits with identical voltage for all components |
| Total Circuit Power | $P_{\text{total}} = \sum P_i$ | Total power is always the sum of individual component powers |

## What's next

This topic is the foundational energy framework for all circuit analysis in AP Physics 2. Mastering consistent sign conventions for Kirchhoff's Loop Rule and correct power calculation is critical for all subsequent circuit topics on the exam. Next, you will combine the loop rule with Kirchhoff's Junction Rule to analyze complex multi-loop circuits with multiple branches, then extend these core rules to RC circuits, where you will derive and interpret exponential charging and discharging behavior. A solid grasp of the concepts in this module is required to correctly solve any circuit problem on both the multiple-choice and free-response sections of the AP Physics 2 exam.

- [AP Physics 2 Unit 4 Electric Circuits Overview](https://www.owlsprep.com/study/ap-physics-2-u4-overview/)
- [Kirchhoff's Junction Rule and Conservation of Charge](https://www.owlsprep.com/study/ap-physics-2-u4-kirchhoff-s-junction-rule-and/)
- [Steady-State DC Circuits with Resistors and Capacitors](https://www.owlsprep.com/study/ap-physics-2-u4-steady-state-dc-circuits-with/)

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