Question

derive the Joule's law of heating effect on the basis of electric power

Answer 1

Derivation of Joules’s Law of Heating

When current is passed through a resistance some work is done.In a passive resistance the work done is equal to the amount of heat evolved in a resistance.

Consider a battery having terminal p.d ‘V’ is connected with a resistance as shown in figure. Let ‘I’ be the amount of current in the circuit. Let ‘q’ be the charge passed through the resistance in time‘t’ second.

We  know that,

Current (I)=Charge (q)Time (t)Current (I)=Charge (q)Time (t)

q = It……(i)q = It……(i)

If the potential difference between A and B is V. Then,

V = IR……(ii)V = IR……(ii)

Let ‘W’ be the amount of work done to move ‘q’ amount of charge through the resistance.

 So, Electric potential(V)=Workdone (W)q  So, Electric potential(V)=Workdone (W)q 

 or, W = qV or, W = qV

 W=VIt……(ii) W=VIt……(ii)

Now, from equation (i) and (ii)  and (iii),

W = It.IRW = It.IR

or, W = I2 RTor, W = I2 RT

Heat produced, H= RT  joule

In CGS system,

H=I2RTJCalH=I2RTJCal$

Verification of Joule’s law:

Apparatus setup for the verification of Joule’s Law of heating

Take a glass container containing some amount of water (about 1/3rd). Take a coil, dip it into the water and connect the coil in a circuit containing a cell, ammeter, rheostat and switch as shown in figure. Thermometer is dipped into the water to measure the temperature of water. As we pass current, the coil will be heated and water absorb that amount of heat.

∴ The amount of heat produced by the coil, H = heat absorbed by the water.

or, H=ms(θ2 – θ1)

Here, θ1 and θ2 are initial and final temperature of water. ‘m’ & ‘s’ are the mass and specific heat capacity of water respectively.

1. Verification of H∝I2

Dip a coil into the water and pass certain amount of current for certain time(say 5 minute).Then calculate the amount of heat produced by the coil by using H=ms(θ2 – θ1)

By the help of the rheostat, pass different amount of current in the coil for the same time (say 5 minute) and calculate the heat produced. Let H1, H2 and H3 be the amount of heat produced for current I1, I2 and I3 respectively.

Here, we find,

H1I21=H2I22=H3I23H1I12=H2I22=H3I32

∴HI2=Constant∴HI2=Constant

On plotting the graph between H and I2 we get a straight line passing through origin.

So, we can conclude that, So, we can conclude that, 

H∝I2H∝I2

2. Verification of H∝R

Pass certain amount of current for certain time (say 5 minute) through different coil and calculate the different amount of heat produced by them as before. Let  be the amount of heat produced by the resistance  respectively.

Here, we find,

H1R1=H2R2=H3R3H1R1=H2R2=H3R3

∴HR=Constant∴HR=Constant

On plotting graph between H and R,we get a straight line passing through origin.

So, we can conclude that, So, we can conclude that, 

H∝RH∝R

3. Verification of H∝T

Pass certain amount of current through a coil for different time and find the amount of heat produced by the coil as before. Let H1, H2 and H3 be the amount of heat produced by the coil in time T1, T2 and T3 respectively.

Here, we find:

H1T1=H2T2=H3R3H1T1=H2T2=H3R3

∴HT=Constant∴HT=Constant

On plotting a graph between H and T, we get straight line passing through origin.

So, we can conclude that, So, we can conclude that, 

H∝TH∝T

Hence, Joule’s law of heating is verified.

Answer 2

╔════════════════╗

...HERE GØΣS UR ANS...

╚════════════════╝

$\huge{\bold{\red{HELLO\:!!}}}$

$<marquee direction="left" height="80" behavior="scroll" >LovePhysics</marquee>$

   

JOULE'S HEATING EFFECT

According to Joules law of heating, when a current I flows through a resistance R then :

Heat produced is

(I) Directly proportional to the square of current for a given resistance

(II) Directly proportional to the resistance of given circuit.

(III) Directly proportional to the time for which the resistance flows through the resistor.

So eqⁿ is $\bold{\huge{\boxed{\red{H=I^2Rt}}}}}$

$\bold{\huge{\boxed{\pink{DERIVATION}}}}$

let us consider a resistance R , in which I amount of current flows.

Work must be done by current to move continuously.

W=Qx V

BUT Q=IXt

W=IxVxt

but from ohms law: V=IR

$\bold{\huge{\boxed{\purple{W=I^2Rt}}}}$

assuming the electrical energy consumed is converted into heat energy .

we write work done as Heat  produced.

so $\bold{\huge{\boxed{\red{H=I^2Rt}}}}}$

$\bold{\huge{\boxed{\pink{APPLICATIONS}}}}$

❶ In electrical heaters.

❷ In electrical iron

❸ In electrical fuse wire

❹ In electrical bulb

$\bold{\huge{\boxed{\red{\mathscr{ThankYou}}}}}$

$\bold{\underline{\huge{\red{\mathscr{PlzMarkBrainliest}}}}}}$