Mechanical Reasoning Guide

Electric Circuits

Current flows only round a complete loop, and shares itself out by the resistances it meets. Mechanical aptitude tests ask what a bulb does next. Can you tell?

Which bulb lights?

Eleven rules are all it takes. Read on, then try the test.

Where it appears

Questions about batteries, bulbs and switches belong to mechanical reasoning (or mechanical comprehension) tests, beside the gears, levers and pulleys. Each shows a picture and asks what happens next. You will meet them in:

Bennett Mechanical Comprehension TestThe classic test of mechanical understanding, used in hiring for technical and engineering roles. Electricity is among its everyday-physics topics.

Differential Aptitude Tests (DAT)Its Mechanical Reasoning subtest, used in school and career guidance.

ASVABThe Mechanical Comprehension and Electronics Information sections of the US military's entrance test.

Technician and apprenticeship testsMechanical and electrical aptitude tests for maintenance, manufacturing and trade roles, and the sections of employers' aptitude batteries.

Those tests usually give a picture with no numbers to count and no time to spare. Here every picture carries its numbers, so each answer is exact, and the questions grow harder as you get better. It trains the habit those questions reward: find the path the current takes, then how it is shared.

Bulbs and switches: five rules

A battery pushes current round a loop. A bulb lights when current flows through it. A closed switch is just a wire; an open switch (its lever up) is a gap. The question is which bulb lights, how many are lit, which is brightest, or what happens to a bulb when something changes.

1A bulb lights only round a complete loop

Follow the wire from the battery's plus, through the bulb, back to its minus. Switches in line with the bulb must all be closed. Switches side by side give the current a choice: one closed switch is enough.

The slip is to look at the bulb and ignore the switches. Count every switch on the path and ask whether it is a wire or a gap.

A: both closed in line, lit. B: one open in line, dark. C: both open side by side, dark. D: one closed side by side, lit

2An open switch cuts its branch

When the loop splits, each branch is its own path. A bulb in a branch with an open switch is dark, but the bulbs on the other paths are not affected: they still have a complete loop.

Counting every bulb whatever the switches do is the slip. Go bulb by bulb and ask: is there a complete path through this one?

The open switch cuts C's branch: C is dark, A and B are lit: 2 bulbs

3A wire across a bulb puts it out

A closed switch side by side with a bulb is an easy path with no resistance, so the current goes round the bulb. The bulb stays dark though the circuit is complete.

A complete circuit is not enough: the bulb itself must have current flowing through it. With a wire across it, none does.

The closed switch is a wire across B: B is dark, A is lit: 1 bulb

4Brightness is current × current × resistance

A bulb's brightness is its power. Work out the current through each bulb: bulbs in line carry the same current; bulbs side by side share it. Then the power is the current squared times the bulb's resistance.

power = current × current × resistance

Do not pick the bulb nearest the battery just because it is nearest: current is not used up on the way round. In the test it is never the answer. When the bulbs differ, the biggest resistance and the smallest are traps too.

C carries all 2 A: 16 W. A and B share it, 1 A each: 4 W each. C is brightest

5When something changes, work the bulb out before and after

A switch closes, a bulb is added beside or in line with another, or a wire is put across one: what happens to a different bulb? Work out the bulb's power before and after. It may get brighter, dimmer, go out, or stay the same; each of the four is right about equally often, so no guess is safe.

A new path lowers the circuit's resistance, so the battery gives more current, but a bulb beside the new path can end up with less. A bulb added in line makes its own path harder, which can brighten a bulb beside it. Do not guess: work the powers out.

Closing the switch gives C's branch a path: B falls from 9 W before to 4 W after, so it gets dimmer

Amps and volts: six rules

Here the parts are resistors, in ohms (Ω), on a battery of so many volts (V). The question is the current in amps (A), the resistance of a block, the voltage across a marked part, or the power in watts (W).

6Ohm's law

The current through a resistor is the voltage across it divided by its resistance.

current = volts ÷ ohms

Divide, never multiply, and do not take the ohms away from the volts: a bigger resistance means less current.

20 V across 10 Ω: 20 ÷ 10 = 2 A

7In line, the resistances add

Resistors one after another make the current's way harder. Add them.

R = R + R + R

Using the side-by-side rule for resistors in line is the slip, and so is taking only the largest.

3 + 6 + 1 = 10 Ω

8Side by side, the total is smaller than any of them

Resistors side by side give the current more ways round, so the total is less than the smallest. Add the 1 ÷ R of each, then turn the sum back over. For two, multiply them and divide by their sum.

1 ÷ R = 1 ÷ R + 1 ÷ R

Adding them, taking only the smallest, averaging and multiplying are the slips. A total bigger than the smallest resistor is always wrong.

15 Ω and 10 Ω side by side: 15 × 10 ÷ 25 = 6 Ω

9In line, the voltage is shared in proportion

First find the current from the total resistance. Then the voltage across a part is the current times that part's resistance. A bigger resistance takes a bigger share.

The whole battery voltage is a slip (the other parts share it), and so is sharing it equally when the resistances differ.

9 Ω in all, so 9 ÷ 9 = 1 A; across the marked 5 Ω: 1 A × 5 Ω = 5 V

10Side by side, the current splits

Find the whole current, then the voltage across the pair. Both branches have that voltage, so each branch's current is the pair's voltage divided by its own resistance.

The battery's whole current is a slip (it splits between the branches), and so is splitting it equally or giving the other branch's current.

3 A through the pair's 2 Ω makes 6 V; the marked 3 Ω: 6 ÷ 3 = 2 A

11Power is current squared times resistance

Find the current through the marked resistor (rules 6 to 10). Its power is that current times itself times its resistance.

power = current × current × ohms

Current times resistance is a voltage, not a power. The battery's voltage times this resistor's current, and the whole circuit's power, are slips too.

8 A through the marked 4 Ω: 8 × 8 × 4 = 256 W

Two questions, solved

1A question about bulbs

How many bulbs are lit?

The question

Try it first, then open the answer.

Show the answer: 3

Follow the current from the battery: a closed switch is a wire, an open switch a gap.
Bulb B is dark: an open switch in its path cuts the loop.
Bulb E is dark: an open switch in its path cuts the loop.
Bulb C is on a complete path: lit.
Bulb D is on a complete path: lit.
Bulb A is on a complete path: lit.
3 bulbs are lit.

The other options, and the slip each one is:

0 Not the number of lit bulbs2 Not the number of lit bulbs5 Every bulb counted, whatever the switches do

The working, on the picture

2A question about amps and volts

What is the voltage across the marked part?

The question

Try it first, then open the answer.

Show the answer: 25 V

First reduce the network to one resistance, the way the battery sees it.
Total resistance 10 + 8 = 18 Ω; current 45 ÷ 18 = 2.5 A
Across the marked part: 2.5 A × 10 Ω = 25 V

The other options, and the slip each one is:

20 V The voltage across another part23 V Not what Ohm's law gives24 V Not what Ohm's law gives

The working, on the picture

How the practice test runs

3:00 on the clock, and it only runs while a question is on screen.
Levels: each question is about bulbs or about amps and volts, drawn half and half. Bulbs climb from which of four little circuits lights its bulb, to how many of four bulbs are lit, which of four identical bulbs is brightest, what happens to a bulb when a switch closes, how many of five bulbs are lit with three switches, which of four different bulbs is brightest, and what happens when a bulb is added or a wire put across one. Amps and volts climb from the current through one resistor, to resistors in line, resistors side by side, the battery's current, the voltage across a marked part, the current in one of a pair of branches, and the power a resistor uses.
The ladder: until your first miss, every right answer jumps a whole level; after it, three right answers in a row jump a level and a miss drops one. Your level carries over to your next session.
Scoring: a right answer earns the question's points (harder questions pay more), a wrong one costs half as many, so guessing does not pay.
Every question is new, and the review after a session works every answer out on its picture and names the slip behind each wrong option.

Try the Electric Circuits test

Free and adaptive, no account needed. Start with which bulb lights and climb to resistors in line and side by side, voltage and power.

Start the Electric Circuits test