State that, for a parallel circuit, the current from the source is larger than the current in each branch

4.3.2 Series and Parallel Circuits

1. Learning Objective ^{(Syllabus 4.3.2)}

State that, for a parallel circuit, the current supplied by the source is larger than the current flowing in any individual branch.

2. Key Facts (AO1 – quick recall) ^{(Syllabus 4.3.2)}

3. Conceptual Understanding ^{(Syllabus 4.3.2)}

• Series circuit: Only one path for charge; therefore the same current flows through every component.
• Parallel circuit: Each branch is connected directly across the source, so every branch experiences the same potential difference (voltage).
• Kirchhoff’s Junction (Current) Law: At any junction, the sum of currents entering equals the sum of currents leaving.

  
  → In a parallel network the source current equals the algebraic sum of the branch currents.

4. Quantitative Relationships ^{(Syllabus 4.3.2)}

• Series resistance (combined): \(R{\text{eq}} = R1 + R_2 + \dots\)
• Parallel resistance (combined): \(\displaystyle\frac{1}{R{\text{eq}}}= \frac{1}{R1}+ \frac{1}{R_2}+ \dots\)
• Series e.m.f. of cells: \(E{\text{total}} = E1 + E_2 + \dots\)
• Parallel e.m.f. of cells (optional extension): When identical cells are placed in parallel, the e.m.f. remains the same as a single cell, but the available current‑capacity adds.

5. Construction Checklist – Building Simple Circuits ^{(Syllabus 4.3.2)}

1. Identify the source (battery or cell) and note its polarity.
2. For a series connection, join components end‑to‑end so there is only one continuous path.
3. For a parallel connection, connect each component across the same two points of the source (identical voltage on every branch).
4. Use a breadboard or secure all joints with solder/twisted leads.
5. Before powering the circuit, verify connections with a multimeter (continuity test, correct voltage at each point).

6. Quick‑Reference Tables ^{(Syllabus 4.3.2)}

6.1 Series Circuits

6.2 Parallel Circuits

7. Why the Source Current Is Larger in a Parallel Circuit ^{(Syllabus 4.3.2)}

Consider a parallel circuit with two resistors, \(R1\) and \(R2\), across a battery of e.m.f. \(V\).

1. Each branch sees the full battery voltage \(V\).
2. Branch currents (Ohm’s law):

   \[

   I1 = \frac{V}{R1}, \qquad I2 = \frac{V}{R2}

   \]

   Unit check: V (volt) ÷ Ω (ohm) = A (ampere).

3. The source must supply both currents simultaneously (Kirchhoff’s junction law):

   \[

   I{\text{total}} = I1 + I2 = \frac{V}{R1} + \frac{V}{R_2}

   \]

   Unit check: A + A = A.

4. Both terms are positive, so \(I{\text{total}}\) is inevitably larger than either \(I1\) or \(I_2\) alone.

Common mistake: Adding currents from series branches. In a series circuit the same current flows through every component; only in parallel does the total current equal the sum of branch currents.

8. Worked Example – Parallel Circuit ^{(Syllabus 4.3.2)}

Calculate the total current supplied by a 12 V battery connected to two parallel resistors, \(R1 = 6\ \Omega\) and \(R2 = 12\ \Omega\).

1. Branch currents:

   \[

   I_1 = \frac{12\ \text{V}}{6\ \Omega}=2\ \text{A},\qquad

   I_2 = \frac{12\ \text{V}}{12\ \Omega}=1\ \text{A}

   \]

   Unit check: V ÷ Ω = A.

2. Total current:

   \[

   I{\text{total}} = I1 + I_2 = 2\ \text{A} + 1\ \text{A}=3\ \text{A}

   \]

   Unit check: A + A = A.

3. Since \(3\ \text{A} > 2\ \text{A}\) and \(3\ \text{A} > 1\ \text{A}\), the source current is larger than the current in either branch, confirming the principle.

9. Worked Example – Series Circuit (including series e.m.f.) ^{(Syllabus 4.3.2)}

Two 6 V cells are connected in series with a \(4\ \Omega\) resistor. Find the current through the circuit.

1. Combined e.m.f.: \(E_{\text{total}} = 6\ \text{V} + 6\ \text{V} = 12\ \text{V}\).
2. Series resistance is just the resistor: \(R_{\text{eq}} = 4\ \Omega\).
3. Current (Ohm’s law):

   \[

   I = \frac{E{\text{total}}}{R{\text{eq}}}= \frac{12\ \text{V}}{4\ \Omega}=3\ \text{A}

   \]

   Unit check: V ÷ Ω = A.

4. Because the circuit is series, the same \(3\ \text{A}\) flows through each cell and the resistor.

10. Optional Extension – Cells in Parallel ^{(Supplementary material)}

If identical cells (same e.m.f. \(E\) and internal resistance \(r\)) are placed in parallel, the overall e.m.f. remains \(E\) but the effective internal resistance becomes \(r_{\text{eq}} = r/n\) (where \(n\) is the number of cells). Consequently the maximum current the arrangement can supply increases proportionally to \(n\).

11. Practice Questions ^{(Syllabus 4.3.2)}

11.1 Parallel

Three resistors \(R1 = 4\ \Omega\), \(R2 = 6\ \Omega\), and \(R_3 = 12\ \Omega\) are connected in parallel across a 9 V source. Calculate the current supplied by the source and state whether it is larger than the current in each branch.

Hint – unit check after each step (V in volts, R in Ω, I in A).

11.2 Series

Four identical \(2\ \Omega\) resistors are connected in series with a 12 V battery. Determine the total current and the voltage drop across each resistor.

Remember: use \(R{\text{eq}} = R1+R2+R3+R_4\) and check units.

12. Suggested Diagram ^{(Syllabus 4.3.2)}