CDS (II) 2021 General Knowledge · Question 20 of 120

Correct — C, 22 W. Electric power is the rate at which electrical energy is consumed, and for a device carrying a steady current it is the product of the potential difference across it and the current through it — P equals V times I. Substituting the two numbers the stem supplies, 110 volts multiplied by 0·2 ampere gives 22 watts. The watt is defined so that this multiplication works directly: one watt is the power consumed by a device carrying one ampere at a potential difference of one volt. Nothing else in the question needs computing, and the only place to slip is the decimal in the current. Two useful cross-checks are available. The resistance implied is V divided by I, that is 550 ohms, which is a sensible figure for a small lamp. And the answer options are all the same digits with the decimal moved, which is a standing invitation to multiply by 0·2 rather than by 2 or by 0·02.

• (a)0·22 W — A hundred times too small. It comes from treating the supply as 1·1 V, or from misplacing two decimal places in the multiplication.
• (b)2·2 W — Ten times too small — the value obtained if the current is read as 0·02 A instead of 0·2 A.
• (d)220 W — Ten times too large. It appears if the current is taken as 2 A, or if the supply is confused with the familiar 220 V mains figure.

Power is energy per unit time, and in an electric circuit it is given by P equal to VI. Combining that with Ohm's law produces two further forms, P equal to I squared R and P equal to V squared divided by R, and which of the three is easiest depends on what the question supplies. The SI unit is the watt, with the kilowatt equal to a thousand watts. Energy is power multiplied by time, which gives the watt hour and, in commercial use, the kilowatt hour — the unit on an electricity bill.

The item tests a single substitution, so the marks are really being awarded for care with units and decimals. Reading 0·2 A as two hundred milliamperes helps, because milliamperes are the units small lamps and electronic parts are usually rated in. The four options form a decimal ladder, which is the examiner's way of catching a misplaced point rather than a misunderstood formula; running the sanity check of V divided by I for the resistance filters out any answer that would imply an absurd device. Note the printing style in the paper: the decimal point is set as a raised dot, so 0·2 A is two tenths of an ampere and 0·22 W is twenty-two hundredths of a watt. That is a typographic convention of the booklet, faithfully reproduced here, and not a symbol for anything else.

• Electric power P equals VI, and also I squared R and V squared divided by R.
• One watt is the power consumed by a device that carries 1 A of current when operated at a potential difference of 1 V.
• 110 V multiplied by 0·2 A gives 22 W; the implied resistance, V divided by I, is 550 ohms.
• The kilowatt equals 1000 watts, and electrical energy is power multiplied by time.
• The commercial unit of electrical energy is the kilowatt hour, equal to 3·6 × 10⁶ joules.

• Misplacing the decimal point in a current given in tenths of an ampere or in milliamperes.
• Assuming the supply must be 220 V because that is the familiar mains value.
• Using P equal to V squared divided by R when the resistance has not been given, instead of the direct product VI.

As a direct substitution into P equal to VI, as a rated-bulb question asking for the current drawn, or as an energy-consumption sum in kilowatt hours.

• practice — not a real PYQ

  A 60 W bulb operates on a 240 V supply. The current drawn by it is

  • (a)0·25 A
  • (b)0·5 A
  • (c)4 A
  • (d)14400 A

  Answer(a) 0·25 A — current is power divided by voltage, that is 60 ÷ 240.

• practice — not a real PYQ

  Which one of the following expressions does not represent electric power in a circuit?

  • (a)I²R
  • (b)IR²
  • (c)VI
  • (d)V²/R

  Answer(b) IR² — the other three follow from P equal to VI combined with Ohm's law, but IR² does not.