Question

The charge flowing in a conductor varies time as $Q=at-\frac{1}{2}b{t}^2+\frac{1}{6}c{t}^3$, where a, b, c are positive constants. Then, the current

• A. has an initial value a
• B. reaches a minimum value after time b/c
• C. reaches a maximum value after time b/c
• D. has either a maximum or a minimum value $(a-\frac{b^2}{2c})$

Answer 1

Answer: (A), (B), (D)

The correct options are
A has an initial value a
B reaches a minimum value after time b/c
D has either a maximum or a minimum value $(a-\frac{b^2}{2c})$

Here, the charge flowing in a conductor varies time as $Q=at-\frac{1}{2}b{t}^2+\frac{1}{6}c{t}^3$.
Now, by definition current is rate of flow of charge, hence
$I=\frac{dQ}{dt}$
$I=\frac{d}{dt}(at-\frac{1}{2}b{t}^2+\frac{1}{6}c{t}^3)$
$I=a-bt+\frac{1}{2}c{t}^2$...................................................(1)
For t = 0,
$I=a-b\left(0\right)+\frac{1}{2}c(0{)}^2=a$
$\Rightarrow$ The initial value of I is 'a'.
Now, from(1) we can write
$\frac{dI}{dt}=\frac{d}{dt}(a-bt+\frac{1}{2}c{t}^2)$
$\frac{dI}{dt}=0-b+ct=b+ct$
Since, current is varying with time, it will have a minimum value when $\frac{dI}{dt}=0$, i.e. $I$ becomes constant
$0=b+ct$
$t=\frac{b}{c}$
Using this value in equation (1) we get
$I=a-b\left(\frac{b}{c}\right)+\frac{1}{2}c(\frac{b}{c}{)}^2$
$I=a-\frac{b^2}{c}+\frac{1}{2}\frac{b^2}{c}$
$I=a-\frac{b^2}{c}$
Hence, for $t=\frac{b}{c}$ current is minimum than for $t=0$.
$\Rightarrow$ Current will have a minimum value for $t=\frac{b}{c}$.