Question

A heating coil is labelled 100 W, 220 V. The coil is cut in half and the two pieces are joined in parallel to the same source. The energy now liberated per second is

• A. 200 J
• B. 400 J
• C. 25 J
• D. 50 J

Answer 1

Answer: (B)

400 J

Answer 2

Let resistance of the heating coil be R, when coil cut in two equal parts, resistance of each part will be $\frac{R}{2}$. When these two parts are corrected in parallel, $R_{eq} = \frac{R}{4}$ i.e. resistance becomes, so according to $P \propto \frac{1}{R};$ Power becomes 4 times i.e. P′ = 4P = 400 J/sec