An electric bulb is marked 100 W, 230 V. If the supply voltage drops to 115 V, what is the total heat and light energy produced by the bulb in 20 min?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 30 000 J
The rating 100 W, 230 V gives the resistance of the filament through \( P=V^{2}/R \). Taking this resistance as unchanged, the power at the lower voltage follows from the same relation, and the energy is power × time.
\( \begin{aligned} R &= \frac{V^{2}}{P} \\ &= \frac{(230\,\mathrm{V})^{2}}{100\,\mathrm{W}} \\ &= 529\,\Omega \end{aligned} \)
\( \begin{aligned} P' &= \frac{V'^{2}}{R} \\ &= \frac{(115\,\mathrm{V})^{2}}{529\,\Omega} \\ &= 25\,\mathrm{W} \end{aligned} \)
\( t=20\,\mathrm{min}=1200\,\mathrm{s} \)
\( \begin{aligned} W &= P't \\ &= 25\,\mathrm{W}\times1200\,\mathrm{s} \\ &= 30\,000\,\mathrm{J} \end{aligned} \)