A heater coil is rated 100 W, 200 V. It is cut into two identical parts. Both parts are connected together in parallel to the same source of 200 V. The energy liberated per second in the new combination is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 400 W
The power dissipated by a heater coil on a supply V is \( P = V^{2}/R \), so the rated values give the resistance of the coil, and cutting and re-joining the coil changes R.
\( \displaystyle R = \frac{V^{2}}{P} = \frac{(200\,\mathrm{V})^{2}}{100\,\mathrm{W}} = 400\,\Omega \)
Each of the two identical parts has half the length, hence half the resistance, \( 200\,\Omega \). Joined in parallel:
\( \displaystyle R' = \frac{200\times200}{200+200} = 100\,\Omega \)
\( \displaystyle P' = \frac{V^{2}}{R'} = \frac{(200\,\mathrm{V})^{2}}{100\,\Omega} = 400\,\mathrm{W} \)
Shortcut: Two halves in parallel give \( R/4 \) (each half is \( R/2 \), and two equal resistors in parallel halve it again). At the same voltage \( P\propto1/R \), so the power becomes 4 times, \( 4\times100\,\mathrm{W} = 400\,\mathrm{W} \).