Class 1 Electrician 第一種電気工事士 practice test with answers – free sample questions

Correct answer: A

For two capacitors C₁ = 20 μF and C₂ = 30 μF in series, the equivalent capacitance is C = (20 × 30) ÷ (20 + 30) = 600 ÷ 50 = 12 μF = 12 × 10⁻⁶ F. The electrostatic energy stored at 100 V is W = 0.5 × C × V² = 0.5 × 12 × 10⁻⁶ × 100² = 0.06 J. Value 0.12 J omits the 0.5 factor; 0.25 J treats the capacitors as parallel (50 μF); 0.50 J makes both errors.

💡 Series capacitance C = (C₁ × C₂) ÷ (C₁ + C₂) = 12 μF. Electrostatic energy W = 0.5 × C × V² at 100 V equals 0.06 J.

Correct answer: A

Converting L = 40 mH = 0.04 H, the stored magnetic energy is W = 0.5 × L × I² = 0.5 × 0.04 × 5² = 0.5 J. The magnitude of the self-induced EMF when current changes by ΔI = 5 A in Δt = 0.1 s is e = L × (ΔI ÷ Δt) = 0.04 × (5 ÷ 0.1) = 2.0 V.

Correct answer: B

Let V_ab be the voltage between nodes a and b. By Kirchhoff's current law at node a, I₁ + I₂ = I₃, so (24 − V_ab) ÷ 4 + (8 − V_ab) ÷ 4 = V_ab ÷ 6. Multiplying by 12 gives 3(24 − V_ab) + 3(8 − V_ab) = 2V_ab, or 96 = 8V_ab, which yields V_ab = 12 V. Thus, the current through center resistor R₃ = 6 Ω is I₃ = 12 V ÷ 6 Ω = 2.0 A.

💡 Use node voltage V_ab with Kirchhoff's current law at node a: (24 − V_ab)/4 + (8 − V_ab)/4 = V_ab/6 gives V_ab = 12 V and I₃ = 2.0 A.

Correct answer: C

Since period T = 20 ms = 0.02 s, the frequency is f = 1 ÷ T = 1 ÷ 0.02 = 50 Hz. The angular frequency is ω = 2πf = 2 × 3.14 × 50 = 314 rad/s.

Correct answer: D

Resolving each current into in-phase and quadrature components: I₁ has active component I₁p = 20 × 0.5 = 10 A and reactive component I₁q = 20 × (√3 ÷ 2) = 10√3 A, whereas resistive load I₂ has I₂p = 20 A and I₂q = 0 A. In the neutral wire, active components subtract: I_N = √((10 − 20)² + (10√3 − 0)²) = √(100 + 300) = √400 = 20 A.

💡 In a single-phase 3-wire line with lagging power factor cosθ₁ = 0.5 (60°) and cosθ₂ = 1.0, neutral wire current I_N = √((I₁p − I₂p)² + I₁q²) = 20 A.

Correct answer: C

The current in section A−B is the sum of both loads: I_AB = 20 + 10 = 30 A, while the current in section B−C is I_BC = 10 A. For a single-phase 2-wire line, the voltage drop in A−B is ΔV_AB = 2 × 0.1 × 30 = 6.0 V, and in B−C is ΔV_BC = 2 × 0.15 × 10 = 3.0 V. Thus, V_C = 108 − (6.0 + 3.0) = 99.0 V.

Correct answer: C

The current in each outer wire is I = P ÷ (V × cosθ) = 1600 W ÷ (100 V × 0.8) = 20 A. Because the loads are balanced, the neutral current is 0 A. Thus, the total line power loss in the two outer wires is P_l = 2 × I² × r = 2 × 20² × 0.15 = 120 W.

Correct answer: B

Because the resistive load has 100% power factor, its reactive power is Q₂ = 0 kvar. The total active power is P = P₁ + P₂ = 16 + 14 = 30 kW, and the total reactive power is Q = 16 + 0 = 16 kvar. Therefore, the combined apparent power is S = √(P² + Q²) = √(30² + 16²) = √(900 + 256) = √1156 = 34 kVA.

Correct answer: C

The synchronous speed of a 4-pole motor at 60 Hz is N_s = (120 × f) ÷ p = (120 × 60) ÷ 4 = 1800 min⁻¹. At a slip of s = 5% = 0.05, the rotor speed is N = N_s × (1 − s) = 1800 × (1 − 0.05) = 1710 min⁻¹.

Correct answer: D

At no load, output power is zero so efficiency is 0%. A transformer achieves maximum efficiency at the load where variable copper loss equals constant iron loss (copper loss = iron loss).