D.C. Circuits: Complete O Level Physics Cheatsheet
O Level Physics, Chapter 17 · Read time: ~10 minutes
These notes come from TGC Academy's O Level Physics tuition in Singapore.
Key Formulas
Series and parallel circuits
Series: the current is the same at every point. The p.d.s across the components add up to the p.d. across the whole circuit. Effective resistance R = R₁ + R₂ + ...
Parallel: the p.d. across each branch is the same. The branch currents add up to the current from the source. Effective resistance: 1/R = 1/R₁ + 1/R₂ + ...
Adding a resistor in series increases the total resistance. Adding one in parallel decreases it, because it opens another path for current.
Trick: Sanity check: resistors in parallel always have a combined resistance smaller than the smallest one. If your answer is bigger, recheck the 1/R step.
Trap: Students often add parallel resistances directly. Use 1/R = 1/R₁ + 1/R₂, then remember to flip the answer to get R.
Potential dividers
Two resistors in series across a supply share the supply p.d. in proportion to their resistances. This is a potential divider.
A potentiometer (variable potential divider) has a sliding contact, so the output p.d. can be varied smoothly from zero to the full supply p.d.
Thermistors and LDRs as sensors
An NTC thermistor's resistance decreases as temperature increases.
A light-dependent resistor (LDR) has a resistance that decreases as light intensity increases.
Placed in a potential divider, they act as input transducers: as temperature or light changes, their resistance changes, so the p.d. across them changes. This output p.d. can switch on a fan, heater or street lamp.
Trap: Read carefully which component the output p.d. is taken across. If the LDR's resistance falls in bright light, the p.d. across the LDR falls, but the p.d. across the fixed resistor rises.
Worked Example
A 12 V battery is connected to a 4.0 Ω resistor in series with a parallel combination of a 6.0 Ω resistor and a 3.0 Ω resistor.
(a) Calculate the combined resistance of the parallel section. [2]
1/R = 1/6.0 + 1/3.0 = 3/6.0
R = 2.0 Ω
(b) Calculate the current from the battery. [2]
Total R = 4.0 + 2.0 = 6.0 Ω
I = V / R = 12 ÷ 6.0 = 2.0 A
(c) Calculate the current in the 3.0 Ω resistor. [2]
P.d. across the parallel section = 2.0 × 2.0 = 4.0 V
Current in 3.0 Ω resistor = 4.0 ÷ 3.0 = 1.3 A
Why this question is a good test of the topic: it requires reducing the parallel section to one resistance, treating the circuit as series to find the current, then going back into the parallel section to split the p.d. and current, the full whole-circuit method K323 expects.
Frequently Asked Questions
What stays the same in a series circuit?
The current is the same at every point. The p.d. is shared between the components.
What stays the same in a parallel circuit?
The p.d. across each branch is the same. The current splits between the branches.
How does an NTC thermistor behave?
Its resistance decreases as its temperature increases.
How does an LDR behave?
Its resistance decreases as the light intensity on it increases.
What is a potential divider used for?
To supply a chosen fraction of the supply p.d., or, with a thermistor or LDR, to give an output p.d. that changes with temperature or light.
Still mixing up series and parallel circuits?
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