Question

) A cylindrical resistor of radius 5.5 mm and length 1.8 cm is made of material that has a resistivity of 3.50×10^-5Ω m. What is the current density when the energy dissipation rate in the resistor is 1.0 W?

What is the potential difference?

Answer 1

   given powwer dissipation rate at the resistor is 1 W
we know that P = i^2*R so current
   I= sqrt(P/R)
   = sqrt(PA/rho*L)
   = sqrt(1*3.14*5.5*10^-3*5.5*10^-3/3.50*10^-5*1.8*10^-2)
   = 12.278 A
now
   current density J = i/A
           = 12.278/3.14*5.5*10^-3*5.5*10^-3
           = 0.129262515*10^6
           = 129.262515 kA/m2

potential difference V= ir
   v= i*R
   = i*rho*l/A
   = 12.278*3.5*10^-5*1.8*10^-2/(3.14*5.5*10^-3*5.5*10^-3)
    = 81.4276 mV