Two identical resistors connected in parallel to the same fixed-voltage supply draw four times the total power they draw in series. With two 6 Ω resistors and a 6 V supply, the total is 3 W in series and 12 W in parallel.
Decide whether the question wants the power supplied to the whole circuit or the power dissipated by one resistor. Use the voltage, current and resistance that describe that same part of the circuit.
- Find equivalent resistance before calculating total current.
- In series, the same current passes through both resistors. In parallel, each branch has the full supply voltage.
- Use the circuit’s equivalent resistance for total power and an individual resistor’s resistance for its own power.
CHAPTER 1 OF 4 · Back to contents
Choose the right circuit quantities
Assume an ideal 6 V supply, negligible wire resistance and two fixed 6 Ω resistors. In series, the resistances add to 12 Ω. In parallel, the two identical branches give an equivalent resistance of 3 Ω.
Use V = IR to find current, then P = VI to calculate power. For a resistor you can also use P = I²R or P = V²/R. The voltage and current in each formula must refer to the same component or equivalent circuit.
CHAPTER 2 OF 4 · Back to contents
Work out the power in each connection
In series, the supply current is 6 ÷ 12 = 0.5 A. Total power is 6 × 0.5 = 3 W. Each resistor carries 0.5 A and has a potential difference of 3 V, so each dissipates 3 × 0.5 = 1.5 W.
In parallel, each branch has 6 V, so each carries 6 ÷ 6 = 1 A. Total current is 2 A and the supply delivers 6 × 2 = 12 W. Each resistor dissipates 6 × 1 = 6 W.
Check by adding component powers. In series, 1.5 + 1.5 = 3 W. In parallel, 6 + 6 = 12 W. With the supply voltage fixed, reducing the equivalent resistance from 12 Ω to 3 Ω makes the total current and total power four times as large.
CHAPTER 3 OF 4 · Back to contents
Practise with unequal resistors
Keep the 6 V ideal supply and replace one resistor with a 3 Ω resistor. In parallel, the 3 Ω branch carries 2 A and dissipates 12 W. The 6 Ω branch carries 1 A and dissipates 6 W. The supply delivers 18 W in total.
Now connect them in series. Both carry 6 ÷ 9 = ⅔ A. Using P = I²R, the 3 Ω resistor dissipates 4/3 W and the 6 Ω resistor dissipates 8/3 W. Total power is 4 W. In parallel the lower resistance dissipates more power because both branches have the same potential difference. In series the higher resistance dissipates more because the same current passes through both components.
CHAPTER 4 OF 4 · Back to contents
Frequently asked questions
Why can’t I use 6 V for each series resistor?
The 6 V is across the pair. The two identical series resistors each have a 3 V potential difference, and the two potential differences add to 6 V.
Does the fourfold rule always work?
The rule here assumes two identical fixed resistors and the same ideal voltage across both arrangements. For unequal resistances, calculate each equivalent resistance and use P = V²/R.
How do I get energy from power?
Multiply power by time. At 12 W for 5 s, the circuit transfers 60 J. A watt is one joule per second.
Contents · Official syllabus and next practice
Official syllabus and next practice
Sketch both circuits and label the current through each resistor and the voltage across it before doing the arithmetic.