Question

If the current in the coil of resistance r decreases according to i-t graph shown in the figure then find the total heat produced in the coil

Answer 1

Given that,

If the current in the coil of resistance r decreases according to i-t graph.

We need to write the line of equation

$y=mx+c$

We need to write the slope

Using i-t graph

$m=-\dfrac{i_{0}}{t_{0}}$

Negative sign shows the decrement of current

Now, write the equation of line again

Using to graph

$i=(-\dfrac{i_{0}}{t_{0}})t+i_{0}$......(I)

We need to find the charge flow

Using formula of charge

$\dfrac{dq}{dt}=i$

$dq=idt$

We know that,

charge flow = area of i

The charge flown is

$q=\dfrac{1}{2}\times base\times height$

Put the value into the formula

$q=\dfrac{1}{2}\times t_{0}\times i_{0}$

So, the current will be,

$i_{0}=\dfrac{2q}{t_{0}}$

Put the value of $i_{0}$ in equation (I)

$i=-(\dfrac{2q}{t_{0}^2})t+i_{0}$

Again put the value of $i_{0}$ in equation (I)

$i=(-\dfrac{2q}{t_{0}^2})t+\dfrac{2q}{t_{0}}$....(II)

We need to calculate the total heat produced in the coil

Using formula of heat

$dh=i^2Rdt$

Put the value of i

$dh=((\dfrac{2q}{t_{0}})-\dfrac{2qt}{t_{0}^2})^2Rdt$

On integrating

$h=\dfrac{4q^2R}{3t_{0}}$

Put the value of q into the formula

$h=\dfrac{4\times t_{0}^2\times i_{0}^2R}{4\times3t_{0}}$

$h=\dfrac{i_{0}^2t_{0}R}{3}$

Hence, The total heat produced in the coil is $\dfrac{i_{0}^2t_{0}R}{3}$.

(3) is correct option.