A. A.d.c. circuit comprises three closed loops. Applying Kirchhoff's laws to the closed loops gives the following equations for current flow in milliamperes: 2I 1 + 31 41, = 26 -71, + 21, + 61, = 12 Use determinants to solve for I₁, 1₂ and 13

The supply voltage is $v = 30\sin(100\pi t)$, so its phasor is $V = 30 \angle 0 \text{ rad}$.
The voltage drop across one component is $v_1 = 20\sin(100\pi t - 0.59)$, so its phasor is $V_1 = 20 \angle -0.59 \text{ rad}$.

$V_1 = 20(\cos(-0.59) + j\sin(-0.59))$
$V_1 \approx 20(0.8310 - j0.5564)$
$V_1 \approx 16.62 - j11.13$

$V_R = V - V_1 = (30 + j0) - (16.62 - j11.13)$
$V_R = (30 - 16.62) + j(0 - (-11.13))$
$V_R = 13.38 + j11.13$

Amplitude $A = \sqrt{(13.38)^2 + (11.13)^2} = \sqrt{179.0244 + 123.8769} = \sqrt{302.9013} \approx 17.404$ V.
Phase angle $\beta = \arctan\left(\frac{11.13}{13.38}\right) = \arctan(0.8318) \approx 0.694 \text{ rad}$.

The voltage drop across the remainder of the circuit is \boxed{\text{17.40\sin(100\pi t + 0.694) V}}.

5.a.ii) Supply frequency

$\omega = 100\pi \text{ rad/s}$.

$100\pi = 2\pi f$
$f = \frac{100\pi}{2\pi} = 50 \text{ Hz}$.

The supply frequency is $\boxed{\text{50 Hz}}$.

5.a.iii) Periodic time of the supply

$T = \frac{1}{f}$.

$T = \frac{1}{50} = 0.02 \text{ s}$.

The periodic time of the supply is $\boxed{\text{0.02 s}}$.

5.a.iv) r.m.s. value of the supply voltage

$V_m = 30$ V.

$V_{rms} = \frac{V_m}{\sqrt{2}}$.
$V_{rms} = \frac{30}{\sqrt{2}} = \frac{30\sqrt{2}}{2} = 15\sqrt{2} \text{ V}$.
$V_{rms} \approx 15 \times 1.4142 = 21.213$ V.

The r.m.s. value of the supply voltage is \boxed{\text{15\sqrt{2} V (or approx. 21.21 V)}}.

5.b) Solve for $I_1, I_2, I_3$ using determinants

The currents are \boxed{\text{I_1 = 10 mA$,$I_2 = 14 mA$,$I_3 = 9 mA}}.

5.c) Acceleration and tensions for three masses

Given: $m_A = 4 \text{ kg}$, $m_B = 3 \text{ kg}$, $m_C = 1 \text{ kg}$, $\mu = 0.6$, $g = 10 \text{ m/s}^2$.

Assume $m_A$ moves down, $m_B$ moves right, and $m_C$ moves up. Let $a$ be the acceleration.
For mass A: $m_A g - T_1 = m_A a$
$4(10) - T_1 = 4a \Rightarrow 40 - T_1 = 4a$ (Eq. 1)

For mass C: $T_2 - m_C g = m_C a$
$T_2 - 1(10) = 1a \Rightarrow T_2 - 10 = a$ (Eq. 2)

For mass B:
Normal force $N = m_B g = 3(10) = 30$ N.
Friction force $f_k = \mu N = 0.6 \times 30 = 18$ N.
The equation of motion for B (moving right): $T_1 - T_2 - f_k = m_B a$
$T_1 - T_2 - 18 = 3a$ (Eq. 3)

From (Eq. 1): $T_1 = 40 - 4a$
From (Eq. 2): $T_2 = 10 + a$
Substitute $T_1$ and $T_2$ into (Eq. 3):
$(40 - 4a) - (10 + a) - 18 = 3a$
$40 - 4a - 10 - a - 18 = 3a$
$12 - 5a = 3a$
$12 = 8a$
$a = \frac{12}{8} = 1.5 \text{ m/s}^2$

$T_1 = 40 - 4a = 40 - 4(1.5) = 40 - 6 = 34 \text{ N}$
$T_2 = 10 + a = 10 + 1.5 = 11.5 \text{ N}$

The acceleration of the masses is \boxed{\text{1.5 m/s^2}}.
The tension in the string between A and B is \boxed{\text{34 N}}.
The tension in the string between B and C is \boxed{\text{11.5 N}}.

5.d) Atwood machine

Given: $m_1 = 4 \text{ kg}$, $m_2 = 3 \text{ kg}$, $g = 9.81 \text{ m/s}^2$. Released from rest.

Let $T$ be the tension in the string and $a$ be the acceleration.
For $m_1$ (moving downwards): $m_1 g - T = m_1 a$
$4(9.81) - T = 4a \Rightarrow 39.24 - T = 4a$ (Eq. 1)

For $m_2$ (moving upwards): $T - m_2 g = m_2 a$
$T - 3(9.81) = 3a \Rightarrow T - 29.43 = 3a$ (Eq. 2)

Add (Eq. 1) and (Eq. 2):
$(39.24 - T) + (T - 29.43) = 4a + 3a$
$9.81 = 7a$
$a = \frac{9.81}{7} \approx 1.4014 \text{ m/s}^2$

i) The acceleration of each mass is \boxed{\text{1.4014 m/s^2}}. (4 kg mass downwards, 3 kg mass upwards).

Using the kinematic equation $v = u + at$, with initial velocity $u=0$.
$v = 0 + (1.4014)(2)$
$v = 2.8028 \text{ m/s}$

ii) Their speed after 2 seconds is \boxed{\text{2.8028 m/s}}.

5.e) Regression analysis

Given data:
Electrical (x): 3, 6, 4, 5, 7, 8, 9, 7, 10, 9
Electronics (y): 5, 6, 5, 6, 7, 9, 7, 8, 9, 6

$n = 10$
$\sum x = 68$
$\sum y = 68$
$\sum x^2 = 510$
$\sum y^2 = 482$
$\sum xy = 485$
$\bar{x} = \frac{68}{10} = 6.8$
$\bar{y} = \frac{68}{10} = 6.8$

i) Scatter diagram for the data:
A scatter diagram would plot the given points (x, y) on a Cartesian plane. The x-axis would represent Electrical marks and the y-axis would represent Electronics marks. The points would be: (3,5), (6,6), (4,5), (5,6), (7,7), (8,9), (9,7), (7,8), (10,9), (9,6). Visually, these points would show a general upward trend, indicating a positive correlation between Electrical and Electronics marks.

ii) Equation of best fit and prediction:
To predict the Electrical Installation mark (x) given the Electronics mark (y), we need the regression line of x on y, which has the form $x = a + by$.

Calculate the slope $b$:

Calculate the y-intercept $a$:

The equation of best fit (x on y) is \boxed{\text{x = -1.0411 + 1.1531y}}.

Predict the Electrical Installation mark (x) for a student who scored 7 in Electronics (y=7):

The predicted Electrical Installation mark is $\boxed{\text{7.03}}$.

iii) Regression line of y on x:
The regression line of y on x has the form $y = a' + b'x$.
Calculate the slope $b'$:

Calculate the y-intercept $a'$:

The equation of the regression line of y on x is $y = 3.5714 + 0.4748x$.
To draw this line on the scatter diagram, you would plot two points from this equation (e.g., for $x=3, y \approx 5.0$; for $x=10, y \approx 8.3$) and draw a straight line connecting them. This line represents the best linear fit for predicting Electronics marks based on Electrical marks.