100+ Free Tanzania ERB Electrical Practice Questions

Explanation: To convert the generator's per-unit reactance to a new system base, the formula X_new_pu = X_old_pu * (S_base_new / S_base_old) * (V_base_old / V_base_new_at_element)^2 is used. The new system base MVA is 100 MVA. The base voltage on the generator side remains 13.8 kV, as it is derived from the system's 132 kV base through the transformer ratio (132 kV * (13.8/132) = 13.8 kV). Therefore, X_gen_new_pu = 0.2 * (100/50) * (13.8/13.8)^2 = 0.2 * 2 * 1 = 0.4 pu.

Explanation: The conservator tank, located above the main transformer tank, serves as an expansion vessel for the insulating oil. As the transformer heats up during operation, the oil expands; when it cools down, the oil contracts. The conservator tank accommodates these volume changes, preventing excessive pressure buildup and minimizing exposure of the main tank oil to the atmosphere.

Explanation: For a long transmission line with negligible resistance and shunt admittance (lossless line), the characteristic impedance is Zc and the propagation constant is γ. For an open-circuited receiving end, the receiving end current Ir is zero. The sending end voltage Vs is given by the hyperbolic cosine relationship: Vs = Vr cosh(γL) + Zc * Ir sinh(γL). Since Ir = 0, the equation simplifies to Vs = Vr cosh(γL).

Explanation: First, calculate the synchronous speed (Ns) using the formula Ns = (120 * f) / P, where f is the frequency (50 Hz) and P is the number of poles (4). Ns = (120 * 50) / 4 = 1500 rpm. The slip (s) is then calculated as s = (Ns - Nr) / Ns, where Nr is the rotor speed (1440 rpm). So, s = (1500 - 1440) / 1500 = 60 / 1500 = 0.04. Expressed as a percentage, the slip is 4%.

Explanation: Distance relays are specifically designed for transmission line protection. They measure the impedance between the relay location and the fault point. Since impedance is proportional to distance, the relay operates if the fault is within a predetermined zone, providing fast and selective protection against both phase and ground faults on long lines.

Explanation: For unity power factor operation, the excitation voltage E_f and power angle δ are calculated using P = (E_fV_t/X_s)sinδ and Q = (E_fV_tcosδ - V_t^2)/X_s. For P=1, Q=0, V_t=1, X_s=1, we find E_f = √2 pu and δ = 45°. When E_f increases by 20% to 1.2√2 ≈ 1.697 pu, and P remains constant at 1 pu, the new power angle δ' can be found from sinδ' = P / (E_f'V_t/X_s) = 1 / 1.697, giving δ' ≈ 36.1°. The new reactive power Q' = (E_f'V_tcosδ' - V_t^2)/X_s = (1.697*1*cos(36.1°) - 1^2)/1 ≈ 0.371 pu. The new power factor is cosφ = P / √(P^2 + Q'^2) = 1 / √(1^2 + 0.371^2) ≈ 0.938 lagging.

Explanation: First, calculate the initial reactive power (Q1) from P=1.2MW and cos(φ1)=0.75. φ1 = arccos(0.75) ≈ 41.41°, so Q1 = P * tan(φ1) = 1.2 MW * tan(41.41°) ≈ 1.2 MW * 0.8819 = 1.0583 MVAr. Next, calculate the target reactive power (Q2) for cos(φ2)=0.95. φ2 = arccos(0.95) ≈ 18.19°, so Q2 = P * tan(φ2) = 1.2 MW * tan(18.19°) ≈ 1.2 MW * 0.3287 = 0.3944 MVAr. The required reactive power from the capacitor bank is Qc = Q1 - Q2 = 1.0583 - 0.3944 = 0.6639 MVAr, or approximately 664 kVAr.

Explanation: A DC series motor's speed is highly dependent on its field flux, which is in series with the armature current. The diverter method involves connecting a variable resistance in parallel with the series field winding. By diverting some armature current away from the field, the field flux is weakened, leading to an increase in motor speed.

Explanation: Substation earthing is critical for safety, to minimize step and touch potentials, and to provide a low impedance path for fault currents. Industry standards and regulations (e.g., IEC 61936-1) generally recommend an earth resistance of less than 1 Ohm for major substations, especially in areas with significant human activity, to ensure adequate safety and fault current dissipation.

Explanation: For satisfactory parallel operation of transformers, several conditions are essential: identical voltage ratios (to avoid circulating currents), same percentage impedance (to ensure proper load sharing), and same polarity (to avoid short-circuiting windings). While having identical kVA ratings is desirable for optimal utilization and load sharing, transformers with different kVA ratings can operate in parallel, provided the essential conditions are met, albeit with the load sharing proportional to their kVA ratings if their percentage impedances are equal. Thus, identical kVA ratings are NOT strictly essential.