AP® Physics C: Electricity and Magnetism review sheet from Aim for Five (aimforfive.com/physics-c-em/units/11/11-5)
Unit 11 · Topic 11.5
11.5 Compound Direct Current Circuits
Most circuits mix series and parallel connections. This topic shows how to replace groups of resistors with one equivalent resistance, how a real battery's internal resistance lowers its terminal voltage, and how to connect ammeters and voltmeters without changing what you're measuring.
Key terms
- series connection
- parallel connection
- equivalent resistance
- internal resistance
- terminal voltage
- ammeter and voltmeter
Series connections
Elements are in series when they sit one after another on a single path with no branches between them. The same current passes through each. Their potential differences add up to the total across the group, and their resistances add: .
Adding a resistor in series always raises the equivalent resistance. The equivalent resistance is bigger than any single resistor in the group.
Parallel connections
Elements are in parallel when both their ends connect to the same two points, so each sits on its own branch. They all have the same potential difference across them. The currents in the branches add up to the current entering the group, and the reciprocals add: .
Adding a branch in parallel always lowers the equivalent resistance, because it gives the charge another path. The equivalent resistance is smaller than the smallest branch. For just two resistors, , and n identical resistors R in parallel give .
To simplify a compound circuit, find the innermost group that's purely series or purely parallel, replace it with its equivalent, redraw, and repeat. Then work back out to find each current and potential difference.
Real batteries and terminal voltage
A real battery behaves like an ideal emf in series with a small internal resistance r. When current I flows out of it, some potential is lost inside, so the potential difference you measure across its terminals, the terminal voltage, is .
With no current, the terminal voltage equals the emf. The more current you draw, the lower it gets. That's why car headlights dim while the starter motor draws a big current. The current from the battery is , where is the equivalent resistance of everything outside the battery.
Wires are usually treated as having no resistance, because a good conducting wire's resistance is tiny next to the other elements. That only works when the circuit has other elements with resistance. A bare wire straight across a battery is a short circuit (11.2), and then the wire's own resistance and the battery's internal resistance are what limit the current.
Ammeters and voltmeters
An ammeter measures current, so the current must pass through it: connect it in series. An ideal ammeter has zero resistance, so it doesn't change the current.
A voltmeter measures the potential difference between two points: connect it in parallel, across the element. An ideal voltmeter has infinite resistance, so it draws no current.
Real meters aren't ideal. An ammeter's small resistance slightly lowers the current it measures. A voltmeter with a resistance comparable to the element it's across draws noticeable current and lowers the reading. A good meter has a resistance far from the element's: much smaller for an ammeter, much larger for a voltmeter.
| Connection | Same for each element | Adds up | Equivalent resistance |
|---|---|---|---|
| Series | current | potential differences | sum of R |
| Parallel | potential difference | currents | sum of 1/R, then flip |
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
A series-parallel circuit with internal resistance
A battery with emf 12 V and internal resistance 1.0 Ω is connected to a 3.0 Ω resistor in series with a parallel pair of 6.0 Ω and 3.0 Ω resistors. Find the current from the battery, the terminal voltage, and the current in each parallel branch.
Show the solutionHide the solution
- Step 1: Parallel pair: .
- Step 2: Everything outside the battery: 3.0 + 2.0 = 5.0 Ω. Include the internal resistance: .
- Step 3: Terminal voltage: .
- Step 4: Potential difference across the parallel pair: (2.0 A)(2.0 Ω) = 4.0 V. Branch currents: 4.0/6.0 ≈ 0.67 A and 4.0/3.0 ≈ 1.33 A, which add back to 2.0 A.
Answer: 2.0 A from the battery, 10 V terminal voltage, 0.67 A in the 6.0 Ω branch and 1.33 A in the 3.0 Ω branch
- Example 2Calculator allowed
A voltmeter that isn't ideal (classic trap)
Two 10 kΩ resistors are in series across an ideal 10 V battery. A voltmeter with an internal resistance of 10 kΩ is connected across one of the resistors. What does it read?
Show the solutionHide the solution
- Step 1: Without the meter, each resistor would have 5.0 V across it.
- Step 2: The meter is in parallel with one resistor: .
- Step 3: The circuit is now 10 kΩ in series with 5.0 kΩ, so the current is 10 V / 15 kΩ and the meter's section gets 5.0/15 of the 10 V, which is about 3.3 V.
- Step 4: The trap is answering 5.0 V. A voltmeter whose resistance is close to the element's changes the circuit it's measuring.
Answer: About 3.3 V instead of the 5.0 V an ideal meter would show
Common mistakes
- Calling two resistors series just because they're next to each other on the page. Series means nothing branches off between them; parallel means both ends share the same two points.
- Forgetting to flip the sum of reciprocals at the end of a parallel calculation.
- Leaving out the internal resistance when finding the current, or calling the emf the terminal voltage when current flows.
- Connecting an ammeter in parallel. With almost no resistance, it would short out the element.
On the exam
- Free-response questions often ask what happens to a meter reading or a bulb when a branch is added or a switch opens. Reason from the change in equivalent resistance to the change in total current, then to each part.
- Internal resistance often appears in a lab: you're given terminal voltage at different currents. Graph against I; the line's vertical intercept is and its slope is .
Connected topics
Videos
Check yourself
5 questions on 11.5 Compound Direct Current Circuits. Pick an answer to see if you got it, and why.
A 6.0 Ω resistor is connected in series with a parallel combination of a 12 Ω resistor and a 4.0 Ω resistor. What is the equivalent resistance of the network?
An ideal 24 V battery is connected to resistor R₁ = 4.0 Ω. R₁ is in series with a parallel combination of R₂ = 6.0 Ω and R₃ = 12 Ω.
Described circuit
What is the current in R₃?
What is the potential difference across R₁?
Resistor R₃ is removed, leaving a gap in its branch. What happens to the current in R₂?
Where should an ideal ammeter be connected to measure the current in R₂ alone?
0 of 5 answered