List FOUR advantages of a DC series motor.

$$E_b = V_T - I_a (R_a + R_{se})$$ $$E_b = 240 \text{ V} - 0 \text{ A} \times (0.20 \Omega + 0.15 \Omega) = 240 \text{ V}$$ $$E_b = \boxed{240 \text{ V}}$$

1.2.2 The torque developed at full load
Full-load: $I_L = 40 \text{ A}$, $N = 1000 \text{ r/min}$.

$$E_b = V_T - I_a (R_a + R_{se})$$ $$E_b = 240 \text{ V} - 38 \text{ A} \times (0.20 \Omega + 0.15 \Omega) = 240 \text{ V} - 38 \text{ A} \times 0.35 \Omega$$ $$E_b = 240 \text{ V} - 13.3 \text{ V} = 226.7 \text{ V}$$

$$T_d = \frac{P_d}{\omega} = \frac{8614.6 \text{ W}}{104.72 \text{ rad/s}} \approx 82.26 \text{ Nm}$$ $$T_d = \boxed{82.26 \text{ Nm}}$$

1.2.3 The shaft power at full load
Assuming rotational losses are negligible or not provided, the shaft power is approximately equal to the mechanical power developed in the armature.

$$P_{shaft} = P_d = E_b I_a = 8614.6 \text{ W}$$ $$P_{shaft} = \boxed{8.61 \text{ kW}}$$

1.3 A DC generator with a differential compound winding has its series field winding connected such that its magnetic flux opposes the magnetic flux of the shunt field winding. This connection results in a decreasing terminal voltage with increasing load.

\begin{tikzpicture}[scale=0.8] % Armature \draw[thick] (0,0) circle (1.5cm); \node at (0,0) {Armature}; \draw[thick] (0,1.5) -- (0,2); \draw[thick] (0,-1.5) -- (0,-2); % Brushes \draw[fill=black] (-0.5,1.5) rectangle (0.5,1.7); \draw[fill=black] (-0.5,-1.5) rectangle (0.5,-1.7); % Shunt Field \draw[thick] (0,2) -- (2,2) -- (2,0.5); \draw[thick] (0,-2) -- (2,-2) -- (2,-0.5); \draw[thick, decorate, decoration={coil,aspect=0.3,segment length=3mm,amplitude=2mm}] (2,0.5) -- (2,-0.5) node[right, midway] {$R_{sh}$}; % Series Field \draw[thick] (0,2) -- (-2,2) -- (-2,0.5); \draw[thick] (0,-2) -- (-2,-2) -- (-2,-0.5); \draw[thick, decorate, decoration={coil,aspect=0.3,segment length=3mm,amplitude=2mm}] (-2,0.5) -- (-2,-0.5) node[left, midway] {$R_{se}$}; % Load \draw[thick] (2,2) -- (3,2) node[right] {$V_T$}; \draw[thick] (2,-2) -- (3,-2); \draw[thick] (3,2) -- (3,-2); \draw[thick] (3.5,0) node[right] {Load}; % Arrows for current direction (indicating differential connection) \draw[->] (2,1.5) -- (2,0.5); % Shunt current down \draw[->] (-2,1.5) -- (-2,0.5); % Series current down (opposing shunt) \draw[->] (0,1.7) -- (0,2); % Armature current out \draw[->] (0,-1.7) -- (0,-2); % Armature current out \draw[->] (2.5,2) -- (3,2); % Load current \draw[->] (2.5,-2) -- (3,-2); % Load current \end{tikzpicture}

Note: The diagram shows a long shunt connection, where the series field is in series with the armature, and this combination is in parallel with the shunt field and the load. For differential compounding, the winding direction of the series field is such that its magnetic field opposes the shunt field.

1.4 If the armature voltage of a DC shunt motor is increased while the load torque remains constant and the field flux is constant, the speed of the motor will increase. This is because with constant load torque and constant field flux, the armature current remains constant. An increase in armature voltage leads to an increase in the back EMF ($E_b = V_T - I_a R_a$), and since $E_b$ is proportional to speed (for constant flux), the speed must increase.

QUESTION 2: AC CIRCUIT THEORY

Given: Delta-connected load, $Z_{ph} = (6 - j8) \Omega$ per phase, $V_L = 400 \text{ V}$, $f = 50 \text{ Hz}$.

2.1 The line current

$$I_L = \sqrt{3} I_{ph} = \sqrt{3} \times 40 \text{ A} \approx 69.28 \text{ A}$$ $$I_L = \boxed{69.28 \text{ A}}$$

2.2 The power factor

$$\cos\theta = \frac{R_{ph}}{|Z_{ph}|} = \frac{6 \Omega}{10 \Omega} = 0.6$$

Since the impedance has a negative imaginary part (capacitive reactance), the power factor is leading.

$$\cos\theta = \boxed{0.6 \text{ leading}}$$

2.3 The active power consumed

$$P = 3 I_{ph}^2 R_{ph} = 3 \times (40 \text{ A})^2 \times 6 \Omega = 3 \times 1600 \times 6 = 28800 \text{ W}$$ $$P = \boxed{28.8 \text{ kW}}$$

2.4 The reactive power consumed

$$Q = 3 I_{ph}^2 X_{ph} = 3 \times (40 \text{ A})^2 \times (-8 \Omega) = 3 \times 1600 \times (-8) = -38400 \text{ VAR}$$

The negative sign indicates that the load is supplying reactive power (or consuming negative reactive power), which is characteristic of a capacitive load.

$$Q = \boxed{-38.4 \text{ kVAR}}$$

2.5 The apparent power (kVA)

$$S = \sqrt{3} V_L I_L = \sqrt{3} \times 400 \text{ V} \times 69.28 \text{ A} \approx 48000 \text{ VA}$$

Alternatively, using $P$ and $Q$:

$$S = \sqrt{P^2 + Q^2} = \sqrt{(28800)^2 + (-38400)^2} = \sqrt{829440000 + 1474560000} = \sqrt{2304000000} = 48000 \text{ VA}$$ $$S = \boxed{48.0 \text{ kVA}}$$