Question

A current flow in wire of resistance 5 ohm having potential difference 7 volt in 20 minutes the heat produced is

Answer 1

Given :

• Resistance of wire =5 Ω
• Potential difference =7 V
• Time ,t =20min=1200sec

To Find :

Heat produced in a wire

Solution :

We have to find heat produced in a wire .

According to the Joule's Law of Heating

The heat produced in a wire is directly proportional to:

• Square of current
• Resistance of wire
• time (t) , for which current is passed.

$\sf\implies\:H=I^2RT$

We know that,Ohm's law : $\tt\:I=\dfrac{V}{R}$

$\sf\implies\:H=(\dfrac{V}{R})^2\times\:RT$

$\sf\implies\:H=\dfrac{V^2}{R}\times\:T$

Now put the given values

$\sf\implies\:H=\dfrac{7\times7}{5}\times\:1200$

$\sf\implies\:H=49\times240$

$\sf\implies\:H=11,760J$

We know that

1J = 0.239 Calorie

⇒11,760 Joules = 2810.7 Calories( aprrox)

Therefore, Heat produced in a wire is 2810 calories .

Answer 2

Answer:

Given :

Resistance of wire =5 Ω

Potential difference =7 V

Time ,t =20min=1200sec

To Find :

Heat produced in a wire

Solution :

We have to find heat produced in a wire .

According to the Joule's Law of Heating

The heat produced in a wire is directly proportional to:

Square of current

Resistance of wire

time (t) , for which current is passed.

\sf\implies\:H=I^2RT⟹H=I

2

RT

We know that,Ohm's law : \tt\:I=\dfrac{V}{R}I=

R

V

\sf\implies\:H=(\dfrac{V}{R})^2\times\:RT⟹H=(

R

V

)

2

×RT

\sf\implies\:H=\dfrac{V^2}{R}\times\:T⟹H=

R

V

2

×T

Now put the given values

\sf\implies\:H=\dfrac{7\times7}{5}\times\:1200⟹H=

5

7×7

×1200

\sf\implies\:H=49\times240⟹H=49×240

\sf\implies\:H=11,760J⟹H=11,760J

We know that

1J = 0.239 Calorie

⇒11,760 Joules = 2810.7 Calories( aprrox)

Therefore, Heat produced in a wire is 2810 calories .