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Concepts Of Physics MCQ Edition [Volume 1]

Exam: Concepts Of Physics MCQ Edition [Volume 1]
Subject: Physics
Chapter: Physics and Mathematics

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Physics and Mathematics

In a discharging R – C circuit, the electric current is given by i...

In a discharging

R

–

C

circuit, the electric current is given by

i = i_0 e^{-t/RC}

, where

i_0

,

R

and

C

denote constant parameters and

t

represents time. If

i_0 = 2.00 \text{ A}

,

R = 6.00 \times 10^5 \Omega

and

C = 0.500 \mu\text{F}

, ascertain the rate of change of current at

t = 0.3 \text{ s}

.

Options

A $-\dfrac{20}{3e} \text{ A/s}$
B $\dfrac{20}{3e} \text{ A/s}$
C $-\dfrac{3}{20e} \text{ A/s}$
D $-\dfrac{2.00}{e} \text{ A/s}$

Correct answer

A. $-\dfrac{20}{3e} \text{ A/s}$

Step-by-step solution

Given

i = i_0 e^{-t/RC}

The rate of change of current is given by:

\dfrac{di}{dt} = \dfrac{d}{dt} (i_0 e^{-t/RC}) = -\dfrac{i_0}{RC} e^{-t/RC}

Substituting the given values:

R = 6.00 \times 10^5 \Omega

C = 0.500 \times 10^{-6} \text{ F}

RC = (6.00 \times 10^5) \times (0.500 \times 10^{-6}) = 0.3 \text{ s}

At

t = 0.3 \text{ s}

:

\dfrac{di}{dt} = -\dfrac{2.00}{0.3} e^{-0.3/0.3} = -\dfrac{20}{3} e^{-1} = -\dfrac{20}{3e} \text{ A/s}

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