Electric Circuits and Resistance: Question 6

Syllabus 4.2.4

Multiple choice Core 1 mark

A filament lamp is connected to a 12 V12\text{ V} d.c. supply. When the lamp is operating normally, a current of 0.40 A0.40\text{ A} flows through it. What is the resistance of the lamp at this current?

Choose an answer to check it, then compare with the worked solution below.

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Worked solution

Step 1: Recall the relationship between resistance, p.d. and current

Resistance is defined by:

R=VIR = \frac{V}{I}

Step 2: Substitute the values

The potential difference across the lamp is V=12 VV = 12\text{ V} and the current through it is I=0.40 AI = 0.40\text{ A}:

R=12 V0.40 AR = \frac{12\text{ V}}{0.40\text{ A}}

Step 3: Work out the resistance

R=30ΩR = 30\,\Omega

Why the other options are wrong

OptionHow it arisesError
0.033Ω0.033\,\Omega0.40÷120.40 \div 12Formula inverted (current ÷ p.d.)
4.8Ω4.8\,\Omega12×0.4012 \times 0.40Multiplied instead of dividing
11.6Ω11.6\,\Omega120.4012 - 0.40Subtracted instead of dividing

Final answers

  • Resistance =30Ω= \boxed{30\,\Omega}, option D