BPSC 71st CCE Prelims General Studies · Question 128 of 150

Correct — B, IR². Electric power is the rate at which electrical energy is converted in a circuit, and it has exactly one starting definition. Moving a charge Q through a potential difference V takes work VQ; doing that in time t makes the rate VQ/t, and since current is charge per unit time, I = Q/t, the rate is P = VI. Everything else is that one equation wearing a different coat, obtained by substituting Ohm's law V = IR. Replace V with IR and you get P = (IR) × I = I²R. Replace I with V/R and you get P = V × (V/R) = V²/R. So three of the four options — I²R, VI and V²/R — are the same physical quantity expressed in whichever pair of variables a problem happens to give you, and the fourth, IR², is not a power expression at all. The fastest proof is the units. A watt is a volt-ampere, so any valid expression must reduce to V × A. Check them: I²R = A² × Ω = A × (A × Ω) = A × V ✓; VI = V × A ✓; V²/R = V × (V/Ω) = V × A ✓; but IR² = A × Ω² = (A × Ω) × Ω = V × Ω, which is a volt-ohm and not a watt ✗. A numerical check makes it concrete: pass 2 A through a 10 Ω resistor, so V = IR = 20 V. Then VI = 40 W, I²R = 4 × 10 = 40 W and V²/R = 400/10 = 40 W, all agreeing; IR² gives 2 × 100 = 200, a number that corresponds to no physical quantity in the circuit. Note the provenance, because it tells you how to prepare: this item is NCERT Class X Science, 'Electricity' — Chapter 11 in the book currently in print, Chapter 12 in older editions, exercise question 2, reproduced word for word down to the order of the four options. It is also a negative stem — 'does not denote' — which is where most of the marks are actually lost.

• (a)I²R — This one IS a power expression, so it cannot be the answer. It is Joule's law of heating written as a rate: NCERT gives the heat produced as H = I²Rt, so the power is H/t = I²R. It also follows in one step from the definition, since P = VI and V = IR give P = (IR)I = I²R. This is the form used whenever current and resistance are known — heating elements, fuse wires and transmission-line losses — and it is the option built to catch a candidate who skims past the word 'not' and picks the first formula that looks familiar.
• (c)VI — Also a genuine power expression, and in fact the primary one from which the other two are derived. It is the definition itself: work done per unit time in driving a current I across a potential difference V. It gives the SI unit directly — one watt is the power consumed by a device carrying one ampere at one volt, so 1 W = 1 V × 1 A. It is the form used for every appliance rating on a nameplate, which is why a 220 V, 1 kW electric iron is known at once to draw 1000/220 = 4.54 A.
• (d)V²/R — Equally valid, obtained by substituting I = V/R into P = VI. It is the form to use when the supply voltage and the appliance's resistance are known, and it carries a consequence worth remembering: at fixed resistance the power varies as the SQUARE of the applied voltage, so a bulb rated 220 V, 100 W run on a 110 V supply consumes not 50 W but 25 W — a quarter of the power for half the voltage. That is a separate exercise question in the very same NCERT chapter.

Energy and power are different quantities and a circuit question always wants one or the other. Energy is what an appliance consumes in total, measured in joules; power is the rate at which it consumes it, measured in watts, where 1 W = 1 J s⁻¹ = 1 volt-ampere. For an electric circuit the chain of reasoning runs: a source maintains a potential difference V, work VQ is done in pushing charge Q across it, and since I = Q/t the rate of doing that work is P = VI. If the circuit is purely resistive, all of that energy is dissipated as heat — the heating effect of electric current, quantified by Joule's law, H = I²Rt. That law states three proportionalities at once: the heat produced rises with the SQUARE of the current, rises in direct proportion to the resistance, and rises in direct proportion to the time for which the current flows. The squared dependence on current is the single most consequential fact in domestic electricity, because it means doubling the current quadruples the heat — which is why an electric iron, toaster, kettle and heater work at all, why a bulb filament of tungsten (melting point given as 3380 °C in NCERT's own text; the accepted modern value is 3422 °C) glows, and why a fuse melts when the current runs away. Because power and energy are billed differently, the commercial unit is the kilowatt-hour, the 'unit' on a domestic bill: 1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J.

Two habits solve this question and dozens like it. The first is the unit test, which is faster and safer than recalling a list of formulae. Write the target unit — watt = volt × ampere — and reduce each option to volts and amperes using ohm = volt per ampere. I²R becomes A × (A × Ω) = A × V, correct. VI is already V × A, correct. V²/R becomes V × (V/Ω) = V × A, correct. IR² becomes (A × Ω) × Ω = V × Ω, which is not a watt, and that single line is the whole answer. The second habit is structural: every genuine power formula, once Ohm's law has been used, is quadratic in the voltage-current pair — it contains two factors drawn from {V, I} and never more than one resistance, whether that resistance sits on top or underneath. IR² breaks that pattern by carrying one current and two resistances, and once you have seen the pattern it is unmissable. The single discriminating detail, though, is simpler still and it is typographical: I²R and IR² differ only in where the superscript sits. Under a 150-question, two-hour clock, with −1/3 for a wrong answer, this question is won or lost by reading the exponents carefully and by noticing the negative stem. 'Which term does NOT denote' inverts the task, so the three familiar formulae are the wrong answers here — and a candidate who has been trained to recognise I²R as a correct power expression will tick it precisely because it is correct.

• P = VI is the definition of electric power (work VQ done in time t, with I = Q/t). Ohm's law V = IR converts it into P = I²R and P = V²/R, so all three are the same equation in different variables.
• The SI unit is the watt: one watt is the power consumed by a device carrying 1 ampere at a potential difference of 1 volt, so 1 W = 1 V × 1 A. The commercial unit of electrical energy is the kilowatt-hour, and 1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J.
• The unit test that eliminates IR²: ampere × ohm² = (ampere × ohm) × ohm = volt × ohm, which is not a watt. By contrast ampere² × ohm and volt²/ohm both reduce cleanly to volt-ampere.
• Joule's law of heating, H = I²Rt: the heat produced in a resistor is proportional to the square of the current, to the resistance, and to the time. Doubling the current therefore quadruples the heat — the basis of the electric iron, toaster, kettle, heater and the fuse.
• Because P = V²/R, power at fixed resistance goes as the square of the voltage: a bulb rated 220 V, 100 W draws only 25 W on a 110 V supply, not 50 W.
• The stem is NCERT Class X Science, 'Electricity' — Chapter 11 in the book currently in print, Chapter 12 in older editions, exercise question 2, reproduced verbatim including the order of the options — 'Which of the following terms does not represent electrical power in a circuit? (a) I²R (b) IR² (c) VI (d) V²/R'.

• Missing the negative stem: the question asks which term does NOT denote power, so the three familiar and correct formulae are all wrong answers here
• Reading IR² as I²R — the two differ only in the position of the superscript, and under time pressure the eye supplies the familiar shape
• Assuming power always halves when voltage halves: at fixed resistance P = V²/R, so half the voltage gives one-quarter of the power

BPSC in this paper does not adapt NCERT, it photocopies it: Q128 is exercise question 2 of the Class X 'Electricity' chapter with the four options in their original order, and Q129 is exercise question 5 of the very next chapter. The reward for reading the NCERT exercises themselves is therefore direct and immediate. UPSC has never asked 'which of these is the formula for power'. Its electricity questions are causal and assertion-reason shaped — why a metal wire heats when current flows through it, why domestic wiring must be parallel, what a fluorescent tube's choke does, what happens to resistance when a wire's length, diameter and resistivity all change together. Same chapter, but UPSC tests what the formula explains rather than what it looks like.

• practice — not a real PYQ

  An electric bulb rated 100 W, 220 V is operated on a 110 V supply. Assuming its resistance stays the same, the power it now consumes is

  • (a)100 W
  • (b)50 W
  • (c)25 W
  • (d)12.5 W

  Answer(c) 25 W — the bulb's resistance is R = V²/P = 220²/100 = 484 Ω, and at 110 V the power is V²/R = 110²/484 = 25 W. Since P varies as the square of the voltage at fixed resistance, halving the voltage quarters the power; 50 W is the trap for anyone who assumes power falls in direct proportion to voltage.

• practice — not a real PYQ

  If the current flowing through a resistor is doubled while its resistance and the time of flow remain unchanged, the heat produced in it becomes

  • (a)half
  • (b)double
  • (c)four times
  • (d)unchanged

  Answer(c) four times — Joule's law of heating gives H = I²Rt, so heat depends on the SQUARE of the current; doubling I multiplies H by 2² = 4. This squared dependence is exactly why a short circuit, which sends the current up by a large factor, produces enough heat to melt a fuse wire almost instantly.