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100+ Free Pancyprian Electrical Engineering III (Practical) Practice Questions

Cyprus Pancyprian Access Examination Technical School Electrical Engineering III, Practical Direction (ΗΛΕΚΤΡΟΛΟΓΙΑ ΙΙΙ Τ.Σ. (Π.Κ.), code 509) practice questions are available now; exam metadata is being verified.

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2026 Statistics

Key Facts: Pancyprian Electrical Engineering III (Practical) Exam

2.5 hours

Written paper duration (08:00–10:30) per official timetable

Cyprus Examinations Service

Code 509

Subject code for ΗΛΕΚΤΡΟΛΟΓΙΑ ΙΙΙ Τ.Σ. (Π.Κ.) practical direction

Cyprus Examinations Service exam guide

Code 409

Distinct theoretical-direction counterpart (Θ.Κ.) - separate paper

Cyprus Examinations Service exam guide

0-20

Marking scale; feeds the access/ranking grade with no pass/fail mark

Cyprus Examinations Service

EUR 25

2026 fee per access subject (EUR 50 only for code 032)

Examinations Service / school SEA guidance

16 June 2026

2026 exam date for code 509 per the official timetable

Cyprus Examinations Service timetable

100 free MCQs covering Pancyprian Electrical Engineering III practical direction (code 509): DC/AC circuits, power factor, three-phase and ΑΗΚ distribution, with heavy calculation practice. This is an English-language study adaptation - the official exam is a 2.5-hour Greek written paper with constructed-response tasks, so practise full written solutions separately.

Sample Pancyprian Electrical Engineering III (Practical) Practice Questions

Try these sample questions to test your Pancyprian Electrical Engineering III (Practical) exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1What is the SI unit of electric resistance?
A.Ampere (A)
B.Ohm (Ω)
C.Volt (V)
D.Watt (W)
Explanation: Resistance is measured in ohms (Ω). One ohm is the resistance that produces a one-volt drop when one ampere flows through it.
2Ohm's law relating voltage U, current I and resistance R is:
A.U = I / R
B.U = I × R
C.U = I + R
D.U = R / I
Explanation: Ohm's law states U = I × R (or I = U/R). Voltage across a resistor equals current times resistance.
3An ammeter reads 5 A and a voltmeter across the same resistive load reads 30 V. The resistance is:
A.5 Ω
B.6 Ω
C.30 Ω
D.150 Ω
Explanation: R = U/I = 30 V / 5 A = 6 Ω. This matches the calculation style used in official Pancyprian 509 papers.
4Two resistors R1 = 2 Ω and R2 = 4 Ω are connected in series across a DC source. Which statement is true?
A.They have the same voltage across each resistor
B.They carry the same current
C.They are connected in parallel
D.The smaller resistor dissipates more power
Explanation: In series, the same current flows through every element. Voltages divide in proportion to resistance, so they are not equal.
5Two resistors R1 = 2 Ω and R2 = 4 Ω are connected in parallel across the same DC source. Which statement is true?
A.They carry equal currents
B.They have the same voltage across their terminals
C.Their equivalent resistance is 6 Ω
D.Current flows only through the 2 Ω branch
Explanation: Parallel branches share the same terminal voltage. Currents then split inversely with resistance.
6Three 9 Ω resistors are connected in parallel across 27 V. The total current drawn from the source is:
A.1 A
B.3 A
C.6 A
D.9 A
Explanation: Each branch current is 27/9 = 3 A. Three identical parallel branches give Itotal = 3 × 3 = 9 A. Equivalently Req = 3 Ω, so I = 27/3 = 9 A.
7Kirchhoff's current law (KCL) states that at any node:
A.The sum of voltages around a closed loop is zero
B.The algebraic sum of currents entering equals the algebraic sum leaving
C.Power into a node always equals voltage times resistance
D.Current is constant only in series capacitors
Explanation: KCL is charge conservation at a node: currents in equal currents out (algebraic sum is zero).
8At a node, currents I1 = 7 A and I2 = 8 A enter, while I3 = 6 A and I4 = 4 A leave. The remaining branch current I5 leaving the node is:
A.1 A
B.5 A
C.9 A
D.25 A
Explanation: KCL: currents in = currents out. 7 + 8 = 15 A entering. Outgoing already 6 + 4 = 10 A, so I5 = 15 − 10 = 5 A leaving.
9Kirchhoff's voltage law (KVL) states that around any closed loop:
A.The sum of currents is zero
B.The algebraic sum of voltage rises and drops is zero
C.Resistance must be constant
D.Power factor must equal 1
Explanation: KVL follows from a conservative electric field in lumped circuits: the net voltage around a closed path is zero.
10Resistors 3 Ω and 6 Ω in series are fed by 36 V. The voltage across the 6 Ω resistor is:
A.12 V
B.18 V
C.24 V
D.36 V
Explanation: Voltage divider: U6 = 36 × (6/(3+6)) = 36 × (6/9) = 24 V. Current is 36/9 = 4 A, so U6 = 4 × 6 = 24 V.

About the Pancyprian Electrical Engineering III (Practical) Practice Questions

Verified exam format metadata for Cyprus Pancyprian Access Examination Technical School Electrical Engineering III, Practical Direction (ΗΛΕΚΤΡΟΛΟΓΙΑ ΙΙΙ Τ.Σ. (Π.Κ.), code 509) is pending. The practice questions above remain available while official exam length, timing, passing score, fee, and administrator details are reviewed.