Parallel Circuits Revision Notes
Subject: Physics | Level: GCSE | Exam Board: OCR
Master OCR GCSE Physics Parallel Circuits (3.6) with this comprehensive guide. We break down the core principles of voltage and current, show you how to tackle Higher Tier resistance calculations, and provide examiner-approved techniques to secure every mark. This is your essential resource for turning a tricky topic into guaranteed marks.
Revision Notes & Key Concepts
Key Terms & Definitions
- Parallel Circuit
- A circuit in which the current divides into two or more paths before recombining to complete the circuit.
- Potential Difference (Voltage)
- The work done per unit charge moved between two points. Measured in Volts (V).
- Current
- The rate of flow of electrical charge. Measured in Amperes (A).
- Resistance
- A measure of the opposition to current flow in an electrical circuit. Measured in Ohms (Ω).
- Junction
- A point in a circuit where three or more conductors meet, allowing current to split or recombine.
- Branch
- A section of a parallel circuit containing one or more components, which provides a separate path for current.
Worked Examples
Worked Example
Question: A 12V battery is connected to two resistors in parallel. Resistor R1 has a resistance of 4.0 Ω and resistor R2 has a resistance of 6.0 Ω. Calculate the total current flowing from the battery.
Solution: Step 1: State the voltage across each resistor. Since they are in parallel, the voltage across each is the same as the source. V1 = 12V and V2 = 12V. (1 mark) Step 2: Use Ohm's Law (V=IR, rearranged to I=V/R) to find the current in the first branch (I1). I1 = 12V / 4.0Ω = 3.0 A. (1 mark for substitution, 1 mark for answer) Step 3: Use Ohm's Law to find the current in the second branch (I2). I2 = 12V / 6.0Ω = 2.0 A. (1 mark) Step 4: Use the rule for currents in parallel to find the total current. I_total = I1 + I2 = 3.0A + 2.0A = 5.0 A. (1 mark) Final answer: 5.0 A
Worked Example
Question: (Higher Tier Only) A circuit contains a 3.0 Ω resistor and a 6.0 Ω resistor connected in parallel. Calculate the total resistance of the circuit.
Solution: Step 1: State the formula for total resistance in parallel. 1/R_total = 1/R1 + 1/R2. (1 mark) Step 2: Substitute the values into the formula. 1/R_total = 1/3.0 + 1/6.0. This can be written as 1/R_total = 2/6 + 1/6 = 3/6 = 0.5. (1 mark) Step 3: Invert the result to find R_total. R_total = 1 / 0.5 = 2.0 Ω. (1 mark) Final answer: 2.0 Ω
Worked Example
Question: Explain why adding another light bulb in parallel to an existing circuit causes the total current from the supply to increase.
Solution: Step 1: Adding another bulb in parallel decreases the total resistance of the circuit. (1 mark) Step 2: This is because it provides an additional path for the current to flow through. (1 mark) Step 3: According to Ohm's Law (I = V/R), if the supply voltage (V) stays the same and the total resistance (R) decreases, the total current (I) from the supply must increase. (1 mark)
Practice Questions
Question: A toaster with a resistance of 30 Ω and a kettle with a resistance of 20 Ω are connected in parallel to the 230V mains supply. State the potential difference across the toaster.
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Question: Describe how the current in the wires from the mains supply is related to the currents in the toaster and the kettle from the previous question.
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Question: Calculate the total current drawn from the 230V mains supply by the 30 Ω toaster and 20 Ω kettle connected in parallel.
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Question: (Higher Tier Only) Calculate the total resistance of a 4.0 Ω resistor and a 12.0 Ω resistor connected in parallel. Give your answer to 2 significant figures.
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Question: A student has two identical lamps. They connect them to a battery in parallel. They then add a third identical lamp in parallel with the first two. Explain what happens to the brightness of the first two lamps and the total current drawn from the battery.
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