Question

A heater coil is rated $100W$, $200V$. It is cut into two idential parts. Both parts are connected together in parallel, to the same source of $200V$. The energy liberated per second in the new combination is

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

Answer 1

Answer: (D)

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.
$R^{\prime }=\frac{200\times 200}{200+200}$
$=100\Omega$.
Energy liberated per second when combination is connected to a source of $200V$,
$=\frac{V^2}{R^{\prime }}=\frac{200\times 200}{100}$
$=400J$.