Does the PE — Electrical & Computer 2025 paper come with solutions?

Yes. Every question on this PE — Electrical & Computer past paper includes a step-by-step solution, plus instant AI feedback when you attempt it on Kekkei.

How many marks is the PE — Electrical & Computer PE — Electrical & Computer 2025 paper?

The PE — Electrical & Computer PE — Electrical & Computer 2025 paper carries 80 full marks and is meant to be completed in 480 minutes, across 10 questions.

Is practising this PE — Electrical & Computer past paper free?

Yes — reading and attempting this PE — Electrical & Computer past paper on Kekkei is completely free.

PE — Electrical & Computer PE — Electrical & Computer Practice Test 2025

PE — Electrical & ComputerModel questions

80marks

480min

10questions

This is the official PE — Electrical & Computer PE — Electrical & Computer question paper for 2025, as set in the Model questions examination. It carries 80 full marks and a time allowance of 480 minutes, across 10 questions. On Kekkei you can attempt this PE — Electrical & Computer past paper online with a timer, get instant AI feedback and step-by-step solutions, and track the topics where you lose marks — completely free. Whether you are revising for your PE — Electrical & Computer PE — Electrical & Computer exam or solving previous years' question papers, this 2025 paper is a great way to practise under real exam conditions.

Level	PE — Electrical & Computer
Subject	PE — Electrical & Computer
Year	2025 BS
Exam session	Model questions
Full marks	80
Time allowed	480 minutes
Questions	10, all with step-by-step solutions

Electrical & Computer Engineering

Select the best answer.

10 questions·1 mark each

1Multiple choice1 mark

Synchronous generator: $S = 100\,\text{MVA}$ , $\text{pf} = 0.85$ lagging. Real power:

P = S \times \text{pf}

a
$70\,\text{MW}$
b
$85\,\text{MW}$
c
$100\,\text{MW}$
d
$117.6\,\text{MW}$

Correct answer: b

$85\,\text{MW}$

P = 100 \times 0.85 = 85\,\text{MW}

power-generation

2Multiple choice1 mark

345 kV line, $Z_c = 300\,\Omega$ . Surge impedance loading:

\text{SIL} = V^2/Z_c

a
$199\,\text{MW}$
b
$297\,\text{MW}$
c
$397\,\text{MW}$
d
$500\,\text{MW}$

Correct answer: c

$397\,\text{MW}$

\text{SIL} = \frac{(345{,}000)^2}{300} = 3.967 \times 10^8\,\text{W} \approx 397\,\text{MW}

transmission

3Multiple choice1 mark

Transformer: $500\,\text{kVA}$ , $13.2\,\text{kV}/480\,\text{V}$ . Primary current:

I_1 = S/V_1

a
$18.9\,\text{A}$
b
$37.9\,\text{A}$
c
$75.8\,\text{A}$
d
$1{,}042\,\text{A}$

Correct answer: b

$37.9\,\text{A}$

I_1 = 500{,}000 / 13{,}200 = 37.9\,\text{A}

distribution

4Multiple choice1 mark

Three-phase fault: $I_f = 10\,\text{kA}$ , $V_{LL} = 13.8\,\text{kV}$ . Fault MVA:

\text{MVA}_{fault} = \sqrt{3} \times V_{LL} \times I_f

a
$138\,\text{MVA}$
b
$239\,\text{MVA}$
c
$345\,\text{MVA}$
d
$414\,\text{MVA}$

Correct answer: b

$239\,\text{MVA}$

\text{MVA} = 1.732 \times 13{,}800 \times 10{,}000 = 239\,\text{MVA}

protection

5Multiple choice1 mark

Induction motor: $N_s = 1{,}800\,\text{rpm}$ , $s = 3\%$ . Rotor speed:

N_r = N_s(1 - s)

a
$1{,}710\,\text{rpm}$
b
$1{,}746\,\text{rpm}$
c
$1{,}764\,\text{rpm}$
d
$1{,}800\,\text{rpm}$

Correct answer: b

$1{,}746\,\text{rpm}$

N_r = 1800 \times 0.97 = 1{,}746\,\text{rpm}

motor-drives

6Multiple choice1 mark

CT ratio 600:5, primary current 450 A. Secondary current:

I_2 = I_1 \times 5/600

a
$2.50\,\text{A}$
b
$3.75\,\text{A}$
c
$5.00\,\text{A}$
d
$7.50\,\text{A}$

Correct answer: b

$3.75\,\text{A}$

I_2 = 450 \times 5/600 = 3.75\,\text{A}

instrumentation

7Multiple choice1 mark

Wind turbine: $D = 80\,\text{m}$ , $v = 12\,\text{m/s}$ , $\rho = 1.225$ , $C_p = 0.40$ :

P = \frac{1}{2}\rho A v^3 C_p

a
$853\,\text{kW}$
b
$1{,}066\,\text{kW}$
c
$1{,}706\,\text{kW}$
d
$2{,}133\,\text{kW}$

Correct answer: c

$1{,}706\,\text{kW}$

A = \pi(80)^2/4 = 5{,}026.5\,\text{m}^2

P = 0.5 \times 1.225 \times 5026.5 \times 1728 \times 0.40 \approx 1{,}706\,\text{kW}

power-generation

8Multiple choice1 mark

Transformer pu impedance base change: $Z_{pu} = 0.10$ at 50 MVA. New base 100 MVA, same voltage:

Z_{new} = Z_{old} \times S_{new}/S_{old}

a
$0.10\,\text{pu}$
b
$0.20\,\text{pu}$
c
$0.05\,\text{pu}$
d
$0.40\,\text{pu}$

Correct answer: b

$0.20\,\text{pu}$

Z_{new} = 0.10 \times 100/50 = 0.20\,\text{pu}

transmission

9Multiple choice1 mark

Distance relay: line $Z_L = 10 + j40\,\Omega$ , Zone 1 at 80%. Reach impedance:

a
$8 + j32\,\Omega$
b
$10 + j40\,\Omega$
c
$12 + j48\,\Omega$
d
$5 + j20\,\Omega$

Correct answer: a

$8 + j32\,\Omega$

Z_1 = 0.80(10 + j40) = 8 + j32\,\Omega

protection

10Multiple choice1 mark

VFD: motor rated 60 Hz / 460 V, operating at 40 Hz. Constant V/Hz voltage:

V = V_{rated} \times f/f_{rated}

a
$230\,\text{V}$
b
$307\,\text{V}$
c
$345\,\text{V}$
d
$460\,\text{V}$

Correct answer: b

$307\,\text{V}$

V = 460 \times 40/60 = 307\,\text{V}

Constant V/Hz maintains constant flux.

motor-drives

Frequently asked questions

Where can I find the PE — Electrical & Computer PE — Electrical & Computer question paper 2025?: The full PE — Electrical & Computer PE — Electrical & Computer 2025 (Model questions) question paper is available free on Kekkei. You can read every question online and attempt the paper under timed exam conditions.
Does the PE — Electrical & Computer 2025 paper come with solutions?: Yes. Every question on this PE — Electrical & Computer past paper includes a step-by-step solution, plus instant AI feedback when you attempt it on Kekkei.
How many marks is the PE — Electrical & Computer PE — Electrical & Computer 2025 paper?: The PE — Electrical & Computer PE — Electrical & Computer 2025 paper carries 80 full marks and is meant to be completed in 480 minutes, across 10 questions.
Is practising this PE — Electrical & Computer past paper free?: Yes — reading and attempting this PE — Electrical & Computer past paper on Kekkei is completely free.