question_answer

A heater coli 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 Calculate the energy liberated per second in the new combination.

Answer:

Given, power, P = 100 W; Voltage, V = 200 V Resistance of heater coil, \[{{R}_{0}}=\frac{{{V}^{2}}}{P}=\frac{{{(200)}^{2}}}{100}=400\Omega \] On cutting it into two identical parts, resistance of each part \[R=\frac{{{R}_{0}}}{2}=\frac{400}{2}=200\Omega \] On joining these two parts in parallel, the equivalent resistance is             \[R'=\frac{200\times 200}{200+200}=100\Omega \] As again voltage, V = 200 V \[\therefore \] Energy liberated per second in new combination             \[=\frac{{{V}^{2}}}{R'}=\frac{200\times 200}{100}=400\,W\]