11.5 Compound Direct Current (DC) Circuits

Compound DC circuits combine resistors in series and parallel, and you simplify them by finding an equivalent resistance. Real batteries also have internal resistance, so the terminal voltage drops below the emf when current flows.

Why This Matters for the AP Physics 2 Exam

This topic builds the circuit-analysis thinking you will reuse across all of Unit 11. Once you can reduce a mixed network to a single equivalent resistance, you can find current, voltage, and power anywhere in the circuit, which is exactly what circuit problems ask you to do. Internal resistance connects circuit math to the conservation of energy, and meter placement connects to experimental design and data collection, both of which show up in free-response questions that ask you to explain or justify your reasoning. The exam also expects you to use terms like current, potential difference, and resistance precisely when you defend a claim, so getting the vocabulary right here pays off later.

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Key Takeaways

• In series, the current is the same through every element and the voltage divides; in parallel, the voltage is the same across each path and the current divides.
• Series equivalent resistance adds: $R_{\text{eq},s}=\sum_i R_i$. Adding resistors in series increases total resistance.
• Parallel equivalent resistance comes from $\frac{1}{R_{\text{eq},p}}=\sum_i \frac{1}{R_i}$. Adding parallel paths decreases equivalent resistance.
• A real battery acts like an ideal battery with emf $\varepsilon$ in series with internal resistance $r$, giving terminal voltage $\Delta V_{\text{terminal}}=\varepsilon-Ir$.
• An ideal ammeter (zero resistance) connects in series; an ideal voltmeter (infinite resistance) connects in parallel. Nonideal meters change the circuit they measure.
• Wire resistance can be neglected only when the circuit has other elements with much larger resistance.

Equivalent Resistance of Multiple Resistors in a Circuit

Circuit elements can be connected in series, in parallel, or in combinations of both. How they are wired changes how charge moves and how voltage is shared.

A series connection is one where any charge passing through one element must pass through all of them, with no other path available.

• The current in each element in series is the same.
• The voltage divides among the resistors in proportion to their resistance values.
• Each resistor produces a voltage drop as current passes through it.

A parallel connection gives charges more than one path to follow.

• The potential difference across each parallel path is the same.
• The current divides among the paths, with more current taking the lower-resistance path.
• The total current equals the sum of the currents through each path.

To make analysis easier, you can replace a group of resistors with a single resistor that has the same effect, called the equivalent resistance ($R_{\text{eq}}$).

For resistors in series, the equivalent resistance is the sum of the individual resistances: $R_{\mathrm{eq}, s}=\sum_{i} R_{i} \quad \text{or} \quad R_{\mathrm{eq}, s}= R_1 + R_2 + R_3 + \dots$

For resistors in parallel, the inverse of the equivalent resistance is the sum of the inverses of the individual resistances: $\frac{1}{R_{\text {eq }, p}}=\sum_{i} \frac{1}{R_{i}} \quad \text{or} \quad \frac{1}{R_{\text {eq }, p}}= \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots$

Adding resistors in series increases total resistance. Adding resistors in parallel increases the number of paths for charge, so the equivalent resistance decreases. A useful check: the parallel equivalent resistance is always smaller than the smallest resistor in the group.

Circuits with Resistive Wires and Batteries with Internal Resistance

In ideal models, batteries have negligible internal resistance and wires have negligible resistance. Real circuits include nonideal parts that change the behavior.

Wire resistance is usually much smaller than other circuit elements, so it can normally be neglected. But the resistance of wires may be neglected only if the circuit also contains other elements that do have resistance. If the wires are the only thing with significant resistance, you cannot ignore them.

A real battery has internal resistance from its chemistry and structure. You can model this as a resistor with resistance $r$ in series with an ideal battery and the rest of the circuit. This internal resistance:

• Causes the terminal voltage to be less than the emf when current flows.
• Converts some electrical energy to thermal energy inside the battery.

The emf ($\varepsilon$) is the potential difference the battery would supply if it were ideal, which equals the terminal voltage measured when there is no current in the battery. When current flows, the terminal voltage is: $\Delta V_{\text {terminal }}=\varepsilon-I r$

As current increases, the terminal voltage decreases because more potential is lost across the internal resistance. When no current flows, the terminal voltage equals the emf.

Measuring Current and Potential Difference in a Circuit

How you connect a meter matters, because a poorly chosen meter can change the very thing you are trying to measure.

Ammeters measure current at a specific point and must be connected in series with the element.

• An ideal ammeter has zero resistance so it does not change the current it measures.
• A nonideal ammeter has some resistance, which reduces the current in that branch.

Voltmeters measure potential difference between two points and must be connected in parallel with the element.

• An ideal voltmeter has infinite resistance so no charge flows through it.
• A nonideal voltmeter has high but finite resistance, so it draws a little current and changes the voltage it reads.

Because nonideal meters have nonideal resistance, the act of measuring can change the circuit's behavior. That is why real meters are designed to minimize their effect.

How to Use This on the AP Physics 2 Exam

Problem Solving

• Simplify step by step. Find purely series or purely parallel groups first, replace each with its $R_{\text{eq}}$, and repeat until one resistor remains. Then work backward to find current and voltage in each part.
• After finding total current from the source, use the rule that series elements share current and parallel branches share voltage to fill in the rest.
• Sanity-check parallel results: the equivalent resistance must be smaller than the smallest resistor in that group.

Free Response

• For internal resistance questions, draw $r$ as a resistor in series with the ideal battery, then apply $\Delta V_{\text{terminal}}=\varepsilon-Ir$.
• When asked to explain, use precise terms: current, potential difference, resistance, and emf each mean something specific, and mixing them up can cost points.

Common Trap

• Be ready to argue in words how a nonideal ammeter or voltmeter would shift a measurement. An ammeter with resistance lowers branch current; a voltmeter with finite resistance draws current and lowers the reading.

Practice Problem 1: Equivalent Resistance

Solution:

a) For resistors in series, add the individual resistances: $R_{eq} = R_1 + R_2 + R_3 = 4\Omega + 6\Omega + 12\Omega = 22\Omega$

b) For resistors in parallel, use the formula: $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{4\Omega} + \frac{1}{6\Omega} + \frac{1}{12\Omega} = \frac{6 + 4 + 2}{24\Omega} = \frac{12}{24\Omega} = \frac{1}{2\Omega}$

Therefore, $R_{eq} = 2\Omega$

c) First, calculate the equivalent resistance of the parallel combination of R₂ and R₃: $\frac{1}{R_{parallel}} = \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{6\Omega} + \frac{1}{12\Omega} = \frac{2 + 1}{12\Omega} = \frac{3}{12\Omega} = \frac{1}{4\Omega}$

So $R_{parallel} = 4\Omega$

Now add this to R₁ in series: $R_{eq} = R_1 + R_{parallel} = 4\Omega + 4\Omega = 8\Omega$

Practice Problem 2: Battery with Internal Resistance

Solution:

a) The total resistance is the sum of the external resistance and the internal resistance: $R_{total} = R_{external} + r = 5.5\Omega + 0.5\Omega = 6\Omega$

Using Ohm's law, the current is: $I = \frac{\mathcal{E}}{R_{total}} = \frac{12V}{6\Omega} = 2A$

b) The terminal voltage is: $\Delta V_{terminal} = \mathcal{E} - Ir = 12V - (2A)(0.5\Omega) = 12V - 1V = 11V$

c) The power dissipated in the internal resistance is: $P = I^2r = (2A)^2(0.5\Omega) = 4 \times 0.5W = 2W$

Common Misconceptions

• Adding resistors does not always increase resistance. Adding resistors in parallel lowers the equivalent resistance because it adds more paths for charge.
• Current is not used up as it flows through a circuit. In a series path the current is the same everywhere; what changes is the potential as charge moves through elements.
• emf is not the same as terminal voltage. emf is the open-circuit value with no current; terminal voltage drops to $\varepsilon - Ir$ once current flows.
• Batteries do not store charge. A battery maintains a potential difference that drives charge around the circuit.
• Wire resistance is not always safe to ignore. You can neglect it only when other elements have much larger resistance.
• An ammeter is not connected in parallel and a voltmeter is not connected in series. Ammeters go in series, voltmeters go in parallel, and reversing this disturbs or even damages the measurement.

Related AP Physics 2 Guides

• 11.2 Simple Circuits
• 11.3 Resistance, Resistivity, and Ohm's Law
• 11.8 Resistor-Capacitor (RC) Circuits
• 11.1 Electric Current
• 11.4 Electric Power
• 11.6 Kirchhoff's Loop Rule