Question

5.Q5. (a) Derive the expression for the heat produced due to a current ‘I’ flowing for a time interval ‘t’ through a resistor ‘R’ having a potential difference ‘V’ across its ends. With which name is this relation known ? (b) The current passing through a room heater has been halved. What will happen to the heat produced by it ? (c) Name the material which is used for making the filaments of an electric bulb.​

Answer 1

Explanation:

(a) When an electric charge Q moves against a p.d. V, the amount of work done is given by

W=Q×V ---------------(1)

We know, current , I=

t

Q

Q=I×t ------------(2)

By ohm's law ,

I

V

=R

V=I×R --------------(3)

Putting eqs. (2) and (3) in eq. (1),

W=I×t×I×R

W=I

2

Rt

Assuming that all the electrical work done is converted into heat energy , we get

Heat produced , H=I

2

Rtjoules

This relation is known as Joule's law of heating.

(b) Given : P=12W,V=12V,t=60sec

using the equation of power

P=VI

I=P/V=12/12=1A

The Ohm's law states that

V=IR

R=V/I=12/1=12ohm

The formula for heat is:

H=I

2

Rt

H=1

2

×12×60

H=720J

(c) The heat produced by the heater will become one - fourth because heat produced is directly proportional to the square of the current.