The electric current in a charging R-C circuit 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)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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
$\frac{d i}{d t} = \frac{- i_{o}}{R C} e^{\frac{- 10 R c}{R C}} = \frac{- i_{o}}{R c e^{10}}$

Suggest Corrections

Similar questions

The electric current in a charging R-C circuit is given by i= $i_{0} e^{- 1 / R C}$ where $i_{0}$ , R and C are constant parameters of the circuit and t is time. Find the rate of change of current at (a) $t = 0$ , (b) t = RC, (c) t = 10 RC.

The electric current in a charging R-C circuit 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

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

Q. The electric current in a charging

R - C

circuit is given by

i = i_{o} e^{- t / R C}

where

i_{o}

R

and

C

are constant parameters of the circuit and t is time. Find the rate of change of current at (a)

t = 0

, (b)

t = R C

, (c)

t = 10 R C

The electric current in a discharging R-C circuit is given by $i = i_{o} e^{\frac{- t}{R C}}$ where $i_{o}$ , R and C are constant parameters and $t$ is time. Let $i_{o}$ = 2.00 A, R= $6 \times 10^{5}$ Ω and C = 0.500 $μ$ F.

(a) Find the current at t = 0.3 s.

(b) Find the rate of change of current at t = 0.3 s

How will the resistance be affected if the various parameters are changed as follows?
$\begin{matrix} i) Length increases & x) Resistance increases ii) Area increases & y) Resistance decreases iii) Resistivity decreases & z) Resistance remains the same iv) Temperature increases v) Velocity increases \end{matrix}$

Join BYJU'S Learning Program

The electric current in a charging R-C circuit 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