Question

Question 1 = calculate heat produced if 2 mpere(a) current is flowing through a 100 om resistance for 2 min. Question 2= what is solenoid how it's strength can be increased​

Answer 1

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Calculate the heat produced if 2 ampere current if flowing through 100 $\Omega$ resistor for 2 minutes

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From the Question,

• Resistance,R = 100 $\Omega$

• Current,I = 2 A

• Time Taken,T = 2 min = 120 seconds

According to Joule's Law of Heating,

$\boxed{ \boxed{ \tt{H = {I}^{2}RT }}}$

Here,

• H is the heat generated

Putting the values,we get :

$\tt{H = 2 {}^{2} \times 120 \times 100} \\ \\ \longmapsto \: \tt{H = 480 \times 100} \\ \\ \longmapsto \: \tt{H = 48000 \: J} \\ \\ \longmapsto \: \underline{ \boxed{\tt{H = 48 \: KJ}}}$

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What is a Solenoid? How can its strength be increased?

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Solenoid

Solenoid is a coil shaped electromagnet which converts electrical energy to mechanical energy

Mechanism

A helix of copper wire is known as a solenoid. When it is wrapped around the conductor and current is passed,a magnetic field formed in the area. Finally,converting electricity energy to mechanical energy

Example,

The engines of vehicles

How can its power be increased?

• By increasing the no.of loops in solenoid

• Increasing the intensity of current

Answer 2

Answer:

$\huge{\boxed{\mathfrak\green{Answer=1)48 \:kgj}}}$

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Solution for question 2 :

What is solenoid? How it's strength can be increased?

Answer:- A solenoid is a type of electromagnet.It's purpose is to generate controlled magnetic field through a coil.

To increase strength of solenoid:-

1. By increasing existing current(I) in the coil.
2. By increasing number of coils in the solenoid and number of turns per unit in the solenoid.
3. By using a soft iron core inside the solenoid.

Refer to the attachment for solenoid reference.

Solution of Question 1 :

Given that,

• I = 2 ampere.
• R = 100 ohm.
• T = 2 min = (2 × 60) sec=120 seconds

Formula used:-

$\huge{\boxed{\tt{H = {I}^{2}Rt}}}$

$\leadsto\tt H = ({2}^{2}\times 100 \times 120)\:j \\ \leadsto\tt H = (4 \times 100 \times 120) \:j \\ \leadsto\tt H = 4800\:j \\ \leadsto\tt H = 48\:kgj$

$\therefore\tt {\underline{Heat\:produced=48\:kgj}}$