The charge flowing in a conductor varies with time as Q=at-bt2, then the current changes at the rate of (-2b) falls to zero after T=( (a/2b) ) (3) reaches a maximum and then decreases (4) will remain constant

Question

The charge flowing in a conductor varies with time as Q=at-bt2, then the current changes at the rate of (-2b) falls to zero after T=( (a/2b) ) (3) reaches a maximum and then decreases (4) will remain constant (A) 1, 2 and 3 are correct (B) 1 and 2 are correct (C) 2 and 4 are correct (D) 1 and 3 are correct

Tardigrade Admin · Accepted Answer

Q=at-bt2 Current, I=(dQ/dt) =(d/dt)(at-bt2) =a-2bt If I=0 then 0=a-2bt or t=(a/2b) Also, (dI/dt)=(d/dt)(a-2bt) =-2b Therefore, current change at the ratio of (-2b) .

Q. The charge flowing in a conductor varies with time as $Q = a t - b t^{2},$ then the current changes at the rate of $(- 2 b)$ falls to zero after $T = (\frac{a}{2 b})$ (3) reaches a maximum and then decreases (4) will remain constant

1, 2 and 3 are correct

1 and 2 are correct

2 and 4 are correct

1 and 3 are correct

$Q = a t - b t^{2}$ Current, $I = \frac{d Q}{d t}$
$= \frac{d}{d t} (a t - b t^{2})$
$= a - 2 b t$ If $I = 0$ then $0 = a - 2 b t$ or $t = \frac{a}{2 b}$ Also, $\frac{d I}{d t} = \frac{d}{d t} (a - 2 b t)$
$= - 2 b$
Therefore, current change at the ratio of $(- 2 b)$ .

Q. The charge flowing in a conductor varies with time as Q=at−bt2, then the current changes at the rate of (−2b) falls to zero after T=(2ba​) (3) reaches a maximum and then decreases (4) will remain constant