Question

320 J of heat is produced in 10 s in a 2 Ω resistor. Find the amount of current flowing through the resistor.

Answer 1

Answer:

Given, H = 320 J,

t = 10 s,

R = 2 ohm,

I = ?

  USING THE RELATION=$I^{2} .RT$

                    $I= sqrt{H/RT}$

                    = $=sqrt {320/2*10}$

                   = 4A

     

Thus, the amount of current flowing through the resistor is 4 A.

Explanation:

Answer 2

Answer:

320 J of heat is produced in 10 s in a 2 Ω resistor. The amount of current flowing through the resistor will be  $4A$

Explanation:

• Let H be the heat produced due to a flow of current then according to Joule's law of heating, it is given by the expression,

       $H = I^{2}R t$       ...(1)

       Or

    $I^{2} =\frac{H}{Rt}$

     That is,

     $I= \sqrt{\frac{H}{Rt}}$        ...(2)  

     

Where R is the resistance, A is the current flowing and t is the time of flow of current

In the question, it is given that,

$R =2\Omega$         $H= 320 J$         $t = 10\ s$

substitute these values into equation(2)

we get,

$I= \sqrt{\frac{320}{2\times 10}}= 4A$