Question

PLZ SOLVE THIS QUESTION




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Answer 1

Given :

The electric current in a charging RC circuit is given by ,

$\displaystyle \star \mathtt i= i •\: {e}^{ - \frac{t}{RC} }$

To find :

The rate of change of current at

• t = 0
• t = RC
• t = 10 RC

Note :

$\displaystyle \bigstar \boxed{ \pink{ \: \frac{d( {e}^{ax}) }{dx} = a {e}^{ax} }}$

Solution :

Differentiating current with respect to time we get ,

$\displaystyle \rightarrow\mathtt{ \frac{di}{dt} = \frac{d(i •\: {e}^{ - \frac{t}{RC} } )}{dt} }$

$\displaystyle \boxed {\: { \red{\mathtt{ \frac{di}{dt} = i•( \frac{ - 1}{RC} ){e}^{ - \frac{t}{RC} } }}}}$

At t = 0

$\displaystyle \rightarrow{\mathtt{ \frac{di}{dt} = i•( \frac{ - 1}{RC} ) {e}^{0} }}$

$\displaystyle \boxed{ \blue{\mathtt{ \frac{di}{dt} = \frac{ - i•}{RC} }}}$

At t = R C

$\displaystyle \rightarrow{\mathtt{ \frac{di}{dt} = i•( \frac{ - 1}{RC} ) {e}^{ \frac{ - R C}{R C} } }}$

$\displaystyle \rightarrow{\mathtt{ \frac{di}{dt} = i•( \frac{ - 1}{RC} ) {e}^{ - 1} }}$

$\displaystyle \boxed{ \blue{\mathtt{ \frac{di}{dt} = \frac{ - i•}{eRC} }}}$

At t = 10 RC

$\displaystyle \rightarrow{\mathtt{ \frac{di}{dt} = i•( \frac{ - 1}{RC} ) {e}^{ \frac{ - 10R C}{R C} } }}$

$\displaystyle \rightarrow{\mathtt{ \frac{di}{dt} = i•( \frac{ - 1}{RC} ) {e}^{ - 10} }}$

$\displaystyle \boxed{ \blue{\mathtt{ \frac{di}{dt} = \frac{ - i•}{10eRC} }}}$