Question

A heater coil is rated 100W , 200V. It is cut into two identical parts. both parts are connected together in parallel to the same source of 200V. Calculate the energy liberated per second in the New combination.​

Answer 1

Answer:

Correct option is

D

400J

P=100W

V=200V

P=

R

V

2

R=

400

(200)

2

=100Ω

Net Resistance:

R

t

1

=

R

′

1

+

R

′

1

=

R

′

2

⟹R

t

=

2

R

′

=

2

200

=100Ω

Net Power=P=

RT

V

2

=

100

40000

=400J

Answer 2

Power , p =100W, voltage , V = 200V

P=VI ,

current ,I = P/V= 100 / 200 = . 5A

The resistance of the coil , R = V / I = 200/ .5 = 400 ohm

When cut it two identical parts resistance becomes halved,

ie , Two 200 ohm resistances ( let , R1 and R2 ) are connected in parallel ,

Current through R1 , I1 = V /R1 = 200 / 200 = 1 A

The energy liberated per second , H = I2 R t = 1*1*20*1 = 20 J

Current through R2 , I2 = V / R2 = 200 / 200 = 1 A

The energy liberated per second , H = I2 R t = 20 J

Total energy liberated = 20 +20 = 40 joules