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10) A heater coil is rated 100W , 200V .If 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.
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


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