The electric current in a charging R-C circuit is given is given by $i = i_{o} e^{\frac{- t}{R C}}$ 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

$(i) \frac{- i_{o}}{R C} (i i) 0 (i i i) \frac{- i_{o}}{R C_{e}} (i v) \frac{- i_{o}}{R C_{e}^{10}} (v) \frac{- i_{o} e^{10}}{R C}$

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

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x - (ii) ; y - (i) ; z - (v)

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

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

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Solution

The correct option is A
x-(i); y-(iii); z-(iv)

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

(y) at t =RC
$\frac{d i}{d t} = \frac{- i_{o}}{R C} e^{- 1} = \frac{- i_{o}}{R C e}$
(z) at t = 10RC
$\frac{d i}{d t} = \frac{- i_{o}}{R C} e^{\frac{- 10 R c}{R C}} = \frac{- i_{o}}{R c e^{10}}$

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The electric current in a charging R-C circuit is given is given by i=ioe−tRC 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 (i)−ioRC (ii)0 (iii)−ioRCe (iv)−ioRC10e (v)−ioe10RC

The correct option is A x-(i); y-(iii); z-(iv) i=i0e−tRC i0 R,Care constants To find the rate of change of i i.e., didt ⇒didt=−ioRCe−tRC (x) at t=0 didt=−ioRC (y) at t =RC didt=−ioRCe−1=−ioRCe (z) at t = 10RC didt=−ioRCe−10RcRC=−ioRce10