Electrician - Kerala State Film Development Corporation Ltd. 2025 — Kerala PSC PYQ Practice with Answers

Correct answer (Option C):\nBoth statements correctly represent fundamental electrical relationships in a DC circuit.\nStatement (i) is the standard mathematical expression of Ohm's law, which states that current is directly proportional to voltage and inversely proportional to resistance.\nStatement (ii) is derived from the power formula P = V × I, which rearranged for current yields I = P / V.\nTherefore, both equations are completely valid expressions for determining current.\n\nWhy others are wrong:\nOption A is incorrect because it ignores the validity of the electrical power formulation.\nOption B is incorrect because it overlooks the fundamental definition of Ohm's law.\nOption D is incorrect because both statements are well-established, accurate physics relations.\n\nStudy tip:\nRemember the power formula wheel relationships where P = V × I = I² × R = V² / R. These algebraic manipulations are essential for solving basic circuit problems.

Correct answer (Option A):\nThe SI unit of electric power is the Watt (W). By definition, electric power represents the rate at which electrical energy is transferred or consumed in an electrical circuit, equivalent to one Joule per second (1 W = 1 J/s).\n\nWhy others are wrong:\nOption B is incorrect because the Joule is the SI unit of energy or work, not power.\nOption C is incorrect because Watts/sec is a meaningless unit in this context; power is already energy divided by time.\nOption D is incorrect because the Kilowatt-hour (kWh) is a commercial unit of electrical energy consumption, not power.\n\nStudy tip:\nAlways distinguish between Power (rate of doing work, measured in Watts) and Energy (total work done over time, measured in Joules or Watt-hours).

Correct answer (Option B):\nWhen lamps are connected in series, the current (I) flowing through both lamps is exactly the same.\nThe brightness of a lamp depends on the actual power dissipated by it, which is given by P = I² × R.\nLet's evaluate their internal resistances from their rated specifications using R = V² / P:\nFor the 100 W lamp: R₁ = 220² / 100 = 484 Ω\nFor the 40 W lamp: R₂ = 220² / 40 = 1210 Ω\nSince the 40 W lamp has a much higher resistance and the series current is identical, it will dissipate more power (I² × R) and thus glow more brightly than the 100 W lamp.\n\nWhy others are wrong:\nOption A is incorrect because a 100 W lamp only glows brighter in a parallel configuration where voltages are equal.\nOption C is incorrect because their unequal resistances lead to completely different power drops under the same current flow.\nOption D is incorrect because the total resistance allows sufficient current to pass to cause illumination.\n\nStudy tip:\nIn series circuits, the component with the highest resistance consumes the maximum power. In parallel circuits, the component with the lowest resistance consumes the maximum power.

Correct answer (Option A):\nIn a parallel electrical circuit, the voltage across each parallel branch remains exactly the same as the source voltage (V₁ = V₂ = V₃ = V_total). Voltages are not additive in parallel; they are only additive in series configurations.\n\nWhy others are wrong:\nOption B is incorrect because total electrical power is always additive in both series and parallel circuits.\nOption C is incorrect because according to Kirchhoff's Current Law, the individual branch currents add up to match the total source current.\nOption D is incorrect because total conductance in a parallel circuit is the direct sum of individual branch conductances (G_total = G₁ + G₂ + ...).\n\nStudy tip:\nRemember the dual properties: Series circuits feature identical current and additive voltages, whereas parallel circuits feature identical voltage and additive currents.

Correct answer (Option D):\nFormula: Total Power (P_total) = P₁ + P₂\nGiven: P₁ = 100 W, P₂ = 60 W, Time (t) = 5 hours\nStep 1: Calculate the total power of the parallel circuit.\nP_total = 100 W + 60 W = 160 W\nStep 2: Convert the total power from Watts to Kilowatts.\nP_total = 160 / 1000 = 0.16 kW\nStep 3: Compute the electrical energy consumed.\nEnergy = Power × Time = 0.16 kW × 5 hours = 0.8 kWh\nHence, Option D is correct.\n\nWhy others are wrong:\nOptions A, B, and C provide mathematically incorrect values because they do not correctly perform the conversion factor from watt-hours to kilowatt-hours or fail to sum up parallel branch powers appropriately.\n\nStudy tip:\nCommercial units of energy are always calculated in Kilowatt-hours (kWh), frequently called units. Double check your metric prefix division by 1000.