Unit 11 · Topic 11.5

11.5 Compound Direct Current (DC) Circuits

Real circuits combine elements in series and in parallel. Series elements carry the same current and their resistances add; parallel elements share the same potential difference and lower the total resistance. This topic also covers a battery's internal resistance and how to connect ammeters and voltmeters.

Key terms

series
parallel
equivalent resistance
internal resistance
terminal voltage
ammeter and voltmeter

Series and parallel

In series, elements sit one after another on a single path, so every bit of charge must pass through all of them. They carry the same current, and their potential differences add up to the total.

In parallel, elements are on separate branches that split and rejoin at the same two points. Charge goes through one branch or another. Every branch has the same potential difference, and the branch currents add up to the total current.

Equivalent resistance

A group of resistors can be replaced by a single equivalent resistance $R_{\text{eq}}$ that draws the same current from the battery.

$R_{\text{eq}} = R_1 + R_2 + \cdots \quad \text{(series)}$

$\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots \quad \text{(parallel)}$

Adding resistors in series raises $R_{\text{eq}}$ . Adding a resistor in parallel always lowers $R_{\text{eq}}$ , because it opens an extra path, like adding a lane to a highway. $R_{\text{eq}}$ of a parallel group is always smaller than its smallest resistor.

To solve a combination circuit: collapse parallel groups and series chains step by step until one $R_{\text{eq}}$ is left, find the total current from I = $\varepsilon/R_{\text{eq}}$ , then work backward, using the shared current in series parts and the shared potential difference in parallel parts.

Internal resistance

An ideal battery has no internal resistance, and ideal wires have no resistance. Real wires do have a little resistance, but it's usually so small next to the other elements that you can ignore it. That only works when something else in the loop has resistance: a bare wire connected straight across a battery is limited only by the wire's own resistance, so the current can be very large.

A real battery behaves like an ideal battery of emf ε in series with a small internal resistance r. With current I, some potential is lost inside the battery, so the potential difference across its terminals is:

$\Delta V_{\text{terminal}} = \varepsilon - Ir$

With no current, the terminal voltage equals the emf. The more current you draw, the lower the terminal voltage, which is why car headlights dim while the starter motor runs.

Meters

Meters are built to measure a circuit without changing it, which only works if they're connected the right way.

Ammeter: measures current, so it goes in series, in the path of the current. An ideal ammeter has zero resistance so it doesn't change the current.
Voltmeter: measures potential difference, so it goes in parallel, across the element. An ideal voltmeter has infinite resistance so no charge flows through it.
Real meters change what they measure: an ammeter's small resistance slightly lowers the current, and a voltmeter's finite resistance draws a little current. You only describe these effects, not calculate them.
Circuits with batteries of different potential differences connected in parallel are not on the exam.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
A combination circuit
A 12 V ideal battery is connected to a 4.0 Ω resistor in series with a parallel pair of 6.0 Ω and 12 Ω resistors. Find the current through each resistor and the potential difference across each.
Show the solutionHide the solution
1. Step 1: Parallel pair: $\dfrac{1}{R_p} = \dfrac{1}{6.0} + \dfrac{1}{12} = \dfrac{3}{12}$ , so $R_p$ = 4.0 Ω.
2. Step 2: Total: $R_{\text{eq}}$ = 4.0 + 4.0 = 8.0 Ω. Total current: I = 12/8.0 = 1.5 A, all through the 4.0 Ω resistor.
3. Step 3: 4.0 Ω resistor: ΔV = (1.5)(4.0) = 6.0 V. That leaves 12 − 6.0 = 6.0 V across the parallel pair.
4. Step 4: 6.0 Ω: I = 6.0/6.0 = 1.0 A. 12 Ω: I = 6.0/12 = 0.50 A. Check: 1.0 + 0.50 = 1.5 A.
Answer: 4.0 Ω: 1.5 A, 6.0 V. 6.0 Ω: 1.0 A, 6.0 V. 12 Ω: 0.50 A, 6.0 V.
Example 2Calculator allowed
A battery with internal resistance
A battery with emf 9.0 V and internal resistance 0.50 Ω is connected to a 4.0 Ω resistor. Find the current and the terminal voltage.
Show the solutionHide the solution
1. Step 1: The internal resistance is in series with the external one: total = 4.5 Ω.
2. Step 2: I = $\varepsilon/R_{\text{total}}$ = 9.0/4.5 = 2.0 A.
3. Step 3: Terminal voltage: ε − Ir = 9.0 − (2.0)(0.50) = 8.0 V. Check: across the 4.0 Ω resistor, (2.0)(4.0) = 8.0 V.
Answer: 2.0 A; 8.0 V
Example 3
Unscrewing a bulb (classic trap)
Identical bulbs B and C are in parallel, and that pair is in series with an identical bulb A and an ideal battery. Bulb C is unscrewed. What happens to the brightness of A and of B?
Show the solutionHide the solution
1. Step 1: Before: $R_{\text{eq}}$ = R + R/2 = 1.5R. The current through A is ε/(1.5R) ≈ 0.67ε/R, and B gets half of that, about 0.33ε/R.
2. Step 2: After C is removed, B is simply in series with A: $R_{\text{eq}}$ = 2R, and both carry ε/(2R) = 0.50ε/R.
3. Step 3: A's current falls from 0.67 to 0.50 (in units of ε/R), so A gets dimmer. B's current rises from 0.33 to 0.50, so B gets brighter.
4. Step 4: The trap is thinking removing a bulb makes everything dimmer, or leaves the others unchanged. Removing a parallel branch raises the total resistance, which shifts the potential difference.
Answer: A gets dimmer; B gets brighter

Common mistakes

Thinking adding a resistor always increases total resistance. Adding one in parallel lowers it.
Connecting an ammeter in parallel (which would short the element) or a voltmeter in series (which would block the current).
Forgetting the internal resistance when a question gives one, or using the emf as the terminal voltage while current flows.
Using the battery's full potential difference across a resistor that shares a series path with others.

On the exam

Expect multi-step circuit problems: find $R_{\text{eq}}$ , the total current, then each element's current and potential difference, often in a table.
Questions on changes (adding, removing or shorting an element) are common. Explain what happens to $R_{\text{eq}}$ , then the total current, then each element.
Lab questions may ask where to place meters to measure a battery's emf and internal resistance; a graph of terminal voltage against current has slope −r and intercept ε.

Connected topics

Videos

AP Physics 2 - Unit 11 - Lesson 8 - Series and Parallel Resistors
Allen Tsao The STEM CoachWatch on YouTube (opens in a new tab)
Resistor Series and Parallel Circuits
Flipping PhysicsWatch on YouTube (opens in a new tab)
Series and parallel circuits | Physics | Khan Academy
Khan AcademyWatch on YouTube (opens in a new tab)
DC Resistors & Batteries: Crash Course Physics #29
CrashCourseWatch on YouTube (opens in a new tab)
EMF, Internal Resistance, and Terminal Voltage of Batteries Explained | Doc Physics
Doc SchusterWatch on YouTube (opens in a new tab)
AP Physics 2 - Unit 11 - Lesson 7 - Circuit Lab Equipment
Allen Tsao The STEM CoachWatch on YouTube (opens in a new tab)

Check yourself

4 questions on 11.5 Compound Direct Current (DC) Circuits. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

Two identical bulbs are in parallel across an ideal battery. A third identical bulb is added in parallel with them. What happens to the brightness of the original bulbs and to the current from the battery?

Question 2 of 4Calculator allowed

A battery with an emf of 9.0 V and an internal resistance of 0.50 Ω is connected to a 4.0 Ω resistor. What is the potential difference across the battery's terminals?

Current (A)	Terminal potential difference (V)
0.20	1.46
0.40	1.42
0.60	1.38
0.80	1.34

Experimental data

Question 3 of 4Calculator allowed

A student connects a battery to a variable resistor and records the terminal potential difference at several currents. The data fall on a straight line. What is the battery's internal resistance?

Question 4 of 4Calculator allowed

What is the emf of the battery?

0 of 4 answered