Which one of the following formulas does not represent electrical power?
- (a)I²R
- (b)IR²
- (c)VI
- (d)V²/R
Answer
Why
Correct — B, IR². Electrical power is the base relation P = VI; using Ohm's law V = IR to substitute gives the two equivalent forms P = I²R and P = V²/R. The expression IR² is none of these — putting V = IR shows IR² equals (V/R) × R² = VR, which carries units of volt × ohm rather than watts, so it cannot represent power.
Why the others are wrong
- (a)I²R — This is a genuine power formula (the Joule-heating form), obtained by substituting V = IR into P = VI; its units work out to watts.
- (c)VI — This is the fundamental definition of electrical power — potential difference times current (volt × ampere = watt) — so it does represent power.
- (d)V²/R — This is a correct power formula, obtained by substituting I = V/R into P = VI; it also has units of watts.
Concept
Electrical power is the rate at which electrical energy is converted, P = VI (volt × ampere = watt). Combining this with Ohm's law V = IR produces two further equivalent expressions, P = I²R and P = V²/R. The three forms are interchangeable; anything else is not a valid power formula.
The trap sits in the near-identical look of I²R and IR². Only the first is power; in the second the exponent is on the resistance instead of the current, which changes the units entirely. Checking units (must reduce to watts) is the safest test.
Key facts
- Electrical power P = VI, with the watt (W) as SI unit; 1 W = 1 J/s = 1 V·A.
- Using Ohm's law V = IR, the three equivalent power formulas are P = VI = I²R = V²/R.
- IR² is not a power formula — it reduces to VR (volt × ohm), not watts.
- P = I²R is the heat-dissipation (Joule's law) form; P = V²/R is convenient when the voltage is fixed.
Study next
Common traps
- Reading IR² as I²R — the exponent is on the wrong factor.
- Forgetting that P = V²/R is also a valid power formula.
Asked as 'which expression is / is not electrical power?' or as a direct calculation using P = VI, I²R or V²/R.
Related PYQs
An incandescent electric bulb converts 20 % of its power consumption into light, and the remaining power is dissipated as heat. The bulb’s filament has a resistance of 200 Ω and 2 A current flows through it. If the bulb remains ON for 10 h and the rate of electricity charge is ₹5 per unit, then which among the following is the correct amount for the money spent on producing light?
- (a) ₹ 5
- (b) ₹ 6
- (c) ₹ 7
- (d) ₹ 8
Answer(d) ₹ 8
Same concept — the numerical uses the P = I²R power formula (the valid form) to find the bulb's power consumption.
Two equal resistors R are connected in parallel, and a battery of 12 V is connected across this combination. A d.c. current of 100 mA flows through the circuit as shown below. The value of R is
- (a) 120 Ω
- (b) 240 Ω
- (c) 60 Ω
- (d) 100 Ω
Answer(b) 240 Ω
Related — Ohm's law (V = IR) is the substitution that turns P = VI into I²R and V²/R, the forms tested here.
Practice
- practice — not a real PYQ
The power dissipated in a resistor of resistance R carrying a current I is
- (a)I²R
- (b)IR
- (c)I/R²
- (d)IR²
Answer(a) I²R — obtained by substituting V = IR into P = VI. - practice — not a real PYQ
An electric heater draws 4 A when connected to a 250 V supply. Its power rating is
- (a)62·5 W
- (b)1000 W
- (c)254 W
- (d)500 W
Answer(b) 1000 W — P = VI = 250 × 4 = 1000 W.