Sanju
Last Activity: 6 Years ago
Given that the cross section area of wire a & b are in the ratio 1:6
We have, H = I2RT & R = ρL/A
where H = heat produced, I = current, R = resistance, T=time, ρ=resistivity of the material, L = length of the wire and A = area of cross section..
Substituting the relation R = ρL/A in the eqn H = I2RT
H = I2(ρL/A)T
Since A1: A2 = 1:6, the ratio of heat produced in wire a & b is:
H1: H2 = A2:A1 = 6:1
The ratio of heat produced in the wire a & b when same voltage is applied across each wire = 6:1
