About Electrical Engineering
Electrical engineering questions in a placement or recruitment paper are numerical: a few components, one relation, one answer in the right unit. This topic covers the whole of that ground on one page - direct current circuits, capacitors, alternating supplies, transformers, motors and cells - rather than splitting it into a separate quiz for each, so that you can practise the subject as it is actually examined.
What you need to understand
- Ohm's law: V = I R, so the current is the voltage divided by the resistance, and the resistance is the voltage divided by the current.
- Resistances end to end (in series) add: R = R1 + R2 + R3. The current through each is the same.
- Resistances side by side (in parallel) add as reciprocals: 1/R = 1/R1 + 1/R2 + 1/R3. The combined value is always smaller than the smallest of them, and the voltage across each is the same.
- Working a network: reduce the innermost parallel group to a single resistance first, then add the series parts. Never combine a series resistor with a parallel group in one step.
- Power in a direct current circuit is P = V I, which also appears as I squared R and as V squared over R. Energy is power multiplied by time; in kilowatt hours it is the power in kilowatts times the hours.
- Capacitors combine the opposite way round to resistors: in parallel they add (C = C1 + C2), and in series they add as reciprocals. The reason is that two capacitors side by side present the same plate area as the two put together.
- The energy a charged capacitor holds is E = C V squared / 2. The half is the part most often dropped.
- The voltage across a capacitor cannot change instantly, and the current through an inductor cannot change instantly. Transient questions usually turn on one of those two facts.
- For a sine wave the root mean square value is the peak divided by the square root of 2, about 1.414. RMS is the value that gives the same heating effect as a steady voltage of that size, and it is the value meters read.
- Reactance is the opposition a capacitor or an inductor offers to alternating current. For a capacitor Xc = 1 / (2 pi f C), which falls as the frequency rises; for an inductor Xl = 2 pi f L, which rises with frequency.
- In an ideal transformer the volts per turn is the same on both windings, so Vs/Vp = Ns/Np and the currents go the other way: a step-up transformer raises the voltage and lowers the current.
- An induction motor always runs a little slower than the field that drives it. The field turns at the synchronous speed 120 f / P, and the slip is the difference between that and the rotor speed, as a percentage of the synchronous speed.
- Efficiency is output over input, as a percentage. The difference between the two is lost as heat, and a machine can never be above 100 per cent.
- A cell's internal resistance is in series with everything else, so the terminal voltage is the emf less the internal drop: V = E - I r. That is why a battery reads low while it is working.
- In a three phase star connection the line voltage is the square root of 3 times the phase voltage, and the line current equals the phase current. In delta it is the other way round.
How to work through these questions
- Write down what is given, each with its unit, and then the one relation that connects those quantities to what is asked. Most questions in this topic need exactly one relation.
- Convert to base units before substituting. Microfarads to farads and kilowatts to watts are the two conversions where marks are lost most often, and both are factors of a million or a thousand.
- For a network, mark the series parts and the parallel parts on the diagram before doing any arithmetic. Reducing the wrong group first is the commonest cause of a wrong answer.
- Sanity check the answer against the physics: a parallel combination must be smaller than the smallest resistance, an efficiency must be under 100 per cent, an induction motor must run below its synchronous speed, and a terminal voltage must be below the emf.
- Check the unit the question asks for. Power in watts and energy in kilowatt hours are different questions about the same appliance.
- If the answer is not among the options, re-read the question for the word series or parallel - it is the single most common misreading in this topic.
Mistakes that cost marks
- Adding resistances that are side by side, or taking the reciprocal of resistances that are end to end. The two rules are exact opposites.
- Combining a series resistor with a parallel group by adding all three values.
- Using the peak value of an alternating supply where the RMS value is needed, or the other way round. Power calculations use RMS.
- Forgetting the half in the energy of a charged capacitor, which doubles the answer.
- Turning the turns ratio of a transformer the wrong way up. More turns on the secondary means a higher secondary voltage, not a lower one.
- Reporting the rotor speed, or the synchronous speed, as the slip. Slip is the difference between them, as a percentage.
- Leaving a power in watts and dividing by 100 rather than 1000 when the answer is wanted in kilowatt hours.
- Forgetting that the internal resistance of a cell takes part of the emf, so the terminal voltage is lower than the emf whenever current flows.
Worked example
A 6 ohm resistor is connected in parallel with a 3 ohm resistor, and the pair is joined in series with a 4 ohm resistor across a 12 V supply. What current flows through the 4 ohm resistor?
- Reduce the parallel pair first. 1/R = 1/6 + 1/3 = 1/6 + 2/6 = 3/6, so R = 2 ohms.
- Check it against the physics: 2 ohms is smaller than either 6 or 3, which is what a parallel combination must be.
- Add the series resistor: the whole circuit is 4 + 2 = 6 ohms.
- The current follows from Ohm law: I = V / R = 12 / 6 = 2 A.
- That current passes through the 4 ohm resistor on its way, so the answer is 2 A. The same current splits between the 6 ohm and the 3 ohm branches, taking 1.33 A and 0.67 A - and those two add back to 2 A, which is the check that nothing has been missed.
Answer: 2 A
Practice questions with answers
A few Electrical Engineering questions with the full solution shown, so you can see
how the method is applied before you attempt the timed set.
Question 1
Three resistors of 32 ohm, 96 ohm and 48 ohm are joined in parallel. What is the equivalent resistance, in ohms?
Answer: Option D — with explanation
Resistances side by side add as reciprocals: the combined resistance is 1 divided by (1/R1 + 1/R2 + 1/R3).
For 32, 96 and 48 ohms that is 1/32 + 1/96 + 1/48, and the reciprocal of that sum is 16 ohms. Adding the three values as though they were end to end gives 176, which is larger than the smallest of them - impossible for resistances side by side, where the combined value has to be smaller than any one of them on its own.
Common mistakes
- 176/3 is not the answer: 176/3
- 48 is not the answer: 48
- 176 is not the answer: 176
Question 2
A resistor of 13 ohm is joined in series with two resistors of 51 ohm and 102 ohm that are in parallel with each other. What is the total resistance, in ohms?
-
A
7191/64
-
B
47
-
C
166
-
D
442/47
Answer: Option B — with explanation
Work from the inside out. The two resistors side by side combine as 1/(1/51 + 1/102), which is 34 ohms.
That leaves the 13 ohm resistor in series with 34 ohms, and resistances end to end simply add: 13 + 34 = 47 ohms.
Adding all three values together, or combining the single resistor with the pair as though they were all side by side, are the two mistakes that produce the other answers offered.
Common mistakes
- 442/47 is not the answer: 442/47
- 166 is not the answer: 166
- 7191/64 is not the answer: 7191/64
Question 3
An appliance draws a current of 10 A from a 200 V supply. What is its power, in watts?
Answer: Option C — with explanation
Power in a direct current circuit is the voltage across the appliance times the current through it: P = V x I.
Here that is 200 x 10 = 2000 watts. Adding the voltage and the current, or dividing one by the other, gives a number with the wrong units as well as the wrong value.
Common mistakes
- 1/20 is not the answer: 1/20
- 20 is not the answer: 20
- 210 is not the answer: 210
Question 4
A 20 microfarad capacitor and a 60 microfarad capacitor are connected in series with each other. What is their combined capacitance, in microfarads?
Answer: Option B — with explanation
Capacitors in series combine the opposite way to resistors: the combined capacitance is 1 divided by (1/C1 + 1/C2).
For 20 and 60 microfarads that comes to 15 microfarads, which is smaller than either one - as it must be, since capacitors in series hold less charge between them than the smaller of the two on its own. Adding them gives 80, which is what connecting them side by side would give.
Common mistakes
- 20 is not the answer: 20
- 80 is not the answer: 80
- 30 is not the answer: 30
Question 5
An ideal transformer has 400 turns on its primary and 4000 turns on its secondary. With 120 V across the primary, what is the secondary voltage?
Answer: Option A — with explanation
In an ideal transformer the volts per turn is the same on both windings, so the voltages are in the same ratio as the turns: Vs/Vp = Ns/Np.
Here Ns/Np = 4000/400, and 120 x 4000/400 = 1200 V. Turning the ratio the other way round gives 12 V, which is the mistake to watch for: more turns on the secondary means a bigger voltage, not a smaller one.
Common mistakes
- 120 is not the answer: 120
- 12 is not the answer: 12
- 1080 is not the answer: 1080
Frequently asked questions
Which formula should I reach for when a question mentions both voltage and current?
Power, P = V I. If it gives you the resistance instead, the same power appears as I squared R, and if it gives you the voltage and the resistance it appears as V squared over R. All three are the same relation rearranged.
How do I know whether a combination is series or parallel?
Follow the current. If the same current must pass through both components one after the other, they are in series. If the current splits between them at a junction, they are in parallel. Redrawing the circuit with the components laid out clearly usually settles it.
Why is the combined resistance of a parallel pair smaller than either one?
Because you have given the current a second way round. Two paths side by side carry more current than one for the same voltage, and more current for the same voltage means less resistance.
Do I use the peak or the RMS value of an alternating supply?
RMS for anything to do with power or heating, which is most questions. RMS is the steady voltage that would heat a resistor by the same amount, so it is the one that goes into P = V I. Use the peak only when the question asks about insulation or the maximum value.
What is the difference between slip and efficiency in a motor?
Slip is about speed and efficiency is about power. Slip compares the rotor speed with the speed of the field that drives it; efficiency compares the mechanical power delivered with the electrical power taken in. A motor can have a small slip and still be inefficient.
How many of these can I expect in a placement paper?
Usually a handful within the technical section. They are quick marks if the relations are known, and they are the reason this topic is worth practising with a timer rather than reading through.
