Question

A heater coil is rated 100W , 200V . it is cut in to two identical parts . both parts are conencted together in parallel to the same sources of 200V . Calculate the energy released per second in the new combination .

Answer 1

P = 100W. (power rating) V = 200V. ( voltage supply) thus resistance of the coil: P = V2 / R thus R = 40000/100 = 400 ohm hence of the two new parts produced, new resistance of each part, R' = 200 ohm each on parallel connection, net resistance: 1/Rt = 1/R' + 1/R' hence Rt = R'/2 = 100 ohm and the new power rating, P = V2 /Rt = 40000/100 = 400 W

Answer 2

The energy released per second (power) in the new combination is 400 W.

Explanation:

Given for heater coil power, P = 100 W.

The voltage, V = 200 V.

The power, voltage and resistance relation is given as

$P=\frac{V^2}{R}$

Here R is the resistance.

substitute the given values, we get

$100W=\frac{(200V)^2}{R}$

$R=\frac{(200V)^2}{100}=400\Omega$

As this resistance cut into two identical parts, so

$R_1=R_2=\frac{400}{2} =200\Omega$

and they are connected in parallel, so equivalent resistance is

$\frac{1}{R_e_q_t} =\frac{1}{200\Omega} +\frac{1}{200\Omega}$

$R_{eqt} =100\Omega$

Therefore, power for the combination,

$P=\frac{(200V)^2}{100\Omega}$

P = 400 W.

Thus, the energy released per second (power) in the new combination is 400 W.

#Learn More: power.

https://brainly.in/question/2101413