Calculate the combined resistance of two resistors in parallel

4.3.2 Series and Parallel Circuits

Learning Objectives (AO1‑AO3)

• State the defining features of series and parallel connections (current and voltage behaviour).
• Calculate the equivalent resistance of resistors in series and in parallel.
• Calculate the combined emf of ideal voltage sources connected in series.
• Explain why lamps (or other appliances) are normally connected in parallel in domestic circuits.
• Interpret circuit diagrams and predict how changes affect total current, voltage and resistance.
• Design and carry out a simple investigation to verify the parallel‑resistance formula (AO3).

Key Concepts

Standard IGCSE Circuit Symbols

Symbol	Name	Typical Use in Questions
	Wire	Shows connections between components.
	Resistor	Represents a fixed resistance (e.g., 120 Ω).
	Battery (source of emf)	Provides the voltage that drives the current.
	Ammeter	Measures total current (connected in series).
	Voltmeter	Measures potential difference (connected in parallel with the component whose voltage is required).

1. Series Circuits

1.1 Defining Features

• All components are connected end‑to‑end so that there is only one path for charge to flow.
• Current: The same current flows through every component. “The current is the same at every point in a series circuit.”
• Voltage: The total supplied voltage is divided among the components: \(V{\text{total}} = V{1}+V_{2}+…\).

1.2 Equivalent Resistance

Because the same current passes through each resistor, the resistances simply add:

\(R{\text{eq}} = R{1}+R{2}+…+R{n}\)

1.3 Combined EMF of Ideal Sources in Series

If ideal batteries (no internal resistance) are placed in series, their emfs add algebraically (taking polarity into account):

\(E{\text{total}} = E{1}+E{2}+…+E{n}\)

Internal resistances (if given) would be added in the same way as ordinary resistors.

1.4 Worked Example – Series Resistance

Three resistors \(R{1}=30\;\Omega\), \(R{2}=50\;\Omega\) and \(R_{3}=70\;\Omega\) are connected in series. Find the equivalent resistance.

1. Apply the series formula: \(R_{\text{eq}} = 30+50+70 = 150\;\Omega\).
2. Interpretation: The series network offers more resistance than any individual resistor, as expected.

2. Parallel Circuits

2.1 Defining Features

• Each resistor (or other component) is connected across the same two nodes; therefore the voltage across every branch is identical.
• Voltage: The same across each branch. “The voltage across each resistor in a parallel circuit is the same.”
• Current: The total current supplied by the source is the sum of the branch currents: \(I{\text{total}} = I{1}+I_{2}+…\).

2.2 Equivalent Resistance for Two Resistors

Starting from Ohm’s law for each branch:

\[

I{1}= \frac{V}{R{1}}, \qquad I{2}= \frac{V}{R{2}}

\]

The total current is

\[

I{\text{total}} = I{1}+I{2}= V\!\left(\frac{1}{R{1}}+\frac{1}{R_{2}}\right)

\]

Define \(R{\text{eq}}\) by \(I{\text{total}} = \dfrac{V}{R_{\text{eq}}}\) and solve:

\[

\frac{1}{R{\text{eq}}}= \frac{1}{R{1}}+\frac{1}{R_{2}}

\qquad\Longrightarrow\qquad

R{\text{eq}} = \frac{R{1}R{2}}{R{1}+R_{2}}

\]

2.3 General Formula (n ≥ 2)

For any number of parallel branches:

\(\displaystyle \frac{1}{R{\text{eq}}}= \sum{i=1}^{n}\frac{1}{R_{i}}\)

In practice, candidates often combine the resistors two at a time using the two‑resistor formula, then treat the result as a single resistor for the next step.

2.4 Worked Example – Two Resistors

Given \(R{1}=120\;\Omega\) and \(R{2}=80\;\Omega\) in parallel, find \(R_{\text{eq}}\).

\[

R_{\text{eq}} = \frac{120 \times 80}{120+80}

= \frac{9600}{200}

= 48\;\Omega

\]

The parallel combination offers less resistance than either individual resistor.

2.5 Worked Example – Three Resistors (successive combination)

Resistors: \(R{1}=100\;\Omega\), \(R{2}=200\;\Omega\), \(R_{3}=300\;\Omega\).

1. Combine \(R{1}\) and \(R{2}\):

   \[

   R_{12}= \frac{100 \times 200}{100+200}= \frac{20000}{300}=66.7\;\Omega

   \]

2. Combine \(R{12}\) with \(R{3}\):

   \[

   R_{\text{eq}}= \frac{66.7 \times 300}{66.7+300}

   = \frac{20010}{366.7}\approx 54.6\;\Omega

   \]

3. Check with the general formula:

   \[

   \frac{1}{R_{\text{eq}}}= \frac{1}{100}+ \frac{1}{200}+ \frac{1}{300}=0.01833

   \quad\Longrightarrow\quad

   R_{\text{eq}}\approx54.6\;\Omega

   \]

2.6 Mixed Series‑Parallel Example (AO2)

A 150 Ω resistor (\(R_{S}\)) is in series with a parallel pair of 300 Ω and 600 Ω resistors.

1. Parallel pair:

   \[

   R_{P}= \frac{300 \times 600}{300+600}=200\;\Omega

   \]

2. Series addition:

   \[

   R{\text{total}} = R{S}+R_{P}=150\;\Omega+200\;\Omega=350\;\Omega

   \]

3. Why Lamps Are Connected in Parallel (AO1)

• Each lamp experiences the full supply voltage, so all lamps have the same brightness.
• If one lamp burns out, the remaining lamps continue to work because the circuit still provides a complete path for current through the other parallel branches.
• Parallel wiring also reduces the total resistance of the lighting circuit, allowing a larger total current to be drawn safely from the mains.

4. Practice Questions

1. Two resistors, \(R{1}=150\;\Omega\) and \(R{2}=300\;\Omega\), are placed in parallel. Calculate \(R_{\text{eq}}\).
2. A circuit contains a \(220\;\Omega\) resistor in parallel with an unknown resistor \(R{x}\). The measured equivalent resistance is \(88\;\Omega\). Find \(R{x}\).
3. Three resistors \(R{1}=100\;\Omega\), \(R{2}=200\;\Omega\), \(R_{3}=300\;\Omega\) are connected in parallel. Determine the total resistance using the two‑resistor formula twice.
4. In the mixed circuit shown below, a 120 Ω resistor is in series with a parallel combination of 240 Ω and 360 Ω resistors. Find the overall resistance.
   

   

5. Three ideal batteries of 3 V, 4.5 V and 6 V are connected in series. What is the total emf?
6. Explain two practical advantages of connecting household lamps in parallel rather than in series.

Answer Key & Hints

1. Answer: \(R_{\text{eq}} = \dfrac{150 \times 300}{150+300}=100\;\Omega\).
   

   Hint: Use the two‑resistor formula directly.

2. Answer: \(R_{x}= \dfrac{(220)(88)}{220-88}=146.7\;\Omega\).
   

   Hint: Start from \(\frac{1}{R{\text{eq}}}= \frac{1}{220}+ \frac{1}{R{x}}\) and solve for \(R_{x}\).

3. Answer: \(R{12}=66.7\;\Omega\); then \(R{\text{eq}}\approx54.6\;\Omega\).
   

   Hint: Keep extra decimal places in the intermediate step.

4. Answer: Parallel pair: \(R_{P}= \dfrac{240 \times 360}{240+360}=144\;\Omega\).
   

   Total: \(R_{\text{total}} = 120\;\Omega + 144\;\Omega = 264\;\Omega\).
   

   Hint: Treat the parallel part first, then add the series resistor.

5. Answer: \(E_{\text{total}} = 3 + 4.5 + 6 = 13.5\;\text{V}\).
6. Answer: (i) Each lamp receives the full mains voltage, giving uniform brightness.
   

   (ii) Failure of one lamp does not affect the others; the circuit remains complete.

Common Mistakes to Avoid

• Adding resistances directly for a parallel network (the series rule only).
• Assuming the same current flows through each branch of a parallel circuit; it is the voltage that is common.
• Using the series formula \(R{\text{eq}} = R{1}+R_{2}\) for a parallel arrangement.
• Neglecting polarity when adding emfs in series – opposite‑polarity batteries subtract.
• Rounding too early in successive‑combination problems; keep at least three significant figures until the final answer.

Practical Investigation (AO3) – Verifying the Parallel‑Resistance Formula

Objective: Show experimentally that the measured equivalent resistance of two resistors in parallel matches the theoretical value given by \(\displaystyle R{\text{eq}} = \frac{R{1}R{2}}{R{1}+R_{2}}\).

Equipment: Breadboard, two known resistors (e.g., 100 Ω and 220 Ω), 6 V DC supply, digital multimeter (voltmeter/ammeter), connecting wires, single‑pole switch.

Method (outline):

1. Assemble the circuit: battery → switch → parallel pair of the two resistors → return to battery (see diagram below).
2. Close the switch and measure the voltage across the parallel network (should be ≈6 V).
3. Place the ammeter in series with the switch and record the total current \(I_{\text{total}}\).
4. Calculate the experimental equivalent resistance: \(R{\text{exp}} = \dfrac{V}{I{\text{total}}}\).
5. Compare \(R_{\text{exp}}\) with the theoretical value obtained from the formula.
6. Discuss possible sources of error (internal resistance of the battery, meter tolerances, contact resistance on the breadboard).

Summary Checklist (AO1)

• Series: same current, voltages add, \(R_{\text{eq}} = \sum R\).
• Parallel: same voltage, currents add, \(\displaystyle \frac{1}{R_{\text{eq}}}= \sum \frac{1}{R}\).
• Series emf: \(E_{\text{total}} = \sum E\) (polarity considered).
• Parallel lamps: full voltage on each lamp; one lamp failing does not affect the others.