11its easy,we hav to only equate the powers
kA∂T/∂x=V2/R
=>R=V2/(kA∂T/∂x )
puting the values i m geting=402*4.2*10-3/(4*10-4*6*.25*100 )80/3 * 4.2=112
The walls of a closed cubical box of edge 50 cm are made of a material of thickness 1 mm and thermal conductivity 4 * 10^-4 cal per sec per cm per °C. The interior of the box maintained at 100°C above the outside temperature by a heater placed inside the box and connected across a 40 Vd.c. source. Calculate the resistance of the heater
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3 Answers
11