Question

The electric current in a charging R-C circuit is given by $i=i_o{e}^{\frac{-t}{RC}}$ where io, R and C are constant parameter of the circuit and t is time. Find the rate of change of current at (x)t=0, (y)t=RC, (z)t=10 RC

$\left(i\right)\frac{-i_o}{RC}\left(ii\right)0\left(iii\right)\frac{-i_o}{R{C}_e}\left(iv\right)\frac{-i_o}{R{C}_e^{10}}\left(v\right)\frac{-i_o{e}^{10}}{RC}$

• A. x-(i); y-(iii); z-(iv)
• B. x - (ii) ; y - (i) ; z - (v)
• C. x - (iii) ; y - (ii) ; z - (iv)
• D. x - (i) ; y - (iii) ; z - (ii)

Answer 1

The correct option is A

x-(i); y-(iii); z-(iv)

$i=i_0{e}^{\frac{-t}{RC}}$
$i_{0}R,C$are constants
To find the rate of change of i
i.e.,
$\frac{di}{dt}$
$\Rightarrow \frac{di}{dt}=\frac{-i_o}{RC}{e}^{\frac{-t}{RC}}$
(x) at t=0
$\frac{di}{dt}=\frac{-i_o}{RC}$

(y) at t =RC
$\frac{di}{dt}=\frac{-i_o}{RC}{e}^{-1}=\frac{-i_o}{RCe}$
(z) at t = 10RC
$\frac{di}{dt}=\frac{-i_o}{RC}{e}^{\frac{-10Rc}{RC}}=\frac{-i_o}{Rc{e}^{10}}$