What is the exam pattern and passing mark for GTU Numerical Methods for Electrical Engineering (BE03000191)?

The GTU theory examination carries 70 total marks with a duration of 2.5 Hours. Minimum passing criteria is 23 marks (33%). The paper consists of 5 main questions with official marking breakdown and internal choices.

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Gujarat Technological UniversityWinter 2025 examination

GTU BE03000191 Numerical Methods for Electrical Engineering Winter 2025 Solved Paper & Solutions

B.E. · Electrical Engineering · Semester 3 · Subject code BE03000191

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Total: 70 marks
To pass: 23 marks (33%)
Duration: 2.5 Hours 10:30 AM – 1:00 PM
Questions: 5 with internal choice

2 of 25 questions on this paper were also asked in other years. 14 more are types of question GTU sets again with new data. Each is marked below with the years it appeared.See all 16 repeated BE03000191 questions, most asked first

Q.1

14 Marks

(a)

3 marks

Define initial value problem. Give an example.

(b)

4 marksSame type asked in 2 of 2 papers

Solve the following system of linear equations using Gauss-Jacobi Method $10x + 2y + z = 9$ $2x + 20y - 2z = -44$ $-2x + 3y + 10z = 22$

Also asked in Summer 2026 Q3c OR

Step 1 Rearrange.

In the Gauss-Jacobi method, the new values come only from the old values of the previous iteration.

x = \frac{9 - 2y - z}{10}, \qquad y = \frac{-44 - 2x + 2z}{20}, \qquad z = \frac{22 + 2x - 3y}{10}

Step 2 Iterate

from $x = y = z = 0$ .

Iteration	$x$	$y$	$z$
1	0.9000	-2.2000	2.2000
2	1.1200	-2.0700	3.0400
3	1.0100	-2.0080	3.0450
4	0.9971	-1.9965	3.0044
5	0.9989	-1.9993	2.9984
6	1.0000	-2.0000	2.9996

Answer

$x \approx 1,\ y \approx -2,\ z \approx 3$ . Check: $10 - 4 + 3 = 9$ .

(c)

7 marksSame type asked in 2 of 2 papers

Apply Secant method to find the root of the equation

x^3 + x^2 - 3x - 3 = 0

correct up to five decimal places.

Also asked in Summer 2026 Q2c

Step 1 Formula.

Let $f(x) = x^3 + x^2 - 3x - 3$ . Since $f(1) = -4$ and $f(2) = 3$ , a root lies between 1 and 2. Take $x_0 = 1$ and $x_1 = 2$ . The secant method uses the last two points and no derivative:

x_{n+1} = x_n - \frac{f(x_n)\,(x_n - x_{n-1})}{f(x_n) - f(x_{n-1})}

Step 2 Iterations.

n	$x_{n-1}$	$x_n$	$f(x_{n-1})$	$f(x_n)$	$x_{n+1}$
1	1.00000	2.00000	-4.00000	3.00000	1.57143
2	2.00000	1.57143	3.00000	-1.36443	1.70541
3	1.57143	1.70541	-1.36443	-0.24775	1.73514
4	1.70541	1.73514	-0.24775	0.02926	1.73200
5	1.73514	1.73200	0.02926	-0.00052	1.73205
6	1.73200	1.73205	-0.00052	0.00000	1.73205

Step 3 Check.

Iterations 5 and 6 agree to five decimal places. Also, $x^3 + x^2 - 3x - 3 = (x + 1)(x^2 - 3)$ , so this root is $\sqrt{3} = 1.73205$ .

Answer

the root is $x \approx 1.73205$ .

Q.2

14 Marks

(a)

3 marks

State the primary criterion for selecting the next interval in the False position Method and write down the iteration formula for the Newton-Raphson Method.

(b)

4 marksSame type asked in 2 of 2 papers

Find the root of the equation

x^3 - 4x - 1 = 0

using the Bisection Method correct up to two decimal place.

Also asked in Summer 2026 Q2c OR

Let $f(x) = x^3 - 4x - 1$ . Since $f(2) = -1$ and $f(3) = 14$ , a root lies in $[2, 3]$ . In the bisection method, $c = \dfrac{a+b}{2}$ and the half with the change of sign is kept.

n	$a$	$b$	$c$	$f(c)$
1	2.00000	3.00000	2.50000	4.62500
2	2.00000	2.50000	2.25000	1.39062
3	2.00000	2.25000	2.12500	0.09570
4	2.00000	2.12500	2.06250	-0.47632
5	2.06250	2.12500	2.09375	-0.19644
6	2.09375	2.12500	2.10938	-0.05191
7	2.10938	2.12500	2.11719	0.02151
8	2.10938	2.11719	2.11328	-0.01530
9	2.11328	2.11719	2.11523	0.00308
10	2.11328	2.11523	2.11426	-0.00612
11	2.11426	2.11523	2.11475	-0.00152
12	2.11475	2.11523	2.11499	0.00078

After 12 iterations the root lies in $[2.11475, 2.11499]$ , and both ends round to $2.11$ .

Answer

the root is $x \approx 2.11$ .

(c)

7 marksSame type asked in 2 of 2 papers

Consider a 3-mesh ladder equations (A.I=V) using mesh current $I_1, I_2, I_3$ : $(Z_1 + Z_{12})I_1 - Z_{12}I_2 = V_1$ $-Z_{12}I_1 + (Z_2 + Z_{12} + Z_{23})I_2 - Z_{23}I_3 = V_2$ $-Z_{23}I_2 + (Z_3 + Z_{23})I_3 = V_3$ Where impedance $(\Omega)$ $Z_1 = 10, Z_2 = 8, Z_3 = 6, Z_{12} = 4, Z_{23} = 5$ , phasor voltages (volts) $V_1 = 10, V_2 = 5, V_3 = 8$ . Apply L-U decomposition method to solve the above system of linear equations.

Also asked in Summer 2026 Q1b

Step 1 Put the numbers in.

The equations are $A\mathbf{I} = \mathbf{V}$ . With $Z_1 = 10,\ Z_2 = 8,\ Z_3 = 6,\ Z_{12} = 4,\ Z_{23} = 5$ , the diagonal terms are $Z_1 + Z_{12} = 14$ , $Z_2 + Z_{12} + Z_{23} = 17$ and $Z_3 + Z_{23} = 11$ :

\begin{bmatrix} 14 & -4 & 0 \\ -4 & 17 & -5 \\ 0 & -5 & 11 \end{bmatrix}\begin{Bmatrix} I_1 \\ I_2 \\ I_3 \end{Bmatrix} = \begin{Bmatrix} 10 \\ 5 \\ 8 \end{Bmatrix}

Step 2 Write $A = LU$

with 1s on the diagonal of $L$ (Doolittle's method). The first row of $U$ is the first row of $A$ , which is $14,\ -4,\ 0$ . The other entries are found row by row. The matrix has a zero at the corner, so $l_{31} = 0$ .

l_{21} = \frac{-4}{14} = -0.285714, \qquad u_{22} = 17 - (-0.285714)(-4) = 15.857143, \qquad u_{23} = -5

l_{31} = 0, \qquad l_{32} = \frac{-5}{15.857143} = -0.315315, \qquad u_{33} = 11 - (-0.315315)(-5) = 9.423423

Step 3 Forward substitution

$L\mathbf{z} = \mathbf{b}$ :

z_1 = 10, \qquad z_2 = 5 + 0.285714(10) = 7.857143, \qquad z_3 = 8 + 0.315315(7.857143) = 10.477477

Step 4 Back substitution

$U\mathbf{I} = \mathbf{z}$ :

I_3 = \frac{10.477477}{9.423423} = 1.1119

I_2 = \frac{7.857143 + 5(1.1119)}{15.857143} = 0.8461

I_1 = \frac{10 + 4(0.8461)}{14} = 0.9560

Answer

$I_1 = 0.9560\ \text{A}$ , $I_2 = 0.8461\ \text{A}$ , $I_3 = 1.1119\ \text{A}$ . Check with the first equation: $14(0.9560) - 4(0.8461) = 10.00$ .

(c)

7 marksSame type asked in 2 of 2 papers

Apply Gauss-Siedel method, to solve the following system of linear equations:

$5x + y - z = 10$ $2x + 4y + z = 14$ $x + y + 8z = 20$

Also asked in Summer 2026 Q3c

Step 1 Rearrange.

The diagonal term is the largest in each equation, so the method converges.

x = \frac{10 - y + z}{5}, \qquad y = \frac{14 - 2x - z}{4}, \qquad z = \frac{20 - x - y}{8}

Step 2 Iterate

with $x = y = z = 0$ at the start. In Gauss-Seidel each new value is used at once.

Iteration	$x$	$y$	$z$
1	2.0000	2.5000	1.9375
2	1.8875	2.0719	2.0051
3	1.9866	2.0054	2.0010
4	1.9991	2.0002	2.0001
5	2.0000	2.0000	2.0000

Step 3 Check.

$5(2) + 2 - 2 = 10$ , $2(2) + 4(2) + 2 = 14$ and $2 + 2 + 16 = 20$ .

Answer

$x = 2,\ y = 2,\ z = 2$ .

Q.3

14 Marks

(a)

3 marks

Write the predictor and corrector formula for the Modified Euler Method.

(b)

4 marksSame type asked in 2 of 2 papers

Solve ordinary differential equation

\dfrac{dy}{dx} = x + y

with initial value

y(0) = 1

using Modified Euler Method with step size

h = 0.1

and compute

y\text{ for} x = 0.2

Also asked in Summer 2026 Q4c OR

(c)

7 marksSame type asked in 2 of 2 papers

Apply RK 4 method, to find approximate value of $y$ when $x = 0.2$ , for the ODE $10\dfrac{dy}{dx} = x^2 + y^2, y(0) = 1$ Use step size of $h = 0.1$ .

Also asked in Summer 2026 Q1c

Step 1 Setup.

Divide the equation by 10 to get $y' = f(x, y) = \dfrac{x^2 + y^2}{10}$ . Here $h = 0.1$ , $x_0 = 0$ and $y_0 = 1$ . Since $0.2 = 2h$ , two steps give $y(0.2)$ . In each step the fourth order Runge-Kutta method calculates four slopes and takes their weighted average. The slopes are $k_1 = hf(x_n, y_n)$ , $k_2 = hf(x_n + \frac{h}{2}, y_n + \frac{k_1}{2})$ , $k_3 = hf(x_n + \frac{h}{2}, y_n + \frac{k_2}{2})$ and $k_4 = hf(x_n + h, y_n + k_3)$ . Then $y_{n+1} = y_n + \frac{1}{6}(k_1 + 2k_2 + 2k_3 + k_4)$ .

Step 2 First step, from $x = 0$ to $0.1$ .

k_1 = 0.1\,\frac{0 + 1}{10} = 0.010000

k_2 = 0.1\,\frac{0.05^2 + 1.005^2}{10} = 0.010125

k_3 = 0.1\,\frac{0.05^2 + 1.0050625^2}{10} = 0.010127

k_4 = 0.1\,\frac{0.1^2 + 1.010127^2}{10} = 0.010304

y(0.1) = 1 + \frac{0.010000 + 0.020250 + 0.020254 + 0.010304}{6} = 1.010135

Step 3 Second step, from $x = 0.1$ to $0.2$ .

k_1 = 0.010304, \qquad k_2 = 0.010533, \qquad k_3 = 0.010535, \qquad k_4 = 0.010818

y(0.2) = 1.010135 + \frac{0.010304 + 0.021066 + 0.021070 + 0.010818}{6} = 1.020678

Answer

$y(0.2) \approx 1.0207$ .

(a)

3 marks

Write the formula for Trapezoidal rule to solve initial Value problem.

(b)

4 marksSame type asked in 2 of 2 papers

Solve the initial value problem

\dfrac{dy}{dx} = x + 3y, y(0) = 1

, for

x = 0.1

using Trapezoidal rule with step size

h = 0.05

Also asked in Summer 2026 Q4b OR

(c)

7 marksSame type asked in 2 of 2 papers

Apply Heun Method, to find approximate value of

y

when

x = 1.2

given that

\dfrac{dy}{dx} = 3x + y, y(1) = 1.3

. Use step size of

h = 0.1

Also asked in Summer 2026 Q4c

Q.4

14 Marks

(a)

3 marks

Define Finite Element Method.

(b)

4 marks

For one dimensional two-node linear finite element of length L, derive the shape function

N_1(x)

and

N_2(x)

(c)

7 marksSame type asked in 2 of 2 papers

Apply Finite Element Method (FEM), to solve the boundary value problem

y'' = -2

with boundary conditions

y(0) = 0

and

y'(1) = 0

. (Use a single, two-node linear element)

Also asked in Summer 2026 Q5c OR

(a)

3 marksAsked in 2 of 2 papers

Explain Finite Element Discretization.

Also asked in Summer 2026 Q2a

(b)

4 marks

Apply three nodal constraints at nodes

\zeta = -1, 0, 1

to derive the expression of quadratic shape function

N_2(\zeta) = a\zeta^2 + b\zeta + c

(c)

7 marksAsked in 2 of 2 papers

Consider a simple series RL circuit with a constant DC voltage source $V_s$ .The governing ODE for the current $i(t)$ is given by Kirchhoff’s Voltage Law:

L\dfrac{di}{dt} + Ri = V_s

where

L = 1H

is the induction,

R = 1\Omega

is the resistance and

V_s = 1V

with initial condition

i(0) = 0

. Apply Finite Element Method to find the current

i(t), t \in [0, 1]

. (Use 3 modes and 2 linear elements )

Also asked in Summer 2026 Q5c

Step 1 Weak form.

With $L = R = V_s = 1$ , the equation is $\dfrac{di}{dt} + i = 1$ , $i(0) = 0$ . Multiply by a weight function $N_i$ and integrate on an element:

\int N_i\left(\frac{di}{dt} + i - 1\right)dt = 0, \qquad i = N_1 i_1 + N_2 i_2

Step 2 Mesh.

Two linear elements on $[0, 1]$ have 3 nodes at $t = 0,\ 0.5,\ 1$ and length $h = 0.5$ . The shape functions are $N_1 = \dfrac{t_b - t}{h}$ and $N_2 = \dfrac{t - t_a}{h}$ .

Step 3 Element equations.

[k] = \frac{1}{2}\begin{bmatrix} -1 & 1 \\ -1 & 1 \end{bmatrix} + \frac{h}{6}\begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} -\tfrac{1}{3} & \tfrac{7}{12} \\ -\tfrac{5}{12} & \tfrac{2}{3} \end{bmatrix}, \qquad \{f\} = \begin{Bmatrix} 0.25 \\ 0.25 \end{Bmatrix}

Step 4 Assemble.

The middle node belongs to both elements, so its row adds the two elements:

\begin{bmatrix} -\tfrac{1}{3} & \tfrac{7}{12} & 0 \\ -\tfrac{5}{12} & \tfrac{1}{3} & \tfrac{7}{12} \\ 0 & -\tfrac{5}{12} & \tfrac{2}{3} \end{bmatrix}\begin{Bmatrix} i_1 \\ i_2 \\ i_3 \end{Bmatrix} = \begin{Bmatrix} 0.25 \\ 0.5 \\ 0.25 \end{Bmatrix}

Step 5 Apply $i_1 = 0$

and drop the first row. Rows 2 and 3, multiplied by 12, are

4i_2 + 7i_3 = 6, \qquad -5i_2 + 8i_3 = 3

Step 6 Solve.

Eliminating $i_2$ gives $67\,i_3 = 42$ .

i_3 = \frac{42}{67} = 0.6269 \text{ A}, \qquad i_2 = \frac{27}{67} = 0.4030 \text{ A}

Answer

i(t) = \begin{cases} 0.8060\,t, & 0 \le t \le 0.5 \\ 0.4030 + 0.4478\,(t - 0.5), & 0.5 \le t \le 1 \end{cases}

The exact values $1 - e^{-t}$ are $0.3935$ at $t = 0.5$ and $0.6321$ at $t = 1$ .

Q.5

14 Marks

(a)

3 marks

Explain continuous and discrete optimization.

(b)

4 marksSame type asked in 2 of 2 papers

Find the local minimum of the continuous, unconstrained function

f(x) = x^3 - 6x^2 + 5

using using optimality criterion.

Also asked in Summer 2026 Q5b OR

(c)

7 marksSame type asked in 2 of 2 papers

Apply Fibonacci search method to minimize the function

f(x) = x^2 + 4x

[1, 5]

with

n = 6

Also asked in Summer 2026 Q5b

k	$L_k^{*}$	$x_1$	$x_2$	$f(x_1)$	$f(x_2)$	New interval
2	$\frac{8}{21}(4) = 1.5238$	2.5238	3.4762	16.465	25.989	$(1, 3.4762)$
3	$\frac{5}{21}(4) = 0.9524$	1.9524	2.5238	11.621	16.465	$(1, 2.5238)$
4	$\frac{3}{21}(4) = 0.5714$	1.5714	1.9524	8.755	11.621	$(1, 1.9524)$
5	$\frac{2}{21}(4) = 0.3810$	1.3810	1.5714	7.431	8.755	$(1, 1.5714)$
6	$\frac{1}{21}(4) = 0.1905$	1.1905	1.3810	6.179	7.431	$(1, 1.3810)$

(a)

3 marks

Explain local and global optimization.

(b)

4 marksSame type asked in 2 of 2 papers

Minimize

f(x) = x^2 - x + 1

, on

[0, 1]

by using Interval-Halving method up to first three iteration.

Also asked in Summer 2026 Q1a

Interval-halving method for $f(x) = x^2 - x + 1$ on $[0, 1]$ . At each step $x_m = \dfrac{a+b}{2}$ , $x_1 = a + \dfrac{L}{4}$ , $x_2 = b - \dfrac{L}{4}$ . If $f(x_1) < f(x_m)$ , set $b = x_m$ . Else if $f(x_2) < f(x_m)$ , set $a = x_m$ . Else set $a = x_1$ and $b = x_2$ .

Iteration	$[a, b]$	$x_m$	$x_1$	$x_2$	$f(x_1)$	$f(x_m)$	$f(x_2)$	New interval
1	$[0, 1]$	0.5	0.25	0.75	0.8125	0.75	0.8125	$[0.25, 0.75]$
2	$[0.25, 0.75]$	0.5	0.375	0.625	0.7656	0.75	0.7656	$[0.375, 0.625]$
3	$[0.375, 0.625]$	0.5	0.4375	0.5625	0.7539	0.75	0.7539	$[0.4375, 0.5625]$

In every iteration $f(x_m)$ is smaller than $f(x_1)$ and $f(x_2)$ , so the middle part is kept.

Answer

after three iterations the minimum lies in $[0.4375, 0.5625]$ . The estimate is $x = 0.5$ with $f = 0.75$ , which is the exact minimum.

(c)

7 marksSame type asked in 2 of 2 papers

Apply Exhaustive search method to minimize the function

f(x) = x^2 - 4x + 5

[0, 5]

with

n = 10

. (Final answer to 3 decimal places)

Also asked in Summer 2026 Q5a OR

Exhaustive search: with $n = 10$ on $[0, 5]$ , the step is $\Delta x = \dfrac{5 - 0}{10} = 0.5$ . We calculate $f(x) = x^2 - 4x + 5$ at each point. The minimum lies between the points on both sides of the lowest value, where $f(x_1) \ge f(x_2) \le f(x_3)$ .

$x$	0	0.5	1	1.5	2	2.5	3	3.5	4	4.5	5
$f(x)$	5	3.25	2	1.25	1	1.25	2	3.25	5	7.25	10

The values fall until $x = 2$ and rise after it. At the first group of three points that meet the condition, $f(1.5) = 1.25 \ge f(2) = 1 \le f(2.5) = 1.25$ .

Answer

the minimum lies in $(1.5, 2.5)$ . The best point is $x = 2.000$ and the minimum value is $f = 1.000$ .

More Numerical Methods for Electrical Engineering papers

Questions repeat across sessions. Working through two or three papers shows you which topics GTU asks most.

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Official Gujarat Technological University (GTU) examination paper and step-by-step solutions for Numerical Methods for Electrical Engineering (Winter 2025, B.E. · Electrical Engineering, Sem 3). Features complete 70-mark regular & remedial examination pattern, official marking distribution across all 5 questions, and direct 1-click official PDF download.

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