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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: Req=R1+R2+⋯R_{eq} = R_1 + R_2 + \cdots.

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: 1Req=1R1+1R2+⋯\dfrac{1}{R_{eq}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \cdots.

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, Req=R1R2R1+R2R_{eq} = \dfrac{R_1R_2}{R_1 + R_2}, and n identical resistors R in parallel give Rn\dfrac{R}{n}.

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 E\mathcal{E} 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 ΔV=E−Ir\Delta V = \mathcal{E} - Ir.

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 I=ERext+rI = \dfrac{\mathcal{E}}{R_{ext} + r}, where RextR_{ext} 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.

ConnectionSame for each elementAdds upEquivalent resistance
Seriescurrentpotential differencessum of R
Parallelpotential differencecurrentssum of 1/R, then flip

Worked examples

Try each one yourself first, then open the solution.

  1. 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 solution
    1. Step 1: Parallel pair: Rp=(6.0)(3.0)6.0+3.0=2.0 ΩR_p = \dfrac{(6.0)(3.0)}{6.0 + 3.0} = 2.0\ \Omega.
    2. Step 2: Everything outside the battery: 3.0 + 2.0 = 5.0 Ω. Include the internal resistance: I=ERext+r=125.0+1.0=2.0 AI = \dfrac{\mathcal{E}}{R_{ext} + r} = \dfrac{12}{5.0 + 1.0} = 2.0\text{ A}.
    3. Step 3: Terminal voltage: ΔV=E−Ir=12−(2.0)(1.0)=10 V\Delta V = \mathcal{E} - Ir = 12 - (2.0)(1.0) = 10\text{ V}.
    4. 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

  2. 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 solution
    1. Step 1: Without the meter, each resistor would have 5.0 V across it.
    2. Step 2: The meter is in parallel with one resistor: Rp=(10)(10)10+10=5.0 kΩR_p = \dfrac{(10)(10)}{10 + 10} = 5.0\text{ k}\Omega.
    3. 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.
    4. 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 ΔV\Delta V against I; the line's vertical intercept is E\mathcal{E} and its slope is −r-r.

Connected topics

Videos

  • Resistor Series and Parallel Circuits

    Flipping PhysicsWatch on YouTube (opens in a new tab)

  • AP Physics C - Circuit Analysis

    Dan Fullerton (APlusPhysics)Watch 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)

  • Internal Resistance of a Battery, EMF, Cell Terminal Voltage, Physics Problems

    The Organic Chemistry TutorWatch on YouTube (opens in a new tab)

  • Ammeter and Voltmeter - Where Do They Go?

    Flipping PhysicsWatch on YouTube (opens in a new tab)

Check yourself

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

Question 1 of 5Calculator allowed

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

Question 2 of 5Calculator allowed

What is the current in R₃?

Question 3 of 5Calculator allowed

What is the potential difference across R₁?

Question 4 of 5Calculator allowed

Resistor R₃ is removed, leaving a gap in its branch. What happens to the current in R₂?

Question 5 of 5Calculator allowed

Where should an ideal ammeter be connected to measure the current in R₂ alone?

0 of 5 answered