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

Yes. Every question on this FE — 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 FE — Electrical & Computer FE — Electrical & Computer 2025 paper?

The FE — Electrical & Computer FE — Electrical & Computer 2025 paper carries 110 full marks and is meant to be completed in 360 minutes, across 10 questions.

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

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

FE — Electrical & Computer FE — Electrical & Computer Practice Test 2025

Q: Where can I find the FE — Electrical & Computer FE — Electrical & Computer question paper 2025?

The full FE — Electrical & Computer FE — 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.

FE — Electrical & ComputerModel questions

110marks

360min

10questions

This is the official FE — Electrical & Computer FE — Electrical & Computer question paper for 2025, as set in the Model questions examination. It carries 110 full marks and a time allowance of 360 minutes, across 10 questions. On Kekkei you can attempt this FE — 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 FE — Electrical & Computer FE — Electrical & Computer exam or solving previous years' question papers, this 2025 paper is a great way to practise under real exam conditions.

Level	FE — Electrical & Computer
Subject	FE — Electrical & Computer
Year	2025 BS
Exam session	Model questions
Full marks	110
Time allowed	360 minutes
Questions	10, all with step-by-step solutions

Electrical & Computer Engineering

Select the best answer.

10 questions·1 mark each

1Multiple choice1 mark

Three resistors $R_1 = 10\,\Omega$ , $R_2 = 20\,\Omega$ , $R_3 = 30\,\Omega$ in parallel. Equivalent resistance:

\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}

a
$5.45\,\Omega$
b
$10.0\,\Omega$
c
$20.0\,\Omega$
d
$60.0\,\Omega$

Correct answer: a

$5.45\,\Omega$

\frac{1}{R_{eq}} = \frac{6+3+2}{60} = \frac{11}{60} \Rightarrow R_{eq} = \frac{60}{11} \approx 5.45\,\Omega

circuit-analysis

2Multiple choice1 mark

Long straight conductor, $I = 10\,\text{A}$ , distance $r = 0.05\,\text{m}$ . Magnetic field intensity:

H = \frac{I}{2\pi r}

a
$15.9\,\text{A/m}$
b
$31.8\,\text{A/m}$
c
$63.7\,\text{A/m}$
d
$100.0\,\text{A/m}$

Correct answer: b

$31.8\,\text{A/m}$

H = \frac{10}{2\pi \times 0.05} = 31.83\,\text{A/m}

electromagnetics

3Multiple choice1 mark

Signal sampled at $f_s = 8{,}000\,\text{Hz}$ . By Nyquist theorem, maximum representable frequency is:

a
$2{,}000\,\text{Hz}$
b
$4{,}000\,\text{Hz}$
c
$8{,}000\,\text{Hz}$
d
$16{,}000\,\text{Hz}$

Correct answer: b

$4{,}000\,\text{Hz}$

f_{\max} = \frac{f_s}{2} = \frac{8{,}000}{2} = 4{,}000\,\text{Hz}

signal-processing

4Multiple choice1 mark

The Boolean expression $\overline{A \cdot B}$ is equivalent to (De Morgan's theorem):

a
$A + B$
b
$\overline{A} + \overline{B}$
c
$\overline{A} \cdot \overline{B}$
d
$A \cdot B$

Correct answer: b

$\overline{A} + \overline{B}$

De Morgan's theorem: $\overline{A \cdot B} = \overline{A} + \overline{B}$ . The complement of AND is the OR of the complements.

digital-systems

5Multiple choice1 mark

Single-phase load: $P = 5\,\text{kW}$ , $\text{pf} = 0.8$ lagging, $V = 240\,\text{V}$ . Apparent power and current:

S = P/\text{pf}, \quad I = S/V

a
$S = 5.0\,\text{kVA}$ , $I = 20.8\,\text{A}$
b
$S = 6.25\,\text{kVA}$ , $I = 26.0\,\text{A}$
c
$S = 4.0\,\text{kVA}$ , $I = 16.7\,\text{A}$
d
$S = 8.0\,\text{kVA}$ , $I = 33.3\,\text{A}$

Correct answer: b

$S = 6.25\,\text{kVA}$ , $I = 26.0\,\text{A}$

S = 5000/0.8 = 6250\,\text{VA} = 6.25\,\text{kVA}

I = 6250/240 = 26.0\,\text{A}

power-systems

6Multiple choice1 mark

Inverting op-amp: $R_f = 100\,\text{k}\Omega$ , $R_{in} = 10\,\text{k}\Omega$ . Closed-loop gain:

A_v = -R_f/R_{in}

a
$-5$
b
$-10$
c
$-20$
d
$+10$

Correct answer: b

$-10$

A_v = -100/10 = -10

Negative sign indicates phase inversion.

electronics

7Multiple choice1 mark

Transfer function $G(s) = \frac{25}{s^2 + 6s + 25}$ . Natural frequency $\omega_n$ and damping ratio $\zeta$ :

\omega_n^2 = 25,\quad 2\zeta\omega_n = 6

a
$\omega_n = 5\,\text{rad/s}$ , $\zeta = 0.6$
b
$\omega_n = 25\,\text{rad/s}$ , $\zeta = 0.3$
c
$\omega_n = 5\,\text{rad/s}$ , $\zeta = 1.2$
d
$\omega_n = 6\,\text{rad/s}$ , $\zeta = 0.5$

Correct answer: a

$\omega_n = 5\,\text{rad/s}$ , $\zeta = 0.6$

\omega_n = \sqrt{25} = 5,\quad \zeta = 6/(2\times 5) = 0.6

Underdamped system ( $0 < \zeta < 1$ ).

control-systems

8Multiple choice1 mark

Series RLC: $R = 100\,\Omega$ , $L = 0.1\,\text{H}$ , $C = 10\,\mu\text{F}$ . Resonant frequency:

f_0 = \frac{1}{2\pi\sqrt{LC}}

a
$79.6\,\text{Hz}$
b
$159.2\,\text{Hz}$
c
$318.3\,\text{Hz}$
d
$1{,}000\,\text{Hz}$

Correct answer: b

$159.2\,\text{Hz}$

f_0 = \frac{1}{2\pi\sqrt{0.1 \times 10^{-5}}} = \frac{1}{2\pi \times 10^{-3}} = \frac{1000}{2\pi} = 159.2\,\text{Hz}

circuit-analysis

9Multiple choice1 mark

Parallel plate capacitor: $A = 0.01\,\text{m}^2$ , $d = 0.001\,\text{m}$ , $\epsilon_r = 4$ , $\epsilon_0 = 8.854 \times 10^{-12}$ :

C = \frac{\epsilon_r \epsilon_0 A}{d}

a
$88.5\,\text{pF}$
b
$177\,\text{pF}$
c
$354\,\text{pF}$
d
$708\,\text{pF}$

Correct answer: c

$354\,\text{pF}$

C = \frac{4 \times 8.854 \times 10^{-12} \times 0.01}{0.001} = 354\,\text{pF}

electromagnetics

10Multiple choice1 mark

A 10-bit ADC, input range 0-5 V. Resolution:

\Delta V = \frac{V_{range}}{2^n - 1}

a
$2.44\,\text{mV}$
b
$4.88\,\text{mV}$
c
$9.77\,\text{mV}$
d
$19.53\,\text{mV}$

Correct answer: b

$4.88\,\text{mV}$

\Delta V = \frac{5}{1023} = 4.88\,\text{mV}

digital-systems

Frequently asked questions

Where can I find the FE — Electrical & Computer FE — Electrical & Computer question paper 2025?: The full FE — Electrical & Computer FE — 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 FE — Electrical & Computer 2025 paper come with solutions?: Yes. Every question on this FE — 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 FE — Electrical & Computer FE — Electrical & Computer 2025 paper?: The FE — Electrical & Computer FE — Electrical & Computer 2025 paper carries 110 full marks and is meant to be completed in 360 minutes, across 10 questions.
Is practising this FE — Electrical & Computer past paper free?: Yes — reading and attempting this FE — Electrical & Computer past paper on Kekkei is completely free.