Know that the current at every point in a series circuit is the same

4.3.2 Series and Parallel Circuits

Learning Objective

Students will be able to:

• State that the current is the same at every point in a series circuit.
• State that the potential difference (p.d.) across every branch of a parallel circuit is the same.
• Calculate the combined e.m.f. of cells in series and the combined resistance of resistors in series or parallel.
• Explain why lamps are normally connected in parallel.
• Apply Kirchhoff’s current (junction) rule where required.

Key Concepts

1. Series Circuits

• Definition: All components are joined end‑to‑end, giving a single continuous path for charge flow.
• Current: The same current I flows through every component because charge cannot accumulate in a steady‑state circuit.
• Resistance: R_total = R₁ + R₂ + R₃ + …
• Voltage division (p.d. across each component): The source voltage is shared in proportion to the resistances:
  Vₙ = I Rₙ and V_total = V₁ + V₂ + V₃ + …
• Series cells (combined e.m.f.): When ideal cells are connected end‑to‑end,
  E_total = E₁ + E₂ + … (internal resistances add in the same way as ordinary resistors).

2. Parallel Circuits

• Definition: Two or more independent branches join the same two nodes.
• Voltage: Every branch experiences the same potential difference as the source,
  V_branch = V_source.
• Resistance:
  \displaystyle\frac{1}{R_{\text{total}}}= \frac{1}{R₁}+ \frac{1}{R₂}+ \frac{1}{R₃}+ …
• Current division (junction rule): The total current leaving the source equals the sum of the branch currents,
  I_{\text{total}} = I₁ + I₂ + I₃ + ….
• Ohm’s law for each branch: Iₙ = V_{\text{source}} / Rₙ.

3. Why the Current Is Uniform in a Series Circuit

In a steady‑state circuit charge cannot build up at any point. If more charge entered a segment than left it, an increasing electric field would develop, opposing further flow. Because a series circuit has no branching points, the same amount of charge that leaves the source per second must pass through every component, giving a single, uniform current I throughout.

4. Why the Voltage Is Uniform Across Parallel Branches

All branches connect the same two nodes. The electric potential difference between those nodes is fixed by the source, so each branch experiences the same p.d. Even though the resistances differ, the voltage across each branch remains equal to the source voltage.

Comparison: Series vs Parallel

Worked Examples

Example 1 – Series Resistors

Problem: A 12 V battery supplies three resistors in series: R₁ = 2 Ω, R₂ = 3 Ω, R₃ = 5 Ω. Find the current through the circuit and the voltage across each resistor.

1. Total resistance:
   R_{\text{total}} = 2 Ω + 3 Ω + 5 Ω = 10 Ω
2. Uniform current (Ohm’s law for the whole circuit):
   I = V_{\text{total}} / R_{\text{total}} = 12 V / 10 Ω = 1.2 A
3. Voltage drop on each resistor (same I):
   V₁ = I R₁ = 1.2 A × 2 Ω = 2.4 V
   V₂ = I R₂ = 1.2 A × 3 Ω = 3.6 V
   V₃ = I R₃ = 1.2 A × 5 Ω = 6.0 V
4. Check: 2.4 V + 3.6 V + 6.0 V = 12 V (matches the source).

Example 2 – Parallel Resistors

Problem: A 9 V battery feeds three resistors in parallel: R₁ = 3 Ω, R₂ = 6 Ω, R₃ = 9 Ω. Determine (a) the total resistance, (b) the total current drawn, and (c) the current through each resistor.

1. Total resistance:
   \[ \frac{1}{R_{\text{total}}}= \frac{1}{3}+\frac{1}{6}+\frac{1}{9}= \frac{6+3+2}{18}= \frac{11}{18} \]
   R_{\text{total}} = \dfrac{18}{11}\;Ω \approx 1.64 Ω
2. Total current:
   I_{\text{total}} = V_{\text{source}} / R_{\text{total}} = 9 V / 1.64 Ω \approx 5.5 A
3. Branch currents (same voltage 9 V):
   I₁ = 9 V / 3 Ω = 3.0 A
   I₂ = 9 V / 6 Ω = 1.5 A
   I₃ = 9 V / 9 Ω = 1.0 A
   Verify: 3.0 A + 1.5 A + 1.0 A = 5.5 A = I_{\text{total}}.

Example 3 – Series Cells

Problem: Two identical 1.5 V dry cells are connected in series with a 3 Ω resistor. Find the current supplied by the cells (ignore internal resistance).

1. Combined e.m.f.:
   E_{\text{total}} = 1.5 V + 1.5 V = 3.0 V
2. Total resistance (only the external resistor):
   R_{\text{total}} = 3 Ω
3. Current:
   I = E_{\text{total}} / R_{\text{total}} = 3.0 V / 3 Ω = 1.0 A.

Common Misconceptions

• Series: “Current splits in a series circuit.” – No junction exists, so the current cannot split.
• Series: “All components have the same voltage.” – Only parallel circuits share the same voltage; in series the voltage divides according to resistance.
• Parallel: “The total resistance is the sum of the branch resistances.” – In parallel the reciprocal of the total resistance equals the sum of reciprocals.
• Parallel: “The current is the same in each branch.” – Currents differ; the voltage is common.
• Series cells: “Connecting cells in series changes the voltage but not the capacity.” – The e.m.f. adds, while the amp‑hour rating remains that of a single cell.

Supplementary Topics

Potential Divider (Series Resistors)

Two resistors R₁ and R₂ in series can be used to obtain a fraction of the source voltage: \[ V_{\text{out}} = V_{\text{source}} \frac{R₂}{R₁+R₂} \] (where V_out is measured across R₂).

Kirchhoff’s Current (Junction) Rule

At any node, the algebraic sum of currents is zero: \[ \sum I_{\text{in}} = \sum I_{\text{out}} \] This is simply the statement that charge does not accumulate at a junction.

Advantages of Connecting Lamps in Parallel

• Each lamp receives the full supply voltage, so its rating is straightforward.
• If one lamp fails (opens), the others continue to operate.
• The overall power drawn is lower than an equivalent series arrangement for the same number of lamps.

Suggested Practical Investigation

1. Build a series circuit with two identical resistors (e.g., 10 Ω each) and a 6 V battery. Measure the total current and the voltage across each resistor.
2. Re‑arrange the same resistors in parallel on the same battery. Record the total current and the voltage across each resistor.
3. Complete a table of measured values, calculate the theoretical R_{\text{total}} for each arrangement, and compare with the experimental results.
4. Discuss sources of error (internal resistance of the battery, contact resistance, meter loading) and suggest ways to improve accuracy.

Challenge Question (AO2)

A circuit is powered by a 15 V battery. It contains two parallel branches. Branch A has a single resistor R_A = 5 Ω. Branch B consists of three resistors in series: R_{B1}=2 Ω, R_{B2}=3 Ω, R_{B3}=6 Ω. Determine:

1. The equivalent resistance of the whole circuit.
2. The total current supplied by the battery.
3. The current flowing through each resistor (R_A, R_{B1}, R_{B2}, R_{B3}).

Solution outline:

1. Series resistance of Branch B: R_B = 2 Ω + 3 Ω + 6 Ω = 11 Ω.
2. Combine with Branch A in parallel: \[ \frac{1}{R_{\text{eq}}}= \frac{1}{5} + \frac{1}{11}= \frac{11+5}{55}= \frac{16}{55} \] \[ R_{\text{eq}} = \frac{55}{16}\;Ω \approx 3.44\;Ω \]
3. Total current: I_{\text{total}} = 15 V / 3.44 Ω ≈ 4.36 A.
4. Branch currents (same voltage 15 V):
   I_A = 15 V / 5 Ω = 3.0 A
   I_B = 15 V / 11 Ω ≈ 1.36 A (Check: 3.0 A + 1.36 A ≈ 4.36 A).
5. Current through each resistor in Branch B is the same as I_B because they are in series:
   I_{B1} = I_{B2} = I_{B3} = 1.36 A.

Quick Revision Checklist

1. Identify the configuration (series, parallel, or a combination).
2. Series circuits – remember:
   • Same current everywhere.
   • Voltage divides: Vₙ = I Rₙ.
   • R_{\text{total}} = ΣRₙ.
3. Parallel circuits – remember:
   • Same voltage across each branch.
   • Current divides: Iₙ = V_{\text{source}}/Rₙ.
   • \displaystyle\frac{1}{R_{\text{total}}}= Σ\frac{1}{Rₙ}.
4. Series cells – add e.m.f.: E_{\text{total}} = ΣEₙ.
5. Apply Kirchhoff’s current rule at any node.
6. Check your answer:
   • Sum of voltage drops = source voltage (series).
   • Sum of branch currents = total current (parallel).