In the above question , if the heaters are connected in series , the combination will bolt the same water in time

Options

A. `(t_(1)t_(2))/(t_(1) - t_(2))`
B. `(t_(1)t_(2))/(t_(1) + t_(2))`
C. `(t_(1)t^(2) + t_(2)^(2))/(t_(1) +t_(2))`
D. `(t_(1)t^(2) - t_(2)^(2))/((t_(1) -t_(2)))`

Answer

d

Explanation

When the heaters are in series , then current through each heater `I = V//(R_(1) +R_(2))`. If `t` is the time taken to boll the water then
`H = ((V)/(R_(1) + R_(2)))^(2) (R_(1) + R_(2)) t = (V^(2))/(R_(1)) t_(1)`
or `(t)/(R_(1) + R_(2)) = (t_(1))/(R_(1))`
or `t = t_(1)((R_(1) + R_(2))/(R_(1))) = t_(1) (1 + (R_(2))/(R_(1)))`
`t = t_(1) (1+(t_(2))/(t_(1))) = t_(1) ((t_(1) +t_(2)))/(t_(1)) = t_(1) + t_(2) = (t_(1)^(2) - t_(2)^(2))/((t_(1) - t_(2)))`