Learning outcomes
- Apply current rules in a series circuit.
- Apply potential-difference rules.
- Calculate total series resistance.
- Predict effects of adding components.
- Explain energy transfer around a loop.
11.1 One current path
Components are in series when they lie on one unbranched path. Charge cannot accumulate indefinitely at a component, so in a steady circuit the same current passes through every series component and the source.
If an ammeter is moved to different positions in a true series loop, it should give the same reading within experimental uncertainty. A break anywhere stops current everywhere.
11.2 Potential differences add
The source supplies energy per coulomb. Each series component transfers part of that energy, so the sum of p.d.s across the components equals the total e.m.f. of the sources, allowing for internal losses.
For a 9.0 V supply with two series components having 3.0 V and 6.0 V, the energy transferred per coulomb is 3.0 J and 6.0 J respectively, totalling 9.0 J.
11.3 Series resistance
Total resistance of series resistors is Rtotal = R1 + R2 + … because the same current flows and the p.d.s add. Adding a resistor in series always increases total resistance.
For a fixed supply voltage, increased total resistance reduces current. This may make lamps dimmer, reduce motor speed or decrease heating power.
11.4 Voltage sharing
In a series circuit, p.d. divides in proportion to resistance if the resistors are ohmic and current is common. A larger resistance has a larger voltage drop because V = IR.
For two resistors, V1/V2 = R1/R2. This is the basis of a potential divider. The relation is exact for ideal resistors under stable conditions.
11.5 Cells in series
Cells connected in the same direction have e.m.f.s that add. A greater source e.m.f. gives a greater current through the same total resistance, assuming source internal resistance is negligible.
If cells oppose, subtract their e.m.f.s according to direction. Never assume “number of cells × voltage” without inspecting polarity.
11.6 Lamps in series
Identical lamps in series share the supply voltage and each receives less power than a single lamp on the same source. They are usually dimmer. If one lamp fails open circuit, all go out because the path is broken.
Non-identical lamps do not necessarily share voltage equally because their resistances and temperatures differ. Use circuit data rather than assuming equal division.
11.7 Fault finding
Zero current everywhere suggests an open switch, broken wire, failed component or exhausted source. A voltmeter may show the full supply p.d. across a break because almost no current flows and the open section separates source potentials.
Continuity tests must be performed with power removed. A systematic approach checks source, switch, connections and components rather than replacing parts randomly.
Worked examples
Total resistance
Resistors of 4.0 Ω, 7.0 Ω and 9.0 Ω are in series. Rtotal = 20 Ω.
Circuit current
A 12 V supply is connected to 20 Ω. I = V/R = 12/20 = 0.60 A.
Voltage division
A 2 Ω and 4 Ω resistor are in series across 12 V. Current = 12/6 = 2 A; p.d.s are 4 V and 8 V.
Practical focus
Investigation
Connect two resistors in series. Measure current before, between and after them. Measure p.d. across each resistor and across both. Compare total resistance calculated from individual values with Vtotal/Itotal.
Examination guidance
- Series current is the same at all points.
- Series p.d.s add to source e.m.f.
- Series resistances add directly.
- Do not assume equal voltage unless resistances are equal.
Check your understanding
- Find total resistance of 3 Ω and 12 Ω in series.
- A 15 V supply drives 0.50 A. Find total resistance.
- Why do all series lamps go out if one breaks?
Answers
- 15 Ω.
- R = 15/0.50 = 30 Ω.
- The only current path is broken.
