Question

Delive an expression for loule's law of heating.
Grive fuso eadmples tot applications of heating effect
of electric current.​

Answer 1

Answer:

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Explanation:

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Answer 2

Correct QUESTION

Delive an expression for Joule's law of heating. Give few examples too and applications of heating effect of electric current.

SOLUTION

The potential difference (V) across the ends of the wire is directly proportional to the current (I) flowing through that wire provide the temperature remains the same.

V ∝ I

V/I = constant (R)

From Ohm's law, we can say that

V = IR

Also, the current is defined as the ratio of charge and time.

So, I = Q/t

Q = I × t

Now, Potential difference (V) is defined as the work done to move a unit charge from one place to another. i.e.

V = W/Q

Substitute value of V and Q in the above formula

IR = W/(I × t)

IR(I × t) = W

I²Rt = W

Work done = Heat produced

(W = H)

H = I²Rt (Joule's Law of Heating)

Examples:

Electric heater, Electric geyser, Electric iron etc.

Find the heat energy produced in an electric heater of resistance 10Ω when a current of 2A flows in it for 2 minutes.

⇛ H = ?, R = 10Ω, I = 2A and t = 2min = 2*60 = 120 sec

→ H = (2)² × 10 × 120

→ H = 4 × 1200

→ H = 4800J or 4.8kJ

Applications:

The heat produced (H) in a resistor is

- directly proportional to the square of the current (I²).

- directly proportional to resistance (R).

{ for current }

- directly proportional to the time (t).

{ for the current that flows through the resistor }