State that the combined resistance of resistors in parallel is less than that of any single resistor in that circuit

4.3.2 Series and Parallel Circuits

Learning Objective

State that the equivalent resistance of resistors connected in parallel is always less than the resistance of any single resistor in the circuit, and apply this fact together with all core and supplementary requirements of the Cambridge IGCSE 0625 Physics syllabus.

Core Requirements (what the syllabus expects)

• Series circuits

  • Current is the same through every component.
  • Series‑current rule: \(I{\text{total}} = I1 = I2 = \dots = In\)
  • Total resistance is the sum of the individual resistances:
    

    R{\text{series}} = R1 + R2 + \dots + Rn

  • Series‑voltage rule: The total potential difference is the sum of the individual drops:
    

    V{\text{total}} = V1 + V2 + \dots + Vn

  • For ideal emf sources in series the emfs add: \(E{\text{total}} = E1 + E_2 + \dots\)

• Parallel circuits

  • All branches have the same potential difference (voltage) across them.
  • Parallel‑voltage rule: \(V{\text{total}} = V1 = V2 = \dots = Vn\)
  • Parallel‑current rule (junction rule): The total current supplied by the source equals the sum of the branch currents:
    

    I{\text{total}} = I1 + I2 + \dots + In

  • Reciprocal‑resistance rule (core quantitative relationship):
    

    1 / R{\text{eq}} = 1 / R1 + 1 / R2 + \dots + 1 / Rn

  • Two‑resistor shortcut (often useful in exam questions):
    

    R{\text{eq}} = (R1·R2) / (R1 + R_2)

  • Because the sum of positive reciprocals is always larger than the reciprocal of any single resistor, R_eq < the smallest R_i.

• Practical advantage of parallel lamps (core)

  • Each lamp receives the full supply voltage, so all shine with the same brightness.
  • If one lamp fails (opens), the others continue to operate because each has its own independent path.

• Safety (core + supplement)

  • Use wires of adequate gauge to avoid overheating when the total current is large (parallel networks draw more current).
  • Secure all connections; loose contacts can cause local heating.
  • Always switch off the supply before making or altering connections.

Supplementary Requirements

• Current‑division (supplement) – In a parallel network the current in a branch is inversely proportional to its resistance:

  
  I1 = I{\text{total}}·\dfrac{R2}{R1+R_2}\quad\text{(for two branches)}

• Voltage‑division (supplement) – In a series network the voltage across a resistor is proportional to its resistance:

  
  V1 = V{\text{total}}·\dfrac{R1}{R1+R_2+…}

• Kirchhoff’s rules (supplement)

  • Junction (current) rule – Σ I_in = Σ I_out
  • Loop (voltage) rule – Σ ΔV (around a closed loop) = 0

• Power in circuits (core AO2)

  • Instantaneous power: \(P = IV = I^2R = V^2/R\)
  • For identical resistors in parallel the total power is the sum of the powers in each branch, explaining why parallel lamps are brighter but draw more current.

• Internal resistance of the source (supplement)

  • A real battery can be modelled as an ideal emf E in series with an internal resistance r.
  • The terminal voltage is \(V = E - I r\); therefore, as the total current of a parallel network increases, the terminal voltage falls.

Why \(R_{\text{eq}}\) Is Always Less Than Any Single Resistor – Proof by Positive‑Term Argument

1. Each term \(\dfrac{1}{R_i}\) is a positive number because resistance is always positive.
2. Adding another resistor in parallel adds another positive term to the sum \(\displaystyle\sum \dfrac{1}{R_i}\), making the sum larger.
3. The equivalent resistance is the reciprocal of this sum:

   \[

   R{\text{eq}} = \frac{1}{\displaystyle\sum \frac{1}{Ri}}

   \]

   A larger denominator yields a smaller overall value.

4. Hence, adding any resistor in parallel can only decrease the equivalent resistance; it can never increase it. Consequently,

   \[

   R{\text{eq}} < \min\{R1,R2,\dots,Rn\}.

   \]

Formula Summary

Worked Example – Three Resistors in Parallel

Given \(R1 = 12\;Ω\), \(R2 = 18\;Ω\), \(R3 = 30\;Ω\). Find \(R{\text{eq}}\) and comment on its size.

1. Reciprocal sum:

   
   1 / R_{\text{eq}} = 1/12 + 1/18 + 1/30

2. Common denominator 180:

   
   1 / R_{\text{eq}} = 15/180 + 10/180 + 6/180 = 31/180

3. Invert:

   
   R_{\text{eq}} = 180 / 31 Ω ≈ 5.8 Ω

4. Observation: 5.8 Ω is smaller than each of 12 Ω, 18 Ω and 30 Ω, confirming the principle that the parallel equivalent resistance is less than the smallest individual resistor.

Worked Example – Mixed Series‑Parallel Network (uses core & supplement)

Two resistors \(RA = 4\;Ω\) and \(RB = 6\;Ω\) are in series. Their series combination is in parallel with a third resistor \(R_C = 5\;Ω\). Find the overall equivalent resistance and the current through each branch if the network is connected to a 12 V ideal battery.

1. Series pair:

   
   R{AB} = RA + R_B = 4 Ω + 6 Ω = 10 Ω

2. Parallel with \(R_C\):

   
   1 / R{\text{eq}} = 1 / R{AB} + 1 / R_C = 1/10 + 1/5 = 0.1 + 0.2 = 0.3

3. Invert:

   
   R_{\text{eq}} = 1 / 0.3 ≈ 3.33 Ω

4. Total current from the battery (using Ohm’s law):

   
   I{\text{total}} = V / R{\text{eq}} = 12 V / 3.33 Ω ≈ 3.60 A

5. Current division (junction rule):

   
   I{AB} = I{\text{total}}·\dfrac{RC}{R{AB}+R_C} = 3.60·\dfrac{5}{10+5} = 1.20 A


IC = I{\text{total}} - I_{AB} = 3.60 A - 1.20 A = 2.40 A

6. Current through each resistor in the series pair (same current):

   
   IA = IB = I_{AB} = 1.20 A

7. Check with Kirchhoff’s loop rule (optional): the sum of voltage drops around any closed loop equals zero, confirming the values above.

Power Considerations in Parallel Networks

Using the same three‑resistor example, the total power drawn from a 12 V supply is:

• Branch currents: \(I1 = V/R1 = 12/12 = 1.0\;A\), \(I2 = 12/18 = 0.67\;A\), \(I3 = 12/30 = 0.40\;A\).
• Power in each branch: \(Pi = V Ii\) → \(P1 = 12 W\), \(P2 = 8.0 W\), \(P_3 = 4.8 W\).
• Total power: \(P_{\text{total}} = 12 + 8 + 4.8 = 24.8\;W\). This is larger than the power that would be dissipated by any single resistor alone, illustrating why parallel circuits can cause higher current and require careful safety handling.

Internal Resistance of the Source – Effect on Parallel Circuits

If the 12 V battery has an internal resistance \(r = 0.5\;Ω\), the terminal voltage when the mixed network above draws 3.60 A is:

\[

V{\text{terminal}} = E - I{\text{total}} r = 12 V - (3.60 A)(0.5 Ω) = 10.2 V

\]

The reduced terminal voltage slightly lowers the currents in each branch, a point that may be examined in higher‑level (supplementary) questions.

Practical Example – Lamps in Parallel (Re‑examined)

Four identical 2 V, 0.5 A lamps are connected to a 6 V supply.

• Because each lamp is in parallel, each receives the full 6 V, so each lamp operates at its rated brightness (the lamp’s internal resistance adjusts accordingly).
• Current through each lamp: \(I{\text{lamp}} = 6 V / R{\text{lamp}}\). If the lamp’s rated resistance is \(R{\text{lamp}} = V/I = 2 V/0.5 A = 4 Ω\), then \(I{\text{lamp}} = 6 V/4 Ω = 1.5 A\).
• Total current from the supply: \(I_{\text{total}} = 4 × 1.5 A = 6 A\). The large current illustrates the safety point that parallel circuits can draw substantial current, so wire size and fuse rating must be appropriate.
• If one lamp opens, the remaining three still have a complete path to the supply and continue to work.

Kirchhoff’s Rules – Quick Reference

Safety Checklist (When Building Circuits)

• Switch off the supply before making or altering connections.
• Use insulated wires of adequate gauge; parallel networks can draw large total currents, so the wire must be able to carry the summed current without overheating.
• Secure all connections (screw terminals, breadboard clips, or solder joints) to prevent loose contacts that may cause local heating or sparks.
• Never touch exposed conductive parts while the circuit is live.
• Verify that the supply voltage matches the rating of all components (especially lamps and resistors).
• If many branches are added in parallel, check that the total current does not exceed the rating of the power source, wiring, or protective devices (fuse, circuit breaker).

Key Take‑away

In a parallel network the equivalent resistance is always less than the smallest individual resistor because the reciprocal sum of the resistances is larger than the reciprocal of any one resistor. This principle, together with the series‑current rule, parallel‑current rule, Kirchhoff’s laws, power considerations, and practical safety advice, provides a complete framework for analysing all IGCSE series‑parallel circuits.