N2 Electrical Trade Theory: Series, Parallel and the Questions That Follow

by Braintiq Academic Team

N2 Electrical Trade Theory covers a lot of ground, but the circuit calculations that carry the marks reduce to a small number of situations. Work through enough past papers and you stop seeing individual questions and start seeing four shapes.

This is those four shapes, the rules behind them, and the mistakes that cost the most.

The three quantities and the one law

Current is the flow of charge, measured in amperes. Voltage, properly potential difference, is the push that drives it, measured in volts. Resistance opposes it, measured in ohms.

Ohm's law ties them together:

$$V = IR$$

Rearranged as needed:

$$I = \frac{V}{R} \qquad R = \frac{V}{I}$$

Almost every calculation on the paper is this law applied to some part of a circuit. The difficulty is never the algebra. It is knowing which part of the circuit the numbers belong to, and that is what the four shapes are about.

Shape one: series

Components in a line, one path for the current.

$$R_T = R_1 + R_2 + R_3 + \dots$$

Two consequences that get examined directly:

That last point is the check you should always run. If your calculated voltage drops do not sum to the supply, you have made an error and you have found it before the marker does.

Shape two: parallel

Components side by side, more than one path.

$$\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + \dots$$

And the consequences, which are the mirror of series:

Two things about that formula cause most of the lost marks.

The reciprocal must be undone. You calculate $\frac{1}{R_T}$, and then you must invert it to get $R_T$. Students who stop at $\frac{1}{R_T} = 0.25$ and write $R_T = 0.25$ have lost the question. Write the inversion as its own line.

The total is always smaller than the smallest branch. That is a property, not a coincidence: adding another path makes it easier for current to flow, so resistance falls. Use it as a sanity check. If you have two resistors of 10 and 20 ohms in parallel and your answer is bigger than 10, it is wrong.

For the special case of exactly two resistors in parallel, the product over sum shortcut saves time:

$$R_T = \frac{R_1 R_2}{R_1 + R_2}$$

Shape three: series-parallel combination

The one that appears most in exams. A circuit with a parallel group somewhere inside an otherwise series path.

The method is always the same and it is worth following mechanically:

  1. Redraw the circuit. Yes, actually redraw it. Almost nobody does and almost everybody who does gets it right.
  2. Work out the equivalent resistance of the parallel group on its own.
  3. Replace that group with a single resistor of that value, so you now have a simple series circuit.
  4. Find the total resistance and the total current from the supply.
  5. Work backwards to find the voltage across and current through each individual component.

Step 5 is where the marks are, and the thing to hold onto is that the total current flows through the series parts and then splits at the parallel group. So the voltage across the parallel group is found from the total current and the group's equivalent resistance, and only then can you find each branch current.

Shape four: power and energy

Power is the rate of doing electrical work, measured in watts:

$$P = VI$$

and using Ohm's law, the two forms you will need when you do not have both $V$ and $I$:

$$P = I^2 R \qquad P = \frac{V^2}{R}$$

Choose whichever one matches the quantities you already have, rather than calculating a missing quantity first. It is fewer steps and fewer places to slip.

Energy is power multiplied by time:

$$W = Pt$$

In joules when time is in seconds. The electricity account version uses kilowatt hours, where power is in kilowatts and time in hours, and the conversion trips people up. One kilowatt hour is 3,600,000 joules. If a question gives you watts and minutes and asks for kilowatt hours, do both conversions explicitly on their own lines.

Where the marks actually go

Not redrawing the circuit. The single biggest cause of lost marks in combination questions.

Forgetting to invert the parallel formula. Second biggest.

Mixing up which quantity is shared. In series the current is shared, in parallel the voltage is shared. Getting these the wrong way round makes every subsequent number wrong. A way to hold it: in series there is one path so there is one current; in parallel the branches share two connection points so they share the voltage.

Unit prefixes. Milliamps, kilohms and megohms appear constantly. Convert everything to base units before calculating. A resistance in kilohms with a current in milliamps happens to give volts directly, which is convenient and also the reason students get away with sloppiness until the one question where it does not work.

Not using the sanity checks. Series voltage drops sum to the supply. Parallel currents sum to the total. Parallel resistance is less than the smallest branch. Each of these takes seconds and catches real errors.

A worked combination

A 24 V supply feeds a 4 ohm resistor in series with a parallel group of 6 ohms and 12 ohms. Find the total current and the current in each branch.

Parallel group first. Using product over sum:

$$R_p = \frac{6 \times 12}{6 + 12} = \frac{72}{18} = 4 \ \Omega$$

Sanity check: 4 is less than 6, the smallest branch. Good.

Total resistance. Now it is a simple series circuit of 4 and 4:

$$R_T = 4 + 4 = 8 \ \Omega$$

Total current.

$$I = \frac{V}{R} = \frac{24}{8} = 3 \ \text{A}$$

Voltage across the parallel group. The total current flows through the group, whose equivalent is 4 ohms:

$$V_p = 3 \times 4 = 12 \ \text{V}$$

Branch currents. Both branches see 12 V:

$$I_{6} = \frac{12}{6} = 2 \ \text{A} \qquad I_{12} = \frac{12}{12} = 1 \ \text{A}$$

Check. The branch currents sum to 3 A, which is the total current. And the voltage across the series resistor is $3 \times 4 = 12$ V, which added to the 12 V across the group gives 24 V, the supply. Both checks pass.

Those two checks took about fifteen seconds and confirm the entire question.

Preparing for this paper

Work past papers rather than textbook exercises. A textbook exercise announces its topic by which chapter it is in. An exam question does not tell you whether you are looking at a series, parallel or combination circuit, and recognising the shape is the skill being tested.

Then mark yourself against the official memo. In trade theory the memo often awards a mark for the formula stated before substitution, which means writing $R_T = R_1 + R_2$ before putting numbers in is worth doing even when it feels unnecessary.

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The short version