Compound Direct Current Circuits: Study Guide (Physics C: E&M 11.5)

Course note. This guide is for the calculus-based Physics C: Electricity and Magnetism course, Topic 11.5. It turns the loops and junctions of Topic 11.2 and Ohm’s law from Topic 11.3 into a method for solving circuits with several resistors. Unless a question says otherwise, treat every battery, wire and meter as ideal. Circuits with batteries of different emf connected in parallel are outside this course.

Series and parallel: it is about the paths

Whether elements are “in series” or “in parallel” depends on the paths charge can take, not on how the diagram looks.

• Series. Elements are in series if any charge that passes through one of them must go through all of them, with no other route available. Charge is conserved and does not pile up in a steady circuit, so the current is the same in every element in series.
• Parallel. Elements are in parallel if their ends are joined to the same two points, so charge arriving at one point can take any one of the paths to the other. The potential is the same all along an ideal wire, so every parallel path has the same potential difference across it.

Many elements are neither. In Figure 1, R₁ is not in series with R₃ alone, because charge leaving R₁ can go through R₂ instead. R₃ and R₄ are in series with each other, and that pair is in parallel with R₂.

Test for series: is there a junction between the two elements? If yes, they are not in series. Test for parallel: are both ends of each element connected (by wire only) to the same two junctions? If yes, they are in parallel.

Equivalent resistance

A group of resistors can be replaced by a single equivalent resistance R_eq that draws the same current from the same potential difference.

Series. The same current I passes through each resistor, and the potential differences add: ΔV = IR₁ + IR₂ + … = I(R₁ + R₂ + …). So

R_eq = R₁ + R₂ + R₃ + … (series)

Parallel. Each resistor has the same ΔV, and the currents add: I = ΔV/R₁ + ΔV/R₂ + … = ΔV(1/R₁ + 1/R₂ + …). So

1/R_eq = 1/R₁ + 1/R₂ + 1/R₃ + … (parallel)

Three consequences follow.

• Adding a path lowers R_eq. Each new parallel path gives charge another way through, so the total current for the same ΔV rises. The reciprocal sum gets bigger, so R_eq gets smaller.
• A parallel group has less resistance than its smallest member. For example, 4.0 Ω in parallel with 100 Ω gives 3.85 Ω, a little less than 4.0 Ω.
• Two resistors only: R_eq = R₁R₂/(R₁ + R₂), “product over sum”. For n identical resistors R in parallel, R_eq = R/n. Do not use “product over sum” for three at once.

Reducing a network step by step

1. Find the innermost group that is purely series or purely parallel, and replace it with its R_eq.
2. Redraw. A new series or parallel group usually appears.
3. Repeat until one resistor is left, then find the total current from I = ℰ/R_eq.
4. Work back out: use the total current and ΔV = IR to find the potential difference across each group, then the current in each branch.

Worked example 1: currents and potential differences in a network

Question. In Figure 1 the battery is ideal with ℰ = 24 V. (a) Find R_eq. (b) Find the current and potential difference for every resistor. (c) R₂ is removed, leaving a gap. Compare the current from the battery with your answer to (b).

(a) Equivalent resistance.

1. R₃ and R₄ are in series: 6.0 + 9.0 = 15 Ω.
2. That 15 Ω branch is in parallel with R₂ = 30 Ω: 1/R_p = 1/30 + 1/15 = 3/30, so R_p = 10 Ω. (Product over sum: 30 × 15 ÷ 45 = 10 Ω.)
3. R₁ is in series with the group: R_eq = 6.0 + 10 = 16 Ω.

(b) Work back out.

1. Total current: I = ℰ/R_eq = 24 V ÷ 16 Ω = 1.5 A. All of it passes through R₁.
2. ΔV₁ = (1.5 A)(6.0 Ω) = 9.0 V.
3. Across the parallel group: ΔV_PQ = (1.5 A)(10 Ω) = 15 V. Check: 9.0 V + 15 V = 24 V, the full emf.
4. R₂ has the full 15 V across it: I₂ = 15 ÷ 30 = 0.50 A.
5. The R₃–R₄ branch also has 15 V across it: I₃₄ = 15 ÷ 15 = 1.0 A. Then ΔV₃ = (1.0)(6.0) = 6.0 V and ΔV₄ = (1.0)(9.0) = 9.0 V.

Check. The branch currents add up to the total: 0.50 + 1.0 = 1.5 A. The branch with less resistance (15 Ω) carries the larger current, twice as much as the 30 Ω branch.

(c) R₂ removed. Now everything is in series: R_eq = 6.0 + 15 = 21 Ω, and I = 24 ÷ 21 = 1.14 A, less than 1.5 A. Removing a path raised the equivalent resistance and lowered the battery current. The potential difference across R₁ falls to 6.86 V, while the R₃–R₄ branch now has 17.1 V across it, more than before.

Interpretation. Changing one branch changes the potential difference across the others. In this circuit only the battery’s emf stays fixed.

Real batteries and real wires

Ideal wires have negligible resistance. This is a good model because copper leads have far less resistance than the other elements. But the model only works when the circuit contains other resistance. A bare wire across an ideal battery would give I = ℰ/0, which is meaningless. In a real circuit, the wire’s own small resistance and the battery’s internal resistance then set the current.

The emf ℰ of a battery is the potential difference it would supply if it were ideal. You can measure it as the potential difference across the terminals when there is no current in the battery.

A real battery has internal resistance r. Model it as an ideal emf in series with a resistor r, inside the battery’s casing (Figure 2). When there is a current I, there is a potential drop Ir inside, so the potential difference you can measure across the terminals is

ΔV_terminal = ℰ − Ir

The more current the battery delivers, the lower its terminal voltage. The current in a single loop is I = ℰ/(R + r), where R is the external resistance.

Worked example 2: finding internal resistance

Question. With nothing connected, an ideal voltmeter across a battery reads 6.0 V. A 2.5 Ω resistor is then connected across the battery, and the voltmeter reading falls to 5.0 V. (a) Find the current and the internal resistance. (b) Find the terminal voltage when the load is replaced by a 1.0 Ω resistor. (c) A student connects a thick wire of resistance 0.010 Ω across the terminals. Estimate the current, and explain why “ideal wire, ideal battery” fails here.

(a) With no current, the reading is the emf: ℰ = 6.0 V.

1. The voltmeter now reads the potential difference across the 2.5 Ω resistor: I = 5.0 V ÷ 2.5 Ω = 2.0 A.
2. The “missing” 1.0 V is dropped inside the battery: Ir = 6.0 − 5.0 = 1.0 V, so r = 1.0 V ÷ 2.0 A = 0.50 Ω.

(b) I = ℰ/(R + r) = 6.0 ÷ (1.0 + 0.50) = 4.0 A, and ΔV_terminal = 6.0 − (4.0)(0.50) = 4.0 V (the same as IR = 4.0 × 1.0). A smaller load draws more current, so the terminal voltage drops further.

(c) I = 6.0 ÷ (0.010 + 0.50) = 11.8 A, about 12 A. Almost all of the emf is dropped across the internal resistance; the terminal voltage is only about 0.12 V. With an ideal battery and an ideal wire the model would predict an infinite current. Neglecting the wire’s resistance is only allowed when something else in the circuit has resistance; here, that “something else” is the battery’s own r.

Check. In (a) the 2.5 Ω load receives (5.0)(2.0) = 10 W while (2.0)²(0.50) = 2.0 W is dissipated inside the battery, so the battery transfers 12 W = ℰI in total, as energy conservation requires (Topic 11.4).

Measuring current and potential difference

Ammeters measure the current at one point in a circuit. The current must pass through the meter, so an ammeter is connected in series with the element whose current you want. An ideal ammeter has zero resistance, so adding it does not change the current.

Voltmeters measure the potential difference between two points. A voltmeter is connected in parallel with the element, one lead on each side. An ideal voltmeter has infinite resistance, so no charge flows through it and the rest of the circuit is unchanged.

Real meters change the circuit. A real ammeter adds a small resistance in series, so the current falls a little. A real voltmeter provides a parallel path, which lowers the resistance of whatever it is connected across.

Example of meter loading. Two 10 kΩ resistors are in series with an ideal 12 V battery. Each should have 6.0 V across it. Connect a voltmeter of resistance 10 kΩ across one resistor. That resistor and the meter now form a 5.0 kΩ parallel pair, so the meter reads 12 × 5.0 ÷ (10 + 5.0) = 4.0 V, not 6.0 V. A 1.0 MΩ voltmeter would read 5.97 V. A good voltmeter needs a resistance much larger than the resistance it is connected across; a good ammeter needs a resistance much smaller than the resistance of the loop it is in.

Connecting a meter the wrong way round in this sense causes trouble. An ammeter placed in parallel provides a near-zero-resistance path: it shorts the element and may carry a dangerously large current. A voltmeter placed in series adds a huge resistance, so the current almost stops.

Brightness of bulbs in compound circuits

Topic 11.4 showed that bulb brightness rises with power, P = I²R = (ΔV)²/R. To predict brightness changes, follow this order: change in R_eq → change in total current → change in ΔV across each part → change in power of each bulb. With an ideal battery, adding a bulb in parallel directly across the battery leaves the other bulbs on that battery unchanged, because each still has ℰ across it. With a real battery, the extra current raises Ir, the terminal voltage falls, and every bulb dims slightly.

Common misconceptions

• “Side by side on the page means parallel.” Only shared end points make elements parallel. Trace the junctions.
• “Current is used up by the first resistor.” In series, the current is the same in every element. Energy is transferred; charge is not lost.
• “Adding a resistor always increases resistance.” Only in series. Adding one in parallel adds a path and lowers R_eq.
• “The parallel resistance is the average.” It is always less than the smallest resistance in the group.
• Forgetting the last step. 1/R_eq = 0.50 Ω⁻¹ means R_eq = 2.0 Ω, not 0.50 Ω.
• “A battery supplies a fixed current.” An ideal battery fixes the potential difference; the current depends on the circuit.
• “The terminal voltage is always the emf.” Only when no current is drawn from a real battery.
• “Ignore wire resistance in every case.” Not when the wire is the only thing connected across the battery.
• Ammeter in parallel or voltmeter in series. Ammeters go in the path of the current; voltmeters go across the element.

Where this leads

Series and parallel rules work only when a network can be broken into such groups. For other circuits, and for circuits with more than one battery, you need Kirchhoff’s rules: the loop rule (energy conservation) in the next topic and the junction rule (charge conservation) in Topic 11.7. Topic 11.8 then applies the same tools to circuits that contain capacitors. Look back at Topic 11.4, Electric Power for brightness and energy. Practise now with the practice questions, then use the revision notes and the topic checklist. The next study guide is Topic 11.6, Kirchhoff’s Loop Rule.