Question

An electric heater of resistance 10 Ω draws 12 A from the service mains 3 hours. Calculate the rate at which heat is developed in the heater.

Answer 1

Answer:

Given parameters

Given parametersResistance of the electric heater (R) = 8 Ω

Given parametersResistance of the electric heater (R) = 8 ΩCurrent drawn (I) = 15 A

Given parametersResistance of the electric heater (R) = 8 ΩCurrent drawn (I) = 15 AThe rate of heat produced by the device in terms of power is expressed as

Given parametersResistance of the electric heater (R) = 8 ΩCurrent drawn (I) = 15 AThe rate of heat produced by the device in terms of power is expressed as⇒ P= I2R

Given parametersResistance of the electric heater (R) = 8 ΩCurrent drawn (I) = 15 AThe rate of heat produced by the device in terms of power is expressed as⇒ P= I2RSubstitute the given values in the above equation we get

Given parametersResistance of the electric heater (R) = 8 ΩCurrent drawn (I) = 15 AThe rate of heat produced by the device in terms of power is expressed as⇒ P= I2RSubstitute the given values in the above equation we get⇒ P = (15)2 x 8

Given parametersResistance of the electric heater (R) = 8 ΩCurrent drawn (I) = 15 AThe rate of heat produced by the device in terms of power is expressed as⇒ P= I2RSubstitute the given values in the above equation we get⇒ P = (15)2 x 8⇒ P = 1800 J/s

Given parametersResistance of the electric heater (R) = 8 ΩCurrent drawn (I) = 15 AThe rate of heat produced by the device in terms of power is expressed as⇒ P= I2RSubstitute the given values in the above equation we get⇒ P = (15)2 x 8⇒ P = 1800 J/sTherefore, heat is produced by the heater is 1800 J/s.

Explanation:

PLEASE SUBSCRIBE TO MY YOUTUBE CHANNEL NAME KHURAFATI POINT

Answer 2

S O L U T I O N :

As we know that electrical power is the rate at which electric energy is consumed by an appliance .

It's denoted by P .

$\boxed{\bf{P=\frac{W}{t} = Vi = i^{2} R = \frac{V^{2} }{R} }}$

$\underline{\bf{Given\::}}$

• Resistance, (R) = 10 Ω .
• Current, (i) = 12 Ampere .
• Time , (T) = 3  hours

$\underline{\bf{Explanation\::}}$

Rate of heat = Power

$\mapsto\tt{P=\dfrac{E}{t} }$

$\mapsto\tt{P=\dfrac{i^{2}R\cancel{t}}{\cancel{t}} }$

$\mapsto\tt{P=i^{2} R }$

$\mapsto\tt{P=(12)^{2} \times 10}$

$\mapsto\tt{P=144\times 10}$

$\mapsto\bf{P=1440\:watt}$

Thus;

The heat is developed at the rate of 1440 joule/sec.