When an electric current is passed through a conductor, the conductor becomes hot after some time and produce heat. This happens due to the conversion of some electric energy passing through the conductor into heat energy. This effect of electric current is called heating effect of current.The heating effect of current was studied experimentally by Joule in 1941. After doing his experiments Joule came to the conclusion that the heat produced in a conductor is directly proportional to the product of square of current (I2), resistance of the conductor (R) and the time (t) for which current is passed. Thus, H I2Rt Derivation of Formula To calculate the heat produced in a conductor, consider current I is flowing through a conductor AB of resistance R for time t. also consider that the potential difference applied across its two ends is V. Now, total amount of work done in moving a charge q from point A to B is given by: W = q X V (1) Now, we know that charge = current × time or q = I X t and V = I X R (Ohm’s law) Putting the values of q and V in equation (1), we get W = (I X t) X (I X R) Or W = I2Rt Now, assuming that all the work done is converted into heat energy we can replace symbol of ‘work done’ with that of ‘heat produced’. So, H = I2RT Applications of Heating Effect of Current The heating effect of current is used in various electrical heating appliances such as electric bulb, electric iron, room heaters, geysers, electric fuse etc.

1. Heating Effect of Electric
Current Priya Jha KV No3 Delhi cantt

2. Class X
Science
Priya Jha
KV No3 Delhi Cantt

3. Thermoelectric effect
When current flows through a
resistor, some of the electrical
energy delivered to the resistor is
converted into heat energy and it
is dissipated. This heating effect
of current is known as Joule’s
heating effect. • Just as current
produces thermal energy,
thermal energy may also be
suitably used to produce an
electromotive force. This is
known as thermoelectric effect.

4. Joule’s law
• If a current I flows through a
conductor kept across a
potential difference V for a
time t, the work done or the
electric potential energy
spent is W=VIt
• In the absence of any other
external effect, this energy is
spent in heating the
conductor. The amount of
heat(H) produced is H=VIt

5. Q). A battery of 10 volt carries 20,000 C
of charge through a resistance of 20 Ω.
The work done in 10 seconds is
• (a) 2 × 103 joule
(b) 2 × 105joule
(c) 2 × 104 joule
(d) 2 × 102 joule

6. Ans)
(b) W= qV
= 20000 × 10
= 2,00, 000
= 2 × 105 J

7. • For a resistance R , H=I2Rt
• This relation was experimentally
verified by Joule and is known as
Joule’s law of heating. It states
that the heat developed in an
electrical circuit due to the flow
of current varies directly as
(i) the square of the current
(ii) the resistance of the circuit
and
(iii) the time of flow.

8. Q) A boy records that 4000 joule of work is required to transfer
10 coulomb of charge between two points of a resistor of 50 Ω.
The current passing through it is
(a) 2 A
(b) 4 A
(c) 8 A
(d) 16 A

9. Ans)(c) Work done in transferring the charge
W= qV = qlR …….. (V = IR)
•

10. Q) Find out the following in
the electric circuit given
in figure:
• (a) Effective resistance of two 8Ω
resistors in the combination
• (b) Current flowing through 4Ω resistors
• (c) Potential difference across 4Ω
resistances
• (d) Power dissipated in 4 Ω resistors
• (e) Difference in ammeter readings, if
any

11. Ans(a) Here, two 8 Ω resistances are connected
in parallel so their effective resistances are
Here, current through resistor will be
same as current from battery = 1 Ampere

12. (c) Potential difference (V’) across
4resistors (R’) is given by V’ = IR’ =
1 × 4 = 4 V
(d) Power dissipated across 4Ω
resistor, P = I2R =12 × 4 = 4 W
(e) There is no difference in the
readings of ammeters A1 and A2 as
same amount of current flows
through the ammeter.

13. Q) The resistance of hot filament of the bulb is
about 10 times the cold resistance. What will be
the resistance of 100 W-220 V lamp, when not
in use?
(a) 48 Ω
(b) 400 Ω
(c) 484 Ω
(d) 48.4 Ω

15. Application of Joule’s
heating effect
• Electric iron, electric heater,
electric toaster are some of the
home appliances that utilize the
heating effect of current.
• In these appliances, the heating
elements are made of nichrome,
an alloy of nickel and chromium.
• Nichrome has a high specific
resistance and can be heated to
very high temperatures without
oxidation.

16. Electric fuses
• Fuses , are connected in series in a
circuit to protect the electric
devices from the heat developed by
the passage of excessive current.
• It is a short length of a wire made of
a low melting point material. It
melts and breaks the circuit if
current exceeds a certain value.
• Lead and copper wire melts and
burns out when the current
increases above 5 A and 35 A
respectively.

17. • The only disadvantage with the above
fuses is that once fuse wire is burnt
due to excessive current, they need to
be replaced.
• Nowadays in houses, circuit breakers
(trippers) are also used instead of
fuses
• Whenever there is an excessive current
produced due to faulty wire
connection, the circuit breaker switch
opens.
• After repairing the faulty connection,
we can close the circuit breaker
switch.

18. Q)A fuse wire repeatedly gets burnt
when used with a good heater. It is
advised to use a fuse wire of
(a) more length
(b) less radius
(c) less length
(d) more radius

19. Ans)(d) In order to get the working
of heater properly, fused wire of
higher rating must be used.

20. Q) A cooler of 1500 W, 200 volt and a fan
of 500 W, 200 volt are to be used from a
household supply. The rating of fuse to be
used is
(a) 2.5 A
(b) 5.0 A
(c) 7.5 A
(d) 10 A

21. Ans)(d) Total power used, P =
P1 + P1 = 1500 + 500 = 2000
W.
Current drawn from the
supply,

22. THANK YOU
SAVE ELECTRICITY
SAVE ENVIRONMENT