Series and parallel circuits and conservation lawsEdexcel A-Level Physics: Revision notes
Section 1
Conservation of charge and current
Charge cannot be created, destroyed or stored in a wire, so the charge entering any point in a circuit each second equals the charge leaving it each second.
- In a series circuit there is only one path, so the current is the same at every point.
- At a junction in a parallel circuit, the total current entering equals the total current leaving: I = I₁ + I₂.
This is the distribution of current as a consequence of conservation of charge.
Section 2
Conservation of energy and p.d.
The p.d. of a source is the energy it gives to each coulomb of charge. As the charge goes round any complete loop, it transfers all of this energy, so by conservation of energy the sum of the p.d.s across components around a loop equals the p.d. of the source.
- In series, the p.d.s add up to the supply p.d.: V = V₁ + V₂.
- In parallel, each branch is connected across the same two points, so each branch has the same p.d. as the supply.
Saying that the p.d. is shared in a parallel circuit. The p.d. is shared in series; in parallel it is the same across each branch.
Section 3
Resistors in series
In series the current I is the same through each resistor and V = V₁ + V₂. With V = IR:
IR = IR₁ + IR₂, so R = R₁ + R₂
The combined resistance is the sum, and it is larger than any individual resistor. The p.d. across each resistor is in proportion to its resistance.
Section 4
Resistors in parallel
In parallel each resistor has the same p.d. V and the currents add: I = I₁ + I₂. With I = V / R:
V / R = V / R₁ + V / R₂, so 1 / R = 1 / R₁ + 1 / R₂
The combined resistance is smaller than the smallest resistor, because the extra paths allow more current for the same p.d. For N identical resistors R in parallel, the combined resistance is R / N. For two resistors, R = R₁R₂ / (R₁ + R₂).
Forgetting the final reciprocal. The sum 1 / R₁ + 1 / R₂ is 1 / R, not R.
Section 5
Worked example: a mixed circuit
A 9.0 V battery drives a 15 Ω resistor in series with 30 Ω and 60 Ω resistors in parallel.
- Parallel pair: 1 / R = 1 / 30 + 1 / 60 = 3 / 60, so R = 20 Ω.
- Total R = 15 + 20 = 35 Ω.
- Current from the battery I = 9.0 / 35 = 0.26 A.
- p.d. across the 15 Ω resistor = 0.257 × 15 = 3.9 V, so the p.d. across the parallel pair = 9.0 − 3.9 = 5.1 V.
- Current in the 60 Ω resistor = 5.1 / 60 = 0.086 A, and in the 30 Ω resistor = 0.17 A. These add to 0.26 A, as they must.
Work from the inside out: combine parallel groups first, then add series parts. Check the branch currents add up to the total current.
Must know
- Series: same current; p.d.s add to the supply p.d.; R = R₁ + R₂
- Parallel: same p.d. across each branch; currents add at a junction; 1 / R = 1 / R₁ + 1 / R₂
- Charge conservation gives the current rules; energy conservation gives the p.d. rules
- Parallel resistance is smaller than the smallest resistor
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Series and parallel circuits and conservation laws
- A string of decorative lights consists of 20 identical lamps connected in series to a 230 V supply.Explain, using conservation of energy, why the potential differences across the 20 lamps add up to the p.d. of the supply.2 marks
- Two resistors, of resistance 6.0 Ω and 12 Ω, are connected in parallel across a 12 V battery that has negligible internal resistance.Explain why the combined resistance is less than the resistance of the 6.0 Ω resistor alone.2 marks
- A circuit contains a 9.0 V battery of negligible internal resistance. A 15 Ω resistor is connected in series with a parallel combination of a 30 Ω resistor and a 60 Ω resistor.Calculate the total resistance of the circuit.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).