EPFO EO/AO 2023 General Ability Test · Question 86 of 120

Correct — D, (d) 400 W. The quantity that stays with the bulb is its resistance, not its wattage. A rating of 100 W at 110 V is a statement about the bulb only when it is run at 110 V; what the manufacturer has actually fixed is the filament, and the rating is how that filament behaves at its intended voltage. Find the resistance from the rating. Power equals voltage squared divided by resistance, so resistance equals voltage squared divided by power: 110 multiplied by 110 is 12,100, and dividing by 100 gives 121 ohms. Now put that same 121 ohms across 220 volts. Power is again voltage squared divided by resistance: 220 multiplied by 220 is 48,400, and dividing by 121 gives 400. The bulb dissipates 400 watts, which is option (d). The reasoning in one line, and the form to carry into the hall: for a fixed resistance, power varies as the square of the applied voltage. Here the voltage is doubled, so the power is multiplied by two squared, which is four, and four times 100 W is 400 W. No arithmetic beyond that is needed, and the same rule answers the reverse question — halve the voltage on a bulb and it draws a quarter of its rated power, which is why a lamp on a sagging supply is not merely a little dimmer but very much dimmer. One honest qualification, which the paper's model sets aside and a well-taught candidate should still know. A tungsten filament is not an ohmic resistor over this range: its resistance rises steeply as it heats, so a real bulb at double its rated voltage would draw somewhat less than 400 W — and, far more to the point, it would fail almost at once, because the filament would be driven far past its design temperature. The question asks for the power dissipated under the standard constant-resistance model, and under that model the answer is exactly 400 W. Treating resistance as the fixed quantity is the whole method being tested.

• (a)50 W — This is the rated power halved, and it inverts the physics twice over. Halving would follow only if power fell as the voltage rose, which it does not for a fixed resistance, and even an inverse relationship would not give this figure. The number appeals to a candidate who half-remembers that something is halved in a problem involving 110 and 220, but nothing here is halved: the voltage doubles and the power rises.
• (b)100 W — This treats the printed rating as a property the bulb carries with it to any supply, which is the central misconception the item is built to catch. A 100 W marking is shorthand for 'this bulb consumes 100 W when 110 V is applied to it'; the fixed physical quantity is the 121 ohms of the filament, and the power it dissipates changes with whatever voltage is put across it. A bulb is not a constant-power device, and a candidate who chooses this option has read the rating as a guarantee rather than as a measurement at a stated voltage.
• (c)200 W — This is the answer obtained by assuming power is proportional to voltage — double the volts, double the watts. It is the most attractive of the three wrong options because it applies a real relationship in the wrong place: power does equal voltage multiplied by current, but when the voltage across a fixed resistance doubles, the current through it doubles as well, so the product is multiplied by four and not by two. Forgetting that the current also changes is the single commonest slip in this family of problems.

Electrical power in a resistive device can be written three ways, and choosing the right one is most of the work. Power equals voltage multiplied by current; substituting Ohm's law gives power as current squared multiplied by resistance, or as voltage squared divided by resistance. The three forms are equivalent, but each holds a different quantity constant in the reader's mind, and the form to use is the one whose fixed quantity is genuinely fixed in the problem. Here the resistance is the fixed quantity, so the voltage-squared-over-resistance form is the right tool, and it makes the square-law dependence on voltage visible at a glance. A rating plate on any appliance follows the same convention: it states the power the device draws at its design voltage, so it is a pair of numbers and neither means anything without the other. The same square law explains why appliances are so sensitive to supply voltage, why a step-down transformer is used to run equipment designed for a 110 V country on a 230 V supply rather than simply plugging it in, and why the current-squared form is the one used for transmission losses in a cable, where the current is what is fixed and the resistance is the line's own.

The general science block of the EO/AO paper takes its physics from school electricity and mechanics and asks for one clean substitution rather than a derivation. This item is a favourite because it can be answered in five seconds by proportional reasoning and takes a minute by full arithmetic, so it rewards the candidate who has internalised the square law. The habit it trains is asking, before any formula is written, which quantity in the problem is genuinely constant — here the filament's resistance, not the wattage on the box.

• Electrical power can be written as voltage times current, as current squared times resistance, or as voltage squared divided by resistance.
• A bulb rated 100 W at 110 V has resistance 110 squared divided by 100, which is 121 ohms.
• At 220 V that resistance dissipates 220 squared divided by 121, which is 400 W.
• For a fixed resistance, power varies as the square of the applied voltage, so doubling the voltage quadruples the power.
• A wattage rating is meaningless without the voltage at which it was measured; a bulb is not a constant-power device.
• Halving the supply voltage reduces the power to one quarter, which is why lamps dim sharply on a low supply.
• A real tungsten filament is non-ohmic — its resistance rises with temperature — so the constant-resistance model is an approximation.
• The unit of electrical energy billed to consumers is the kilowatt-hour, one kilowatt sustained for one hour.

• Treating the printed wattage as a fixed property of the bulb instead of a value measured at the rated voltage.
• Assuming power is proportional to voltage and doubling it, which gives 200 W.
• Forgetting that the current changes when the voltage changes across a fixed resistance.
• Using the current-squared form when the current is not the fixed quantity in the problem.

Electricity items in EO/AO papers ask for a single substitution: the power drawn at a changed voltage, the resistance implied by a rating, the energy consumed in a stated number of hours, or the comparison of two bulbs in series against parallel. Memorise the three power formulae, and practise the proportional shortcut so that a doubled or halved voltage can be handled without arithmetic.

No directly related past PYQ was found.

• practice — not a real PYQ

  A heater rated 1,000 W at 220 V is operated at 110 V. The power consumed will be :

  • (a)250 W
  • (b)500 W
  • (c)1,000 W
  • (d)2,000 W

  Answer(a) 250 W

• practice — not a real PYQ

  The resistance of a bulb rated 60 W at 120 V is :

  • (a)2 ohms
  • (b)60 ohms
  • (c)120 ohms
  • (d)240 ohms

  Answer(d) 240 ohms