Question

Current passing through a wire decreases linearly from 10 A to 0 in 4s. Find total charge flowing through the wire in the given time interval.

Answer 1

Answer:

Given:

Current in wire decreases from 10 A to zero in 4 seconds.

To find:

Total charge flowing through the conductor.

Concept:

We first need to frame the equation relating current with time.

Since the relationship is linear and Current us decreasing with time , we can say that :

$\: \: \: \: \: \: \: \: \: \: \: \: \huge{ \green{ \sf{I = kt}}}$

k is actually the slope of the line in the graph :

we can calculate k as follows :

$\large{ \sf{ \red{ \bigg \{k = \dfrac{0 - 10}{4 - 0} = - \dfrac{5}{2} \bigg \}}}}$

So the Equation becomes as follows :

$\: \: \: \: \: \: \: \: \: \: \: \: \huge{ \green{ \sf{I = - \dfrac{5}{2} t}}}$

Calculation:

We know the relationship between current and charge :

$\large{I = \dfrac{dq}{dt}}$

$\large{ = > dq = I \: dt}$

Integrating on both sides :

$\large{ = > \int \: dq = \int I \: dt}$

Putting limits :

$\large{ = > \int_{0}^{q} \: dq = \int_{0}^{4} I \: dt}$

$\large{ = > \int_{0}^{q} \: dq = \int_{0}^{4} (\frac{ - 5t}{2}) \: dt}$

$\large{ = > q= \: \dfrac{ - 5}{4} \bigg \{ {t}^{2} \bigg\}_{0}^{4} }$

$= > q = - 20 \: C$

Taking only magnitude into consideration, we can say that :

$\huge{ \red{ \sf{ \bold{q = 20 \: C }}}}$

Answer 2

Given :

Current passing through a wire decreases linearly from 10 A to 0 A in 4s.

To Find :

Total charge flowing through the wire in the given time interval.

Formula :

Formula of current in terms of charge and time is given by...

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \dag \: \underline{ \boxed{ \bold{ \rm{ \pink{I = \frac{dq}{dt}}}}}} \: \dag$

Calculation :

As per formula it is clear that

$\implies \rm \: q = I \times dt \\ \\ \dag \rm \: \: \red{ I \times dt = area \: under \: graph} \\ \\ \therefore \rm \: q = \frac{1}{2} \times base \times perpendicular \\ \\ \therefore \rm \: q = \frac{1}{2} \times 10 \times 4 \\ \\ \therefore \: \underline{ \boxed{ \huge \bold{ \rm{ \orange{q = 20 \: C}}}}} \: \: \huge\purple{ \clubsuit}$