Basic Electrical Theory

I = V ÷ R = 100 ÷ 20 = 5 A. 80 A subtracts the resistance from the voltage, 120 A adds them, and 2,000 A multiplies them (V × R); none of these is Ohm's law. To find current, divide voltage by resistance (Ohm's law, オームの法則).

V = I × R = 3 × 40 = 120 V. 0.075 V is I ÷ R (3 ÷ 40), 13.3 V is R ÷ I (40 ÷ 3), and 43 V is current plus resistance. Voltage is current × resistance.

Resistance is proportional to length and inversely proportional to the cross-sectional area (the square of the diameter). 2 ÷ 2² = 1/2. 1/4 forgets the change in length. 1 wrongly treats resistance as inversely proportional to the diameter itself (2 ÷ 2). 2 ignores the change in diameter.

Area A = π × D² ÷ 4 = 3.14 × 1.6² ÷ 4 ≈ 2.01 mm². R = ρ × L ÷ A = 0.0172 × 100 ÷ 2.01 ≈ 0.86 Ω. 0.21 Ω treats 1.6 mm as the radius. 0.43 Ω calculates the area as π × D² ÷ 2, dividing by 2 instead of 4. 0.67 Ω uses D² (2.56 mm²) as the area and forgets π/4.

Polyvinyl chloride (PVC, ビニル) is an insulator widely used to cover wires and cables. Copper and aluminium conduct current well and are used as the conductor (core) of wires. Iron is also a metal, so it is a conductor.

Metals such as copper have higher resistance at higher temperature. "Decreases as temperature rises" and "increases as temperature falls" state the opposite. "Does not change" is also wrong, because the resistance of a metal changes with temperature.

Conductance (コンダクタンス) is the reciprocal of resistance: G = 1 ÷ R = 1 ÷ 4 = 0.25 S (siemens). 4 S just repeats the resistance value, 2 S is its square root, and 16 S is its square; none of these is the reciprocal.

By convention, current flows out of the positive terminal, through the circuit, and back to the negative terminal. What actually moves in a metal is negatively charged electrons, so they move opposite to the current (from the negative side toward the positive side). The other three statements all contradict this convention.

For resistors in series, the combined resistance (合成抵抗) is the sum: 10 + 20 + 30 = 60 Ω. 5.5 Ω is the parallel result (reciprocal of 1/10 + 1/20 + 1/30), 20 Ω is the average, and 6,000 Ω is the product.

For two resistors in parallel use product ÷ sum: (6 × 3) ÷ (6 + 3) = 18 ÷ 9 = 2 Ω. A parallel combination is always smaller than the smallest resistor (3 Ω). 9 Ω is the series value, 4.5 Ω is the average, and 18 Ω is the product without dividing by the sum.

n equal resistors R in parallel give R ÷ n: 30 ÷ 3 = 10 Ω. 90 Ω is the series value, 30 Ω is just one resistor, and 3.3 Ω divides 30 by 3 twice.

R2 and R3 in parallel: (12 × 6) ÷ (12 + 6) = 4 Ω. Adding R1 (4 Ω) in series gives 8 Ω in total. Current = 24 ÷ 8 = 3 A. 1.1 A treats all three as series (22 Ω). 2 A is the current in R3 only and 1 A is the current in R2 only, not the total.

Parallel part: (6 × 3) ÷ (6 + 3) = 2 Ω; total 6 + 2 = 8 Ω; current 40 ÷ 8 = 5 A. The voltage across the parallel part is 5 × 2 = 10 V, so the current in R3 is 10 ÷ 3 ≈ 3.3 A. 1.7 A is the current in R2 (10 ÷ 6), 2.5 A wrongly splits 5 A equally, and 5.0 A is the total current.

Current = 100 ÷ (20 + 30) = 2 A, so the 20 Ω resistor has 2 × 20 = 40 V. In series, voltage divides in proportion to resistance (100 × 20/50). 60 V is the voltage across 30 Ω, 50 V wrongly splits it equally, and 100 V would apply only in a parallel connection.

Each pair is 4 + 4 = 8 Ω and 2 + 6 = 8 Ω. Two 8 Ω in parallel give 8 ÷ 2 = 4 Ω. 16 Ω puts everything in series, 8 Ω is just one pair, and 0.86 Ω is the value with all four resistors in parallel.

Series gives 4R and parallel gives R ÷ 4, so 4R ÷ (R/4) = 16. In general, for n equal resistors the series value is n² times the parallel value. 4 compares the series value with a single R, 8 is a mistaken 4 × 2, and 64 is a mistaken 4³.

From P = V² ÷ R, R = V² ÷ P = 100² ÷ 500 = 10,000 ÷ 500 = 20 Ω. 5 Ω mistakes P ÷ V (500 ÷ 100) for the resistance; that is actually the current, 5 A. 0.2 Ω is V ÷ P and 50,000 Ω is V × P.

P = I² × R = 5² × 4 = 100 W. 20 W is I × R (that is the voltage, 20 V). 25 W is I² without multiplying by R. 80 W is I × R².

Energy = 1 kW × 2 h = 2 kWh. Since 1 kWh = 3,600 kJ, the answer is 2 × 3,600 = 7,200 kJ. 3,600 kJ is only one hour. 2,000 kJ wrongly treats 1 kW × 2 h as kJ. 120 kJ uses minutes (120 min) without converting to seconds.

Since P = V² ÷ R, halving the voltage gives (1/2)² = 1/4 of the power. 1/2 assumes power is proportional to voltage, forgetting that the current also halves. 1 and 2 do not fit the formula.

Heat absorbed by the water = 10 × 4.2 × (80 − 20) = 2,520 kJ. With 80% efficiency, the electrical energy needed is 2,520 ÷ 0.8 = 3,150 kJ. 1.5 kW = 1.5 kJ per second, so the time is 3,150 ÷ 1.5 = 2,100 s = 35 min. 28 min ignores efficiency, 22 min multiplies by the efficiency instead of dividing, and 53 min uses 1 kW instead of 1.5 kW.

With resistance R each, series gives 2R and power V² ÷ 2R; parallel gives R/2 and power V² ÷ (R/2) = 2V²/R. The ratio is 2 ÷ (1/2) = 4. 2 considers only the number of wires; 1/2 and 1/4 reverse series and parallel.

H = I² × R × t = 2² × 10 × (5 × 60) = 4 × 10 × 300 = 12,000 J = 12 kJ. 6 kJ forgets to square the current (2 × 10 × 300 = 6,000 J). 0.2 kJ leaves the time as 5 instead of converting to seconds (2² × 10 × 5 = 200 J). 0.1 kJ does neither (2 × 10 × 5 = 100 J).

0.8 kW × 3 h × 30 days = 72 kWh. 72,000 kWh forgets to convert W to kW (the correct figure is 72,000 Wh). 2.4 kWh is one day only, and 24 kWh counts only 10 days. Convert to kW and hours before multiplying.

For a sine wave, peak = RMS (実効値) × √2, so 100 × 1.41 ≈ 141 V. 173 V multiplies by √3, confusing this with the three-phase line/phase voltage relation. 200 V simply doubles the value, and 283 V is twice the peak (the peak-to-peak value).

The period is the reciprocal of the frequency: T = 1 ÷ f = 1 ÷ 50 = 0.02 s. 50 s treats the frequency as seconds, and 0.2 s and 0.002 s have the wrong number of decimal places.

Inductive reactance X_L = 2πfL is proportional to frequency, so it becomes 60 ÷ 50 = 1.2 times. About 0.83 (50/60) is how a capacitor's reactance changes, 1 assumes frequency has no effect, and 1.44 assumes it is proportional to the square of the frequency.

Impedance Z = √(8² + 6²) = 10 Ω, so current = 100 ÷ 10 = 10 A. Only the resistor consumes power: P = I² × R = 10² × 8 = 800 W. 1,000 W is the apparent power (100 V × 10 A). 600 W is the reactive power calculated with the reactance (10² × 6, properly in var). 1,400 W simply adds the active power (800 W) and the reactive power (600 var), which cannot be added directly (the relation is S² = P² + Q²).

Z = √(12² + 16²) = √400 = 20 Ω, and the power factor (力率) is R ÷ Z = 12 ÷ 20 = 0.6, i.e. 60%. 75% is R ÷ X (12 ÷ 16), 80% is X ÷ Z (16 ÷ 20, which is sinθ), and 43% wrongly adds R and X (12 ÷ 28).

The resistor current and coil current are 90° out of phase, so the total is √(4² + 3²) = √25 = 5 A. 7 A simply adds them and ignores the phase difference. 12 A multiplies them, and 25 A forgets to take the square root of the sum of squares.

P = V × I × cosθ = 200 × 10 × 0.8 = 1,600 W. 2,000 W is the apparent power in V·A, without the power factor. 2,500 W divides by the power factor instead of multiplying. 1,200 W multiplies by sinθ (0.6) instead of cosθ, which gives the reactive power in var.

In a purely inductive (coil) circuit, current lags voltage by 90°. In a purely resistive circuit, current and voltage are in phase, so "leads by 90°" is wrong. In a purely capacitive circuit, current leads by 90°, so "lags" is wrong. Capacitive reactance 1 ÷ (2πfC) is inversely proportional to frequency, so it decreases as frequency rises.

Apparent power S = 100 × 20 = 2,000 V·A = 2.0 kV·A; active power P = 2.0 × 0.6 = 1.2 kW. From S² = P² + Q², Q = √(2.0² − 1.2²) = √2.56 = 1.6 kvar (equivalently 2.0 × sinθ = 2.0 × 0.8). 1.2 is the active power, 2.0 is the apparent power, and 0.8 is just sinθ.

In a star (Y結線) connection, the phase voltage is 200 ÷ √3 ≈ 115.5 V. The line current equals the phase current: 115.5 ÷ 20 ≈ 5.8 A. 10 A applies the full 200 V to each resistor (the delta phase current), 17.3 A is the delta line current, and 3.3 A wrongly divides 200 by 3.

In a delta (Δ結線) connection, each resistor has the full line voltage of 200 V, so the phase current is 200 ÷ 20 = 10 A. The line current is √3 times the phase current: 10 × 1.73 ≈ 17.3 A. 10 A is the phase current, 5.8 A is the star line current, and 3.3 A wrongly divides 200 by 3 before dividing by the resistance.

In delta, each resistor has 200 V, so one phase uses 200² ÷ 20 = 2,000 W. For three phases, 6,000 W = 6 kW. 2 kW is one phase only (or the total for a star connection), 3.5 kW multiplies 2 kW by √3, and 10.4 kW multiplies 6 kW by √3.

P = √3 × V × I × cosθ = 1.73 × 200 × 30 × 0.85 ≈ 8,830 W, about 8.8 kW. 5.1 kW forgets √3 (200 × 30 × 0.85), 10.4 kW forgets the power factor, and 15.3 kW multiplies by 3 instead of √3.

After the break, the two remaining lines (line voltage V) feed one resistor R in parallel with two resistors in series (2R). Power = V²/R + V²/2R = 1.5V²/R. Before the break it was 3V²/R, so the ratio is 1.5 ÷ 3 = 1/2. 2/3 assumes "two of three resistors remain", 1/3 assumes "only one phase remains", and 1 assumes no effect.

In star, each resistor has V/√3, so power = 3 × (V/√3)² ÷ R = V²/R. In delta, each resistor has V, so power = 3V²/R. The ratio is 3. √3 considers only the voltage ratio, 1 assumes the connection makes no difference, and 1/3 is the inverse.

In star connection, line voltage (線間電圧) = phase voltage (相電圧) × √3, and line current (線電流) = phase current (相電流). "Line voltage equals phase voltage and line current is √3 times" describes delta connection. "3 times" confuses 3 with √3, and "both √3 times" mixes the star and delta relations.