Question

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

• A. 100 J
• B. 200 J
• C. 300 J
• D. 400 J

Answer 1

Answer: (D)

400 J

Answer 2

: Resistance of heater coil,

$R' = \frac{V^{2}}{P} = \frac{200 \times 200}{100} = 400\Omega$

Resistance of either half part $= 200\Omega$.

Equivalent resistance when both parts are connected in parallel,

$R' = \frac{200 \times 200}{200 + 200} = 100\Omega.$

Energy liberated per second when combination is connected to a source of 200 V,

$= \frac{V^{2}}{R'} = \frac{200 \times 200}{100} = 400J.$