Joules Heating.
When some potential difference V is applied across a resistance R then the work done by the electric field on charge q
V 2t
to flow through the circuit in time t will be W = qV = Vit = i2Rt Joule .
R
This work appears as thermal energy in the resistor. i R
2 2 A B
W Vit i Rt V t
Heat produced by the resistance R is H Cal. This relation is called joules heating.
J 4 2 4 2 4 2R
Some important relations for solving objective questions are as follow :
Condition Graph
If R and t are constant H
H i 2 and H V 2
i (or V)
If i and t are constant (series H
grouping)
HR R
If V and t are constant (Parallel H
grouping)
1
H R
R
If V, i and R constant H
Ht
t
Electric Power.
The rate at which electrical energy is dissipated into other forms of energy is called electrical power i.e.
W V2
P Vi i 2 R
t R
(1) Units : It’s S.I. unit is Joule/sec or Watt 40W
220V
Bigger S.I. units are KW, MW and HP, remember 1 HP = 746 Watt
(2) Rated values : On electrical appliances (Bulbs, Heater ……. etc.)
Wattage, voltage, ……. etc. are printed called rated values e.g. If suppose we have a bulb of 40 W, 220 V then rated
power (PR) = 40 W while rated voltage (VR) = 220 V. It means that on operating the bulb at 220 volt, the power dissipated
will be 40 W or in other words 40 J of electrical energy will be converted into heat and light per second.
(3) Resistance of electrical appliance : If variation of resistance with temperature is neglected then resistance of
VR2
any electrical appliance can be calculated by rated power and rated voltage i.e. by using R e.g. Resistance of
PR
220 220
100 W, 220 volt bulb is R 484
100
(4) Power consumed (illumination) : An electrical appliance (Bulb, heater, …. etc.) consume rated power (PR)
only if applied voltage (VA) is equal to rated voltage (VR) i.e. If VA = VR so Pconsumed = PR. If VA < VR then
VA2 V2 V2
Pconsumed also we have R R so Pconsumed(Brightness) A2 . PR
R PR VR
2
110
e.g. If 100 W, 220 V bulb operates on 110 volt supply then Pconsumed 100 25 W
220
(PR Pconsumed)
Note : If VA < VR then % drop in output power 100
PR
For the series combination of bulbs, current through them will be same so they will consume power in
the ratio of resistance i.e., P R {By P = i2R) while if they are connected in parallel i.e. V is constant
1
so power consumed by them is in the reverse ratio of their resistance i.e. P .
R
VR2
(5) Thickness of filament of bulb : We know that resistance of filament of bulb is given by R , also
PR
l 1
R , hence we can say that A PR i.e. If rated power of a bulb is more, thickness of it’s filament is
A ( Thickness) R
also more and it’s resistance will be less.
1 V2
If applied voltage is constant then P(consumed)
(By P A ). Hence if different bulbs (electrical appliance)
R R
1
operated at same voltage supply then Pconsumed PR t hickness
R
Note : Different bulbs
25W 100W 1000W
220V 220V 220V
Resistance R25 > R100 > R1000
Thickness of filament t1000 > t100 > t40
Brightness B1000 > B100 > B25
(6) Long distance power transmission : When power is transmitted through a power line of resistance R, power-
loss will be i 2 R
P2
Now if the power P is transmitted at voltage V P Vi i.e. i (P / V) So, Power loss R
V2
Now as for a given power and line, P and R are constant so Power loss (1 / V 2 )
So if power is transmitted at high voltage, power loss will be small and vice-versa. e.g., power loss at 22 kV is
–4
10 times than at 220 V. This is why long distance power transmission is carried out at high voltage.
(7) Time taken by heater to boil the water : We know that heat required to raise the temperature of any
substance of mass m and specific heat S is H = m.S.
Here heat produced by the heater = Heat required to raise the temp. of water.
J(m.S. )
i.e. p t = J m.S. t {J = 4.18 or 4.2 J/cal)
p
4180( or 4200) m
for m kg water t {S = 1000 cal/kgoC)
p
4180(4200) n
Note : If quantity of water is given n litre then t
p
Electricity Consumption.
(1) The price of electricity consumed is calculated on the basis of electrical energy and not on the basis of
electrical power.
(2) The unit Joule for energy is very small hence a big practical unit is considered known as kilowatt hour (KWH)
or board of trade unit (B.T.U.) or simple unit.
(3) 1 KWH or 1 unit is the quantity of electrical energy which dissipates in one hour in an electrical circuit when
the electrical power in the circuit is 1 KW thus 1 KW = 1000 W 3600 sec = 3.6 106 J.
Total watt Total hours
(4) Important formulae to calculate the no. of consumed units is n
1000
Concepts
When some potential difference applied across the conductor then collision of free electrons with ions of the lattice result’s in
conversion of electrical energy into heat energy
If a heating coil of resistance R, (length l) consumed power P, when voltage V is applied to it then by keeping V is constant if it is cut
R 1
in n equal parts then resistance of each part will be and from Pconsumed , power consumed by each part P' nP .
n R
Joule’s heating effect of current is common to both ac and dc.
Examples
Example: 1 The approximate value of heat produced in 5 min. by a bulb of 210 watt is (J = 4.2
joule/calorie)
(a) 15,000 (b) 1,050 (c) 63,000 (d) 80,000
P t 210 5 60
Solution : (a) By using H 15000Cal
4.2 4.2
Example: 2 A heater coil is cut into two parts of equal length and one of them is used in the heater. The
ratio of the heat produced by this half coil to that by the original coil is
(a) 2 : 1 (b) 1 : 2 (c) 1 : 4 (d) 4 : 1
R
Solution : (a) If suppose resistance of the coil is R so resistance of it’s half will be . Hence by using
2
V 2t 1
H H
R R
HHalf RFull R 2
HFull RHalf R / 2 1
Note : In general if coil is divided in n equal parts then heat produced by each part will be n
times of the heat produced by coil it self i.e. H = nH
Example: 3 If current in an electric bulb changes by 1%, then the power will change by
1
(a) 1% (b) 2% (c) 4% (d) %
2
P i
Solution : (b) By using P = i2R P i2 2 change in power = 2%
P i
Example: 4 A constant voltage is applied on a uniform wire, then the heat is produced. The heat so
produced will be doubled, if
(a) The length and the radius of wire are halved (b) Both length and radius are doubled
(c) Only the length is doubled (d) Only the radius is
doubled
V 2t l l V 2 t r 2 r2
Solution : (b) By using H and R 2 H H ; on doubling both r and l
R A r l l
heat will be doubled.
Example: 5 An electric heater of resistance 6 ohm is run for 10 minutes on a 120 volt line. The energy
liberated in this period of time is
(a) 7.2 10 3 J (b) 14.4 10 5 J (c) 43.2 10 4 J (d) 28.8 10 4 J
V 2t (120)2 10 60
Solution : (b) By using H H 14.4 105 J
R 6
Example: 6 An electric bulb of 100 W is designed to operate on 220 V. Resistance of the filament is
(a) 484 (b) 100 (c) 22000 (d) 242
V2 V 2 (220)2
Solution : (a) By using P R 484
R P 100
Example: 7 An electric bulb is rated 220 V and 100 W. Power consumed by it when operated on 110 volt is
(a) 50 W (b) 75 W (c) 90 W (d) 25 W
2 2
VA 110
Solution : (d) By using Pconsumed PR PConsumed 100 25W
VR 220
Example: 8 A 500 watt heating unit is designed to operate from a 115 Volt line. If the line voltage drops to
110 volt, the percentage drop in heat output will be
(a) 10.20% (b) 8.1% (c) 8.6% (d) 7.6%
2 2
VA 110
Solution : (c) By using Pconsumed PR PConsumed 500 456.6 Watt
VR 115
PActual PConsumed (500 456.6)
So % drop in heat output 100 100 8.6%
PActual 500
Example: 9 An electric lamp is marked 60 W, 230 V. The cost of 1 kilowatt hour of power is Rs. 1.25.
The cost of using this lamp for 8 hours is
(a) Rs. 1.20 (b) Rs. 4.00 (c) Rs. 0.25 (d) Rs. 0.60
Total Watt Total time 60 8 12
Solution : (d) By using consumed unit (n) or KWH n
1000 1000 25
12
So cost 1.25 0.60 Rs
25
Example: 10 How much energy in Kilowatt hour is consumed in operating ten 50 watt bulbs for 10 hours
per day in a month (30 days)
(a) 1500 (b) 15.000 (c) 15 (d) 150
Total Watt Total time (50 10) (10 30)
Solution : (d) By using n n 150
1000 1000
Example: 11 An immersion heater is rated 836 watt. It should heat 1 litre of water from 20o C to 40o C in
about
(a) 200 sec (b) 100 sec (c) 836 sec (d) 418 sec
4180 n 4180 1 (40 20)
Solution : (b) By using t t 100sec
P 836
Example: 12 The power of a heater is 500 watt at 800o C. What will be its power at 200o C if
4 10 4 per C
(a) 484 W (b) 672 W (c) 526 W (d) 611 W
V 2
1 P R (1 t 2 ) 500 (1 4 104 200)
Solution : (d) By using P i 2 R P 1 2
R R P2 R1 (1 t1 ) P2 (1 4 104 800)
500 1.08
611W
P2 1.32
Example: 13 A heater of 220 V heats a volume of water in 5 minute time. A heater of 110 V heats the same
volume of water in
(a) 5 minutes (b) 8 minutes (c) 10 minutes (d) 20 minutes
V 2t
Solution : (d) By using H . Here volume of water is same. So same heat is required in both cases.
R
1
Resistance is also constant so V2t = constant t 2
V
2 2
t1 V2 5 110 1
t 2 20 min
t 2 V1 t 2 220 4
Example: 14 Water boils in an electric kettle in 15 minutes after switching on. If the length of the heating
wire is decreased to 2/3 of its initial value, then the same amount of water will boil with the same
supply voltage in
(a) 15 minutes (b) 12 minutes (c) 10 minutes (d) 8 minutes
V 2t l V 2t A
Solution : (c) By using H where R H . Since volume is constant so H is also constant
R A l
2
t2 l2 t2 l1
so t l which gives 3 t 2 10 min
t 1 l1 15 l1
Tricky Example: 1
If resistance of the filament increases with temperature, what will be power dissipated in a
220 V – 100 W lamp when connected to 110 V power supply
(a) 25 W (b) < 25 W (c) > 25 W (d) None of these
2
2
V 110
Solution : (b) If resistance do not varies with temperature PConsumed A . PR 100 25W.
VR 220
But actually resistance is increasing with temperature so consumed power will be lesser
then 25W.
Combination of Bulbs (or Electrical Appliances).
Bulbs (Heater etc.) are in series Bulbs (Heater etc.) are in parallel
(1) Total power consumed P1 P2
(1) Total power consumed P1
1 1 1 Ptotal P1 P2 P3 ...... Pn
......
Ptotal P1 P2 Supply P2
Supply
P
(2) If ‘n’ bulbs are identical, Ptotal (2) If ‘n’ identical bulbs are in parallel. Ptotal nP
N
1 1
(3) Pconsumed (Brightness) V R i.e. in series (3) Pconsumed (Brightness) PR i R i.e. in parallel
Prated
combination bulb of greater wattage will give more
combination bulb of lesser wattage will give more
bright light and more current will pass through it.
bright light and p.d. appeared across it will be more.
Some Standard Cases for Series and Parallel Combination.
P P
(1) If n identical bulbs first connected in series so PS and then connected in parallel. So PP = nP hence P n 2 .
n PS
(2) To operate a bulb on voltage which is more then it’s rated voltage, a proper resistance is connected in series with
it. e.g. to glow a bulb of 30 W, 6 V with full intensity on 126 volt required series resistance calculated as follows
Bulb will glow with it’s full intensity if applied voltage on it is 6 V i.e. 120 V appears across the series resistance
R current flows through bulb = current flows through resistance
R
30 i
i 5 amp
6 6V 120 V
Hence for resistance V iR i.e. 120 = 5 R 5 R R = 24 126 V
Voperating VR
Note : If you want to learn Short Trick then remember Seriesresistance VR
P R
(3) An electric kettle has two coils when one coil is switched on it takes time t1 to boil water and when the second
coil is switched on it takes time t2 to boil the same water.
If they are connected in series If they are connected in parallel
1 1 1 PP P1 P2
PS P1 P2
H P H1 H2
1 1 1 tp t1 t2
H S / t S H1 / t1 H 2 / t 2
1 1 1
H p H1 H 2 so
HS H1 H2 so t s t1 t 2 t p t1 t 2
i.e. time taken by combination to boil the same i.e. time taken by parallel combination to boil the
quantity of water tS t1 t2 tt
same quantity of water tp 1 2
t1 t2
(4) If three identical bulbs are connected in series as shown in figure then on closing the switch S. Bulb C short
circuited and hence illumination of bulbs A and B increases
A B C
Reason : Voltage on A and B increased.
+ S
V
–
(5) If three bulbs A, B and C are connected in mixed combination as shown, then illumination of bulb A decreases
if either B or C gets fused
A
Reason : Voltage on A decreases.
V
B C
(6) If two identical bulb A and B are connected in parallel with ammeter A and key K as shown in figure.
It should be remembered that on pressing key reading of ammeter becomes twice.
K
A Reason : Total resistance becomes half.
V
A B
Concepts
When a heavy current appliance such us motor, heater or geyser is switched on, it will draw a heavy current from the source so that
terminal voltage of source decreases. Hence power consumed by the bulb decreases, so the light of bulb becomes less.
r
Heater
~
K
If the source is ideal i.e. r = 0, there will be no change in the brightness of the bulb.
Examples
Example: 15 An electric kettle has two heating coils. When one of the coils is connected to ac source the
water in the kettle boils in 10 minutes. When the other coil is used the water boils in 40
minutes. If both the coils are connected in parallel, the time taken by the same quantity of
water to boil will be
(a) 4 min (b) 25 min (c) 15 min (d) 8 min
t1 t 2 10 40
Solution : (d) By using the formula t p (as we discussed in theory) t p 8 min
t1 t 2 (10 40)
Note : In this question if coils are connected in series then the time taken by the same
quantity of water to boil will be ts = t1 + t2 = 10 + 40 = 50 min
Example: 16 If a 30 V, 90 W bulb is to be worked on a 120 V line, a resistance of how many ohms should be
connected in series with the bulb
(a) 10 ohm (b) 20 ohm (c) 30 ohm (d) 40 ohm
Voperating VR
Solution : (c) By using Seriesresistance R VR (As we discussed in theory)
P
R
(120 30)
R 30 30
90
Example: 17 In the circuit shown in figure, the heat produced in 5 ohm resistance is 10 calories per second.
The heat produced in 4 ohm resistance is
4 6
(a) 1 cal/sec
(b) 2 cal/sec
5
(c) 3 cal/sec
(d) 4 cal/sec
i1 10 2 2
Solution : (b) Ratio of currents by using H = i Rt 4 6
i2 5 1 i2
Line (2)
i
2 2
H1 i1 R1 10 2 5
H2 2cal / sec 5
H2 i2 R2 H2 1 4 Line (1)
i1
Example: 18 Two heater wires of equal length are first connected in series and then in parallel. The ratio of
heat produced in the two cases is
(a) 2 : 1 (b) 1 : 2 (c) 4 : 1 (d) 1 : 4
Solution : (d) Both the wires are of equal length so they will have same resistance and by using
V 2t 1
H H
R R
Hs R H R/ 2 1
P; s
HP Rs HP 2R 4
Example: 19 If two bulbs of wattage 25 and 100 respectively each rated at 220 volt are connected in series
with the supply of 440 volt, then which bulb will fuse
(a) 100 watt bulb (b) 25 watt bulb (c) None of them (d) Both of them
1
Solution : (b) In series V A i.e. voltage appear on 25W bulb will be more then the voltage appears on
PR
100 W bulb. So bulb of 25 W will gets fused.
Example: 20 Three equal resistors connected in series across a source of e.m.f. together dissipate 10 watt. If
the same resistors are connected in parallel across the same e.m.f., then the power dissipated
will be
(a) 10 W (b) 30 W (c) 10/3 W (d) 90 W
P
Solution : (d) In series consumed power Ps while in parallel consumed power Pp = nP PP = n2. Ps
n
PP (3)2 10 90W
Example: 21 Forty electric bulbs are connected in series across a 220 V supply. After one bulb is fused, the
remaining 39 are connected again in series across the same supply. The illumination will be
(a) More with 40 bulbs than with 39 (b) More with 39 bulbs than with 40
(c) Equal in both the cases (d) In the ratio of 402 :
2
39
V2
Solution : (b) Illumination = PConsumed . Initially there were 40 bulbs in series so equivalent resistance
R
was 40 R, finally 39 bulbs are in series so equivalent resistance becomes 39 R. Since resistance
decreases so illumination increases with 39 bulbs.
Example: 22 Two bulbs of 100 watt and 200 watt, rated at 220 volts are connected in series. On supplying
220 volts, the consumption of power will be
(a) 33 watt (b) 66 watt (c) 100 watt (d) 300 watt
P1 P2 100 200
Solution : (b) In series PConsumed PConsumed 66W
P1 P2 300
Example: 23 Two wires ‘A’ and ‘B’ of the same material have their lengths in the ratio 1 : 2 and radii in the
ratio 2 : 1. The two wires are connected in parallel across a battery. The ratio of the heat
produced in ‘A’ to the heat produced in ‘B’ for the same time is
(a) 1 : 2 (b) 2 : 1 (c) 1 : 8 (d) 8 : 1
2 2
l l R A l A rB RA 1 1 1
Solution : (d) Resistance R 2 R 2
r r R B l B r
A RB 2 2 8
V 2t H R 8
By using H A B
R H B RA 1
Example: 24 A heating coil is labelled 100 W, 220 V. The coil is cut in half and the two pieces are joined in
parallel to the same source. The energy now liberated per second is
(a) 200 J (b) 400 J (c) 25 J (d) 50 J
Solution : (b) Let resistance of the heating coil be R, when coil cut in two equal parts, resistance of each part
R R
will be . When these two parts are corrected in parallel, Req i.e. resistance becomes, so
2 4
1
according to P ; Power becomes 4 times i.e. P = 4P = 400 J/sec
R
Example: 25 Two identical electric lamps marked 500 W, 220 V are connected in series and then joined to a
110 V line. The power consumed by each lamp is
(a) 125 W (b) 25 W (c) 225 W (d) 125 W
4 4 4
Solution : (a) Both bulbs are identical so voltage across each bulb will be 55V. 500 W 500 W
220 V 220 V
2 2
VA 55 125
Hence power consumed by each bulb is PR 500 W 55 V 55 V
V
R 220 4
110 V
Tricky Example: 2
Electric bulb 50 W – 100 V glowing at full power are to be used in parallel with battery 120
V, 10 . Maximum number of bulbs that can be connected so that they glow in full power
is
(a) 2 (b) 8 (c) 4 (d) 6
i P 50 1 n
Solution : (c) When bulb glowing at full power, current flows through it amp i and
n V 100 2 2
voltage across the bulb is 100 V. If suppose n bulbs are connected in parallel with cell as
n
shown in figure then according to the cell equation E V i r 120 100 10 n 4 .
2
i/n 1
2
i/n
i/n n
i
120 V, 10
