Ohm's Law, Kirchhoff's Loop Rule and Power — AP Physics 2 Study Guide

For: AP Physics 2 candidates sitting AP Physics 2.

Covers: Ohm's Law for ohmic and non-ohmic materials, Kirchhoff's Loop Rule, electric power in DC circuits, sign conventions for loop analysis, and series/parallel resistance combinations tested on AP Physics 2.

You should already know: Basic definitions of current, voltage, and resistance; Conservation of energy for closed systems; The difference between series and parallel circuit connections.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 2 style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Ohm's Law, Kirchhoff's Loop Rule and Power?

This core subtopic within AP Physics 2 Unit 4 (Electric Circuits) makes up roughly 4-8% of the total exam score, per the College Board Course and Exam Description (CED), and appears regularly in both multiple-choice (MCQ) and free-response (FRQ) sections. The topic unifies three key ideas: the empirical relationship between current, voltage, and resistance for circuit components; the application of conservation of energy to closed circuit loops; and the calculation of energy transfer rate in electric circuits.

At its core, this topic allows you to predict how any DC circuit will behave, from simple series-parallel combinations to more complex networks with multiple batteries. It is also the foundation for analyzing RC circuits, which are another key part of Unit 4, and for understanding real-world applications from household wiring to battery operation. All problems in this topic rely on consistent sign and notation conventions: we use $I$ for current (units: amperes, A), $V$ for potential difference (volts, V), $R$ for resistance (ohms, Ω), and $P$ for power (watts, W), with conventional current assumed to flow from the positive terminal of a battery.

2. Ohm's Law

Ohm's Law is an empirical relationship describing the behavior of many common conducting materials. The core relationship for any single component is given by: $$V=IR$$ where $V$ is the potential difference across the component, $I$ is the current through it, and $R$ is its resistance. A critical distinction in AP Physics 2 is between ohmic and non-ohmic materials: an ohmic material has a constant resistance that does not change with varying applied voltage or current (as long as temperature is held constant). Non-ohmic materials have a resistance that changes with voltage or current, so their V-I relationship is non-linear.

Common examples of ohmic materials are most metallic conductors at constant temperature. Common non-ohmic components include diodes, thermistors, and incandescent light bulb filaments (whose resistance increases as temperature rises with increasing current). On a V-I graph, an ohmic material will always produce a straight line passing through the origin, with slope equal to the constant resistance. The formula $V=IR$ can still be used to calculate the instantaneous resistance of a non-ohmic material at a given current, but this does not make the material ohmic.

Worked Example

A student measures current through an unknown component at three different applied voltages, with results: $V=2.0\text{V},I=0.50\text{A}$; $V=4.0\text{V},I=0.90\text{A}$; $V=6.0\text{V},I=1.20\text{A}$. (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: $R=\frac{V}{I}$.
2. Calculate resistance for each measurement:
   • At 2.0 V: $R=\frac{2.0}{0.50}=4.0\Omega$
   • At 4.0 V: $R=\frac{4.0}{0.90}\approx 4.4\Omega$
   • At 6.0 V: $R=\frac{6.0}{1.20}=5.0\Omega$
3. By definition, an ohmic component has constant resistance across all applied voltages (within experimental error).
4. 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.

3. Kirchhoff's Loop Rule

Kirchhoff's Loop Rule is simply a translation of conservation of energy into circuit language. The rule states that for any closed loop in a circuit, the net change in electric potential around the loop is zero: $$\sum \Delta V=0$$ This makes physical sense because if you return to your starting point after traversing the loop, the electric potential at that point is fixed, so you cannot have a net change in potential. For a charge moving around the loop, this means the total energy gained from energy sources (like batteries) equals the total energy lost to components like resistors.

The most important part of applying the loop rule correctly is following a consistent sign convention. The standard AP Physics 2 convention is:

1. Choose any traversal direction (clockwise or counterclockwise; direction does not affect the final result as long as you are consistent).
2. For a battery: if you move from the negative terminal to the positive terminal, $\Delta V=+\epsilon$ (potential gain, where $\epsilon$ is the battery emf). If you move from positive to negative, $\Delta V=-\epsilon$ (potential drop).
3. For a resistor: if you move through the resistor in the same direction as conventional current, $\Delta V=-IR$ (potential drop). If you move opposite the current, $\Delta V=+IR$ (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 positive terminal of the battery. Choose to traverse the loop clockwise starting at the battery's negative terminal.
2. Add up all potential changes according to the sign convention:
   • Moving negative to positive through the battery: $\Delta V=+9.0\text{V}$
   • Moving through 2.0 Ω in the direction of current: $\Delta V=-2.0I$
   • Moving through 4.0 Ω in the direction of current: $\Delta V=-4.0I$
3. Set the sum equal to zero per the loop rule: $9.0-2.0I-4.0I=0$
4. Solve for $I$: $9.0=6.0I⟹I=1.5\text{A}$. The positive sign confirms our 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 as long as magnitude and reasoning are accurate.

4. Electric Power in Circuits

Electric power is the rate at which energy is converted from one form to another in a circuit component. For any circuit component (whether it supplies energy like a battery or dissipates energy like a resistor), the general power formula is: $$P=VI$$ where $V$ is the potential difference across the component, and $I$ is the current through it. Positive $P$ means the component dissipates or absorbs power, while negative $P$ means the component supplies power to the circuit.

For ohmic components, we can substitute Ohm's Law ($V=IR$) into the general formula to get two additional useful forms: $$P=I^{2}R=\frac{V^2}{R}$$ These alternative forms are convenient for different circuit configurations: $P=I^{2}R$ is easiest for series circuits, where all components share the same current, while $P=V^2/R$ is easiest for parallel circuits, where all components share the same voltage. It is critical to remember that $V$ in all power formulas is the potential difference across the specific component you are analyzing, not the total emf of the battery, unless the component is connected directly across the battery with no internal resistance.

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 the current through the bulb and its resistance, so use $P=I^{2}R$ to get actual power: $P=(2.0\text{A}{)}^2(5.5\Omega )=22\text{W}$. This is the true power dissipated by the bulb, regardless of the battery's total emf.
2. Find the actual voltage drop across the bulb using Ohm's Law: $V=IR=(2.0\text{A})(5.5\Omega )=11.0\text{V}$. Confirming with $P=VI$ gives $P=(11.0\text{V})(2.0\text{A})=22\text{W}$, matching the first calculation.
3. The total emf of the battery equals the sum of voltage drops across the internal resistance and the bulb, so voltage across internal resistance is $\epsilon -V=12.0\text{V}-11.0\text{V}=1.0\text{V}$.

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.

5. Common Pitfalls (and how to avoid them)

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

6. Practice Questions (AP Physics 2 Style)

Question 1 (Multiple Choice)

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? A) $4.5\text{W}$ B) $12.0\text{W}$ C) $24.0\text{W}$ D) $36.0\text{W}$

Worked Solution: For parallel-connected resistors, the potential difference across each resistor equals the battery voltage, since internal resistance is negligible. The formula $P=V^2/R$ is convenient here because voltage is the same for all parallel components. For $R_1$, $P_1=(6.0\text{V}{)}^2/2.0\Omega =18\text{W}$. For $R_2$, $P_2=(6.0\text{V}{)}^2/6.0\Omega =6\text{W}$. Total power is the sum of individual powers, so $P_{\text{total}}=18+6=24.0\text{W}$. Option A comes from the common mistake of adding resistors in series instead of parallel, while other options come from equivalent resistance calculation errors. The correct answer is C.

Question 2 (Free Response)

A single-loop circuit has two batteries opposing each other (positive terminals facing one another) in series with two resistors. Battery 1 has emf ${\epsilon }_1=10\text{V}$, Battery 2 has emf ${\epsilon }_2=4\text{V}$, and resistors are $R_1=2\Omega$, $R_2=1\Omega$. All batteries have negligible internal resistance. (a) Use Kirchhoff's Loop Rule to find the magnitude and direction of the current in the loop. (b) Calculate the power dissipated by each resistor. (c) Which battery is being charged, and how much power does it absorb?

Worked Solution: (a) Assume current flows clockwise (driven by the larger ${\epsilon }_1$). Traverse the loop clockwise and sum potential changes: ${\epsilon }_1-{\epsilon }_2-I{R}_1-I{R}_2=0$. Substitute values: $10-4-I(2+1)=0⟹6=3I⟹I=2\text{A}$. The positive result confirms direction is clockwise, so magnitude is $2\text{A}$ clockwise. (b) Power dissipated by a resistor is $P=I^{2}R$. For $R_1$: $P_1=(2{)}^2(2)=8\text{W}$. For $R_2$: $P_2=(2{)}^2(1)=4\text{W}$. (c) Current flows into the positive terminal of ${\epsilon }_2$ (since direction is clockwise), so ${\epsilon }_2$ is being charged. Power absorbed by ${\epsilon }_2$ is $P={\epsilon }_{2}I=(4\text{V})(2\text{A})=8\text{W}$.

Question 3 (Application / Real-World Style)

A portable phone charger is rated to output 5.0 V at a maximum current of 2.4 A. The charger is 85% efficient, meaning 85% of the power it draws from the wall outlet is delivered to the phone, with the rest dissipated as heat in the charger's internal components. Find the maximum power the charger can deliver to the phone, the total power it draws from the wall at maximum output, and the power wasted as heat.

Worked Solution: Maximum power delivered to the phone is output voltage times maximum output current: $P_{\text{out}}=VI=(5.0\text{V})(2.4\text{A})=12\text{W}$. Efficiency is defined as $\eta =P_{\text{out}}/P_{\text{in}}$, so rearrange to find input power drawn from the wall: $P_{\text{in}}=P_{\text{out}}/\eta =12\text{W}/0.85\approx 14.1\text{W}$. Power dissipated as heat is the difference between input and output power: $P_{\text{heat}}=P_{\text{in}}-P_{\text{out}}=14.1\text{W}-12\text{W}=2.1\text{W}$. In context, this small amount of wasted heat is consistent with modern energy-efficient portable electronics.

7. Quick Reference Cheatsheet

8. What's Next

This topic is the foundational energy framework for all circuit analysis in AP Physics 2. Immediately after mastering this topic, you will apply it alongside Kirchhoff's Junction Rule to analyze multi-loop circuits with multiple branches, then extend these rules to RC circuits, where capacitors are charged and discharged through resistors. Without a solid understanding of Ohm's Law, the loop rule, and power, you will not be able to derive or interpret the exponential charging/discharging equations for RC circuits, which are a major tested topic in Unit 4. This topic also underpins the analysis of real-world circuits, including household wiring and semiconductor devices, which are common context-based questions on the AP exam.

• Kirchhoff's Junction Rule
• RC Circuits
• Capacitance
• Household Circuit Safety