Relation between current in a conductor and time is shown in fig.

If the resistance of the conductor is R, then the total heat disspipated across resistance R is

Options

A. `(i_0^2Rt_0)/(2)`
B. `(i_0^2Rt_0)/(4)`
C. `(i_0^2Rt_0)/(3)`
D. `i_0^2Rt_0`

Answer

C

Explanation

`q = int idt =` area of given curve `= (1)/(2)i_0t_0`
`(i)/(i_0) +(t)/(t_0) 1` or `I = i_0(1-(t)/(t_0))`
Heat `= int i^2 Rdt = int_0^(t_0)i_0^2R(1-(t)/(t_0))^2 dt`
`H = (i_0^2R(1-(t)/(t_0))^3)/(3(-(1)/(t_0)))]_0^(t0) or H = (Rt_0i_0^2)/(3)`