4.
Electric Spark
A lightning bolt is an electrical discharge caused
due to potential difference between clouds and
earth. It is three times hotter than Sun.
72
SCHOOL SECTION
MT
Q.I
*1.
*2.
*3.
*4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
EDUCARE LTD.
SCIENCE & TECHNOLOGY
(A) Choose the correct alternatives and rewrite the complete sentences :
1mA = ..................... .
(b) 10–3 A
(a) 103 A
6
(c) 10 A
(d) 10–6 A
To increase the effective resistance in a circuit the resistors are connected
in ..................... .
(a) series
(b) parallel
(c) series and parallel
(d) non of these
1 kilowatt hr = ..................... .
(a) 4.6 × 106 joule
(b) 3.6 × 106 joule
6
(d) 3.6 × 105 joule
(c) 30.6 × 10 joule
The current passing through a 3 resistor if P.D. of 12V is applied across it
is ..................... .
(a) 36
(b) 4 A
(c) 0.25 A
(d) 15 A
The S.I. unit of potential difference is ..................... .
(a) ampere
(b) volt
(c) ohm
(d) joule
The unit of electric charge is ..................... .
(a) ampere
(b) coulomb
(c) volt
(d) ohm
One ..................... is the potential difference when one joule of work is
done to move a charge of one coulomb.
(a) ohm
(b) coulomb
(c) volt
(d) ampere
The S.I. unit of electric current is ..................... .
(a) ampere
(b) coulomb
(c) volt
(d) ohm
According to Ohm’s law, longer the length of wire, ..................... is the
resistance for given cross section of wire.
(a) lower
(b) greater
(c) zero
(d) constant
A substance which does not allow charges to pass through it, easily is
called as ..................... .
(a) metal
(b) conductor
(c) insulator
(d) semiconductor
If a current of 0.1 A is passed through a wire of resistance 20 ohm, the
potential difference across the wire is ..................... .
(a) 20 ohm
(b) 20 volt
(c) 10 volt
(d) 2 volt
Electric current is measured with the help of a device called ..................... .
(a) an ammeter
(b) a volt meter
(c) a thermometer
(d) a calorimeter
The equivalent resistance of a parallel combination is ..................... than
each of the individual resistance.
(a) greater
(b) smaller
(c) stronger
(d) more
If two resistances of 10 ohm and 15 ohm are connected in parallel the
equivalent resistance will be ..................... ohm.
(a) 25
(b) 6
1
(c)
(d) 150
6
In conductors, electrons are always in the state of ..................... motion.
(a) lower to higher
(b) opposite
(c) similar
(d) random
SCHOOL SECTION
73
SCIENCE & TECHNOLOGY
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
74
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A diagram which indicates how different components in a circuit have
been connected by using the electrical symbols for the components is called
..................... .
(a) ray diagram
(b) component diagram
(c) venn diagram
(d) circuit diagram
The electrons flow from negative terminal to positive terminal of the cell
but the ..................... of current is from positive terminal to negative
terminal of the cell.
(a) conventional direction
(b) electron current
(c) real current
(d) negative
The SI unit of resistivity is ..................... .
(a) ohm–metre
(b) ohm-(metre) 2
(c) ohm
(d) joule
When potential difference is applied between two ends of a wire
..................... in it.
(a) heat is produced
(b) charge is set up
(c) charge move randomly
(d) electrons move randomly
The quantity of heat produced depends upon ..................... .
(a) square of the current (I2)
(b) resistance of the conductor (R)
(c) time for which the current flows (t)
(d) I 2Rt
1 cal = ..................... joule.
(a) 10 7
(b) 10 5
(c) 41.8
(d) 4.18
Heat energy produced is expressed in terms of ..................... .
(a) calorie
(b) newton
(c) ampere
(c) coulomb
Tungsten in bulb has ..................... .
(a) high melting point
(b) high charge
(c) high P.D.
(d) high insulation
Fuse is made up of ..................... .
(a) iron and carbon
(b) lead and Tin
(c) copper and Zinc
(d) copper and Aluminum
The ..................... in the fuse is such that it melts when a current passing
through it exceed a certain value.
(a) length
(b) diameter
(c) charge
(d) distance
In surgery finely heated ..................... wire is used for cutting tissues
much more efficiently than a knife.
(a) tungsten
(b) lead and tin
(c) platinum
(d) iron
The charge of an electron is ..................... .
(a) 1.6 × 10–19 C
(b) 1.6 × 10–32 C
–19
(c) 1.6 × 10 A
(d) 1.6 × 10–32 V
1 A (micro-ampere) = ..................... .
(a) 10–5 A
(b) 10–6 A
–3
(d) 10–2 A
(c) 10 A
..................... is the electric discharge travelling from clouds at high
potential to the earth.
(a) Lightning
(b) Tornado
(c) Thunder
(d) Sparks
The resistivity of ..................... is the highest in conductors.
(a) nickel
(b) mercury
(c) chromium
(d) manganese
SCHOOL SECTION
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Answers :
1. 10–3 A
3. 3.6 × 106 joule
5. volt
7. volt
9. greater
11. 2 volt
13. smaller
15. random
17. conventional direction
19. heat is produced
21. 4.18
23. high melting point
25. diameter
27. 1.6 × 10–19C
29. Lightning
Q.I
1.
(i)
(ii)
(iii)
(iv)
Ans.
2.
(i)
(ii)
(iii)
(iv)
Ans.
3.
(i)
(ii)
(iii)
(iv)
Ans.
(B) Match the following :
Column I
Electric current
Electric charge
Electric resistance
Potential difference
series
4A
coulomb
ampere
insulator
an ammeter
6
circuit diagram
ohm-metre
I 2Rt
calorie
lead and tin
platinum
10–6 A
mercury
(a)
(b)
(c)
(d)
(e)
Column II
Joule
Ampere
Ohm
Coulomb
Volt
(a)
(b)
(c)
(d)
(e)
Column II
It
Q/t
mgh
IR
V/I
(a)
(b)
(c)
(d)
(e)
Column II
Wire of high melting point
Prevents oxidation
Wire of low melting point
Surgery
Insulator
(i – b), (ii – d), (iii – c), (iv – e).
Column I
Electric current
Electric charge
Potential difference
Electric resistance
(i – b), (ii – a), (iii – d), (iv – e).
Column I
Electric fuse
Electric bulb
Platinum wire
Argon
(i – c), (ii – a), (iii – d), (iv – b).
*4.
(i)
Column I
Heat generated
(ii)
Resistance in
parallel
(b)
(iii)
Resistivity
(iv)
Ohm’s law
Ans.
2.
4.
6.
8.
10.
12.
14.
16.
18.
20.
22.
24.
26.
28.
30.
1.
Column III
V = IR
Proportional to the square
of current
2.
 =
(c)
Gives relation between
V and I
3.
VIt
cal
4.18
(d)
Depends on the material
4.
1 1
1


R R1 R2
(a)
Column II
Is used to reduced effective
resistance in a circuit
of the conductor
(i – b, 3), (ii – a, 4), (iii – d, 2), (iv – c, 1).
SCHOOL SECTION
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
75
SCIENCE & TECHNOLOGY
Q.I
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12.
Ans.
13.
Ans.
14.
Ans.
(C) State whether the following statements are true or false. If false
write the corrected statement :
The SI unit of charge is volt.
False. The SI unit of charge is coulomb.
Voltmeter is always connected in series with the device.
False. Voltmeter is always connected in parallel with the device.
The conventional direction of flow of current is from positive terminal to
negative terminal.
True.
Silver and copper are good conductors.
True.
Resistivity of pure metals is more than alloys.
False. Resistivity of pure metals is less than alloys.
The electric bulb consists of the filament whose melting point is low.
False. Electric bulb consists of filament whose melting point is high.
Resistance in series arrangement is used to decrease resistance of circuit.
False. Resistance in series arrangement is used to increase resistance of
circuit.
A conducting wire offers resistance to flow of electrons.
True.
Charges are measured in ampere.
False. Charges are measured in coulomb.
The unit of potential difference is ampere.
False. The unit of potential difference is volt.
Resistance of a conductor is inversely proportional to the length of the
conductor.
False. Resistance of a conductor is directly proportional to the length of
the conductor.
Ammeter is connected in parallel to the cell to measure current.
False. Ammeter is connected in series to the cell to measure current.
Fuse is made of wire having high melting point.
False. Fuse is made of wire having low melting point.
Power is measured in joule.
False. Power is measured in watt.
Q.I
1.
Ans.
2.
Ans.
3.
Ans.
(D) Find the odd man out :
Aluminium, Rubber, Gold, Silver.
Rubber. It is an insulator, while the rest are all conductors.
Glass, Hard rubber, Mercury, Paper (dry).
Mercury. It is a good conductor, while the remaining are insulators.
Constantan, Tin, Nichrome, Diamond.
Diamond. It is an insulator and the rest are alloys.
*1.
Ans.
*2.
Ans.
*3.
Ans.
*4.
Ans.
*5.
Ans.
*6.
Ans.
7.
Ans.
8.
Ans.
9.
Ans.
10.
Ans.
11.
Ans.
4.
Ans.
76
I2Rt IRt
VIt
V 2t
,
,
,
.
4.18 4.18 4.18 4.18 R
IRt
. It does not follow Joule’s Law while the remaining follow Joule’s
4.18
Law.
SCHOOL SECTION
MT
5.
Ans.
6.
Ans.
Q.II
*1.
Ans.
2.
Ans.
3.
Ans.
*4.
Ans.
*5.
Ans.
*6.
Ans.
7.
Ans.
*8.
Ans.
9.
Ans.
10.
Ans.
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SCIENCE & TECHNOLOGY
Ammeter, Ampere, Volt, Coulomb.
Ammeter. It is a device used to measure electric current and the remaining
are units.
Electric iron, Electric bulb, Electric fan, Electric toaster.
Electric fan. It is based on magnetic effect of electric current and the rest
is based on the principle of heating effect of electric current.
Define the following :
Resistivity.
Resistivity of a conductor is defined as the resistance of a conductor of
unit length and unit area of cross section.
Conductors.
The substances which have very low electrical resistance are called
conductors.
Insulators.
Those substances which have infinitely high electrical resistance are called
Insulators.
Potential difference.
The potential difference between two points in an electric field is defined
as the amount of work done in moving a unit positive charge from one point
to another point.
1 Volt PD.
The potential difference between two points is said to be one volt, if one joule
of work done in moving 1 coulomb of electric charge from one point to another.
1 ampere.
The current flowing through a conductor is called as one ampere. If one
coulomb of charge is passing through any cross-section of a conductor in
one second.
Ohm’s law.
The electric current flowing in a metallic conductor is directly proportional
to the potential difference across its terminals, provided physical conditions
of the conductor such as length, area of cross section, temperature and
material remain constant.
1 ohm.
If one ampere current flows through the conductor and 1 volt potential
difference is applied across it, then its resistance is one ohm.
1 volt
1 ohm = 1 ampere .
Resistance.
The property of the conductor due to which it opposes flow of current through
it is called resistance.
Joule’s law.
The quantity of heat (H) in a conductor of resistance (R) when a current (I)
flows through it for a time (t) is directly proportional to
1. The square of the current.
2. The resistance of the conductor.
3. The time for which the current flows.
SCHOOL SECTION
77
SCIENCE & TECHNOLOGY
*11.
Ans.
12.
Ans.
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Electric power.
Electric power is defined as the rate at which electrical energy is consumed.
1 watt power.
If one joule of work is done per second the electric power is 1 watt.
Q.III (A) Give scientific reasons :
1. Metals are good conductors of electricity.
Ans. 1. Metals contain a large number of free electrons.
2. When a potential difference (P.D.) is applied between the two ends of
the wire, these free electrons move easily through, the conductor.
3. This motion of free electrons constitutes an electric current.
Hence metals are good conductors of electricity.
*2. The materials used for fuse has low melting point.
Ans. 1. Fuse wire is made of an alloy like lead and tin having low melting point.
2. If excess current passes through the fuse, the fuse wire melts.
3. As the fuse wire melts, the circuit breaks immediately. Fuse protects
the electrical appliances by melting, at low temperature, thus limiting
the current passing through device.
Thus, fuse is made of material having low melting point.
*3. Wood and rubber are good insulators.
Ans. 1. Those substances which have infinitely high electrical resistance are
called insulators.
2. Wood and rubber have high resistance and negligible free electrons for
conduction of electricity.
3. Hence wood and rubber are good insulators.
*4. The melting point of filament of a bulb is very high.
Ans. 1. The filament of a bulb has to get heated to a very high temperature to
emit light.
2. If a filament of low melting point is used, it would melt and the bulb
would become useless.
3. Hence a filament of high melting point is used so that the bulb can be
used for a long time.
*5. Connecting wires in a circuit is made of copper and aluminium.
Ans. 1. Copper and aluminium are good conductors of electricity.
2. They have low electrical resistance.
3. As they are malleable and ductile they can be drawn into thin wires.
Hence connecting wires in a circuit is made of copper or aluminium.
6. A thick wire has a low resistance.
Ans. 1. The resistance (R) of a wire is inversely proportional to the cross
sectional area (A) of a wire. i.e. R  1/A.
2. Thus, greater is the cross sectional area of a conductor (wire), lower is
its resistance.
Hence a thick wire has a low resistance.
7. A series combination of resistances is used to increase the resistance
of a circuit.
Ans. 1. In a series combination, the effective resistance is equal to the sum of
the individual resistances.
2. Thus, any increase in the number of individual resistance will increase
the overall resistance of the circuit.
Hence, a series combination is used to increase the resistance of a circuit.
8. A parallel combination of resistance decreases the effective resistance
of the circuit.
Ans. 1. In a parallel combination, the reciprocal of the effective resistance is
equal to the sum of the reciprocals of the individual resistances.
78
SCHOOL SECTION
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Ans.
10.
Ans.
SCIENCE & TECHNOLOGY
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2. Due to this, any addition of an individual resistance in parallel
combination will decrease the overall resistance of the circuit.
Hence a parallel combination of resistance decreases the effective
resistance of the circuit.
Nichrome wire is used in heating device.
1. Nichrome wire is made of alloy having high resistance.
2. Nichrome wire has high melting point.
3. Thus it produces heat, when current is passed through it.
Thus Nichrome is used in heating device.
Lightning occurs from sky to earth.
1. Lightning is the electric discharge travelling from clouds at high
potential to the earth surface, which is at zero potential.
2. As earth is always at lower potential as compared to the clouds.
3. Hence, lightning occurs from sky to earth.
Q.III (B) Answer the following questions in short :
1. Flow of current through conductors. OR
Discuss how an electric current is set up when charges are in motion.
Ans. 1. If the metal wire is not connected to any cell or a battery then the
electrons present in it move randomly in all the directions between
the atoms of the metal wire.
2. When a source of electricity like a cell or a battery is connected across
the ends of the metal wire, then an electric force acts on the free
electrons present in the wire.
3. Since electrons are negatively charged they start moving from the
negative end to the positive end of the cell or battery through a wire.
4. This flow of electrons constitutes the electric current in the wire. The
electrons in the conductor move with a certain average drift speed.
5. However direction of the conventional current is assumed to be from
the positive terminal to the negative terminal of the cell.
2. Explain potential difference.
Ans. 1. The potential difference between two points in an electric field is defined
as the amount of work done in moving a unit positive charge from one
point to another point.
Potential difference =
3.
Ans.
4.
Ans.
Work done
Quantity of charg e transfered
V
=
W
Q
1V
=
1J
1C
2. The SI unit of potential difference is volt denoted by V.
Explain the working of electric iron.
1. An electric iron consists of coil of high resistance covered by mica
sheets and kept inside heavy metal block.
2. When electric current passes through the coil it gets heated and can
be used for ironing.
3. The mica sheet is a bad conductor of electricity and good conductor of
heat. It prevents the current from entering into the metal and thus
protects the user from getting an electric shock.
Explain the working of electric fuse.
1. An electric fuse protects circuits and appliances by stopping the flow of
any excess electric current.
SCHOOL SECTION
79
MT
SCIENCE & TECHNOLOGY
5.
Ans.
*6.
Ans.
*7.
Fuse holder
2. The fuse is placed in series with the device.
It consist of a piece of wire made of an alloy of
low melting point for example lead and tin.
3. If a current larger than the specified value
Fuse
flows through the circuit, the temperature
wire
of the fuse wire increases.
4. This melts the fuse wire and breaks the
Fuse wire melts and
circuit.
breaks the circuit
5. The fuses used for domestic purpose are
Electric fuse
stated as 1A, 2A, 3A, 5A, 10A etc.
Only
for reference
Explain the concept of resistance.
1. When the free electrons flow from one part to the other part of
conductor, they collide with other electrons and also with the atoms
and ions present in the conductor.
2. Due to these collisions there is some obstruction or opposition to the
flow of electron through the conductor.
3. The property of the conductor due to which it opposes flow of current
through it is called resistance.
4. It is numerically equal to the ratio of potential difference across it
ends to the current flowing through it.
P and Q are the two wires of same length and different cross sectional
areas and made of same metal. Name the property which is same for
both the wires and that which is different for both the wires.
The property which is same for both the wires is resistivity and the property
which is different for both the wires is resistance.
Resistivity of some material is given below. State which one will be the
best conductor.
Material
Copper
Resistivity (m) 1.62 × 10
Ans.
*8.
Ans.
9.
Ans.
Q.IV
*1.
Ans.
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Aluminium
–8
2.63 × 10
–8
Silver
1.60 × 10
Nickel
–8
6.84 × 10–8
Silver has the lowest resistivity (1.60 × 10–8 -m), hence silver offers the
least resistance to the flow of current through it. Thus silver is the best
conductor of them all.
Two dissimilar bulbs are connected in series, which bulb will be brighter ?
(Hint : consider the resistance of the bulb)
1. Bulb works on the principle of heating effect of electric current.
2. Heat generated is directly proportional to the resistance.
3. Hence the bulb having higher resistance will glow brighter.
Explain why a conductor is heated when current flows through it.
1. When current flows through a metallic conductor, the free electrons in
the metal start moving from the end which is at the lower potential.
2. These moving free electrons collide with the atoms of the metal.
3. At each collision, a part of kinetic energy of electron converts into heat.
(A) Distinguish between :
Conductors and Insulators.
Conductors
Insulators
1. The substances which have very 1. Those substances which have
low electrical resistances are
infinitely
high
electrical
called conductors.
resistances are called Insulators.
2. They contain a large number of 2. They contain practically no free
free electrons.
electrons.
3. Conductors are mostly metals.
3. Insulators are mostly non metals.
SCHOOL SECTION
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*2.
Ans.
Voltmeter and Ammeter.
3.
Ans.
Resistance in series and parallel.
*4.
Ans.
Resistance and Resistivity.
*5.
Ans.
High resistance and Low resistance.
Voltmeter
1. It is an instrument to measure
the potential difference between
the two terminals of a cell.
2. It is connected in parallel with
the cell.
3. It has a very high resistance.
Ammeter
1. It is an instrument to measure
the electric current flowing
through the circuit.
2. It is connected in series with the
cell.
3. It has a very low resistance.
Resistance in series
Resistance in parallel.
1. Resistances are connected one 1. Resistances are connected
after another such that the
between two common points
current flowing through its
such
that
the
potential
resistance is the same.
difference
across
each
resistance is the same.
2. This combination increases the 2. This combination decreases the
effective resistance in the
effective resistance in the
circuit.
circuit.
3. This combination decreases the 3. This combination increases the
current in the circuit
current in the circuit.
4. The effective resistance in series 4. The reciprocal of effective
is equal to the sum of the
resistance in parallel is equal to
individual resistors.
the sum of reciprocals of the
individual resistance.
Resistance
Resistivity
1. The property of the conductor due 1. Resistivity of a conductor is
to which it opposes flow of current
defined as the resistance of a
through it is called resistance.
conductor of unit length and unit
area of cross section.
2. The S.I unit of resistance is 2. The S.I unit of resistivity is ohmohm ().
metre ( – m).
3. It depends on temperature, area 3. It depends on material of the
of cross section, length of
conductor.
conductor and material of the
conductor.
High resistance
Low resistance
1. In materials having high 1. In
materials
having
low
resistance the opposition to the
resistance the opposition to the
flow of current is high.
flow of current is low.
2. Insulators have high resistance. 2. Conductors have low resistance.
3. Materials having high resistance 3. Materials having low resistance
produce more heat on passage of
produce less heat on passage of
electric current.
electric current.
SCHOOL SECTION
81
MT
SCIENCE & TECHNOLOGY
Q.IV
1.
Ans.
(B) Draw a neat and labelled diagram of the following :
Symbols commonly used in circuit diagrams.
Sr. No.
82
EDUCARE LTD.
Component
Symbol
+
–
1.
Electric cell
2.
Battery of cells
3.
Tap key or plug key (off)
OR
4.
Tap key or plug key (on)
OR
5.
Wire joint
6.
Wires crossing
7.
Electric bulb
8.
A resistor or resistance
9.
Rheostat or variable
resistance
+
10.
Ammeter
+
11.
Voltmeter
+
–
A
–
V
–
SCHOOL SECTION
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Components of Electrical Circuit
Battery of cells
Plug key (closed)
Resistor
Rheostat
Ammeter
Voltmeter
SCHOOL SECTION
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SCIENCE & TECHNOLOGY
2.
EDUCARE LTD.
Random motion of electrons in a wire and motion of electrons when
potential difference is applied.
Ans.
Electrons
e–
e–
e–
e–
e–
e
e–
–
e–
e–
Metal wire
e–
e–
Random motion of electrons in a wire
Direction of electron flow
+
e–
e–
e–
e–
e–
e–
e–
e–
e–
e–
–
Direction of conventional current
+ –
Motion of electrons when potential difference is applied
3.
Ans.
Circuit diagram for studying Ohm’s Law.
+
–
V
V
V
R
X
V
Y
+ A –
K
+ V –
4.
Ans.
Graph of potential difference and current for Ohm’s Law.
A
Current (I)
O Potential difference (V)
84
SCHOOL SECTION
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5.
Ans.
SCIENCE & TECHNOLOGY
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Heating effect of electric current. (Verification of Joule’s Law)
Battery
E
+
–
Key
K
Ammeter
I
– A+
+ V –
Voltmeter
Thermometer
Stirrer
Nichrome coil
Water
6.
Ans.
Flow of current (Potential difference).
High potential
+
Electric current flows
A
Positively
charged
conductor
Insulated stand
Q.V
1.
Ans.
Low potential
Wire
B
–
Negatively
charged
conductor
Insulated stand
Answer the following in detail :
Find the expression for resistivity of a material.
1. Resistance of a conductor depends on the length ‘’ and area of cross
section ‘A’ of the conductor
R  
1
A

R 
A

R =
A
and R 


2. Where  is called resistivity of the conductor. It is also called as specific
resistance.
If we put  = 1m and A = 1m2 then
 R = 
3. Thus resistivity of a conductor is defined as the resistance of a conductor
of unit length and unit area of cross - section.
4. The S.I. unit of resistivity is ohm – metre (–m).
SCHOOL SECTION
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SCIENCE & TECHNOLOGY
2.
Ans.
3.
Ans.
EDUCARE LTD.
State and explain the factors on which the resistance of a conductor
depends.
The resistance of a conductor depends upon the following factors :
1. Material of the conductor : Conductors of different materials have
different resistance.
2. Length (L) of the conductor : The resistance (R) of a conductor is
directly proportional to the length of the conductor i.e. R  .
3. Area of cross section (A) : The resistance (R) of a conductor is inversely
proportional to the area of cross section (thickness) of the conductor.
i.e. R  1/A.
4. Temperature of conductor : The Resistance of a conductor is directly
proportional to the temperature of conductor.
Find the expression for the resistance connected in series.
+
V –
C
D
R1
R2
R3
–
A
+
+ –
4.
Ans.
86
K
E
1. Let R1, R2 and R3 be three resistances connected in series between C
and D.
2. Let RS be the effective resistance in circuit and V1, V2 and V3 be the
potential difference across R1, R2 and R3 respectively.
3. Let the potential difference across CD be V.
4. In series combination.
V = V1 + V2 + V3
.... (i)
By using Ohm’s law
V = IRS
 V1 = IR1, V2 = IR2 and V3 = IR3
Substituting these values in equation (i) we get
IRS = IR1 + IR2 + IR3
 RS = R1 + R2 + R3
For ‘n’ number of resistors connected in series we get
RS = R1 + R2 + R3 + R4 + R5 + R6 + ........ + Rn
Hence effective resistance in series is the sum of the individual
resistances.
State the characteristics of connecting resistances in series.
If the resistors are connected in series then :
1. In a circuit the current is the same in every part of the circuit.
2. The resistance of the combination of resistors is equal to the sum of
the individual resistors.
3. The total voltage across the combination is equal to the sum of the
voltage drop across the separate resistors.
4. The effective resistance in a series combination is greater than the
individual resistances.
5. This combination is used to increase resistance in a circuit.
SCHOOL SECTION
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Ans.
SCIENCE & TECHNOLOGY
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Find the expression for the resistance connected in parallel.
+
I1
+ A
I2
+ A
I3
+ A
C
V
–
–
R1
–
R2
D
–
R3
–
A
+
+ –
K
E
1. Let R 1 , R 2 and R 3 be the three resistances connected in parallel
combination between points C and D and let R P be their effective
resistance.
2. Let I1, I2 and I3 be the currents flowing through resistances R1, R2 and
R3 respectively.
Let I be the current flowing through the circuit and V be the potential
difference of the cell.
3. For parallel combination of resistances,
I
= I1 + I2 + I3
...... (i)
According to Ohm’s law,
V
I
=
Rp
Therefore,
V
V
V
,
I2 =
,
I3 =
.
R1
R2
R3
4. Substituting the values of I, I1, I2 and I3 in equation (i) we get
I1
=
V
=
RP
V
V
V
+
+
R1
R2
R3


 1 
  V 1  1  1 
R 
R
R
R 
 P
2
3 
 1
V

1
=
RP
1
+
R1
1
+
R2
1
R3
1
1
1
1
1
=
+
+
+ ........ + R
R1
n
RP
R2
R3
Therefore, the reciprocal of the effective resistance in parallel
combination is equal to the sum of the reciprocals of the individual
resistances.
State the characteristics of connecting resistances in parallel.
If the resistors are connected in parallel then :
1. The sum of reciprocals of the individual resistance is equal to the
reciprocal of effective resistance.
For ‘n’ number of resistances
6.
Ans.
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Ans.
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2. The currents in various resistors are inversely proportional to the
resistances (higher is the resistance lower is the current through it).
However the total current is the sum of the currents flowing in the
different branches.
3. The voltage across each resistor is same.
4. The effective resistance of the combination is less than the individual
resistance in the combination.
5. This combination is used to decrease resistance in the circuit.
Derive Joule’s law.
Let V be the potential difference applied between the two terminals of a
conductor of resistance R, then by using Ohm’s law.
V
=R
I
 V = IR
...... (i)
If the current I is passed through a conductor for time t, then
Q = It
...... (ii)
Work done (W) during flow of charge Q is given by
W
V= Q
 W = VQ
...... (iii)
Substituting the values of eq. (i) and (ii) in (iii) we get
W = IR × It
When current flows through conductor, work done is converted into
heat
 W = I2Rt joule
 H = I2Rt joule
Usually we express heat energy in calorie. The relation between joule
and calorie is 4.18 J = 1 cal
Heat produced in terms of calorie is given by
I2Rt
H=
cal
4.18
Joule’s law : The quantity of heat (H) generator in a conductor of
resistance (R) when a current (I) flows through it for a time (t) is directly
proportional to
(a) The square of the current.
(b) The resistance of the conductor.
(c) The time for which the current flows.
Q.VI
1.
Ans.
2.
Ans.
3.
Ans.
4.
Ans.
5.
Ans.
88
Answer the following questions in one sentence each :
What is electric current ?
Electric current is flow of electrons in a conductor or amount of charge
flowing through a particular cross section area in unit time.
What is electric circuit ?
A continuous path consisting of conducting wires and other resistances
between the terminals of a battery, along which an electric current flows
is called electric circuit.
What is the conventional direction of electric current ?
Conventional direction of current is from the positive terminal to negative
terminal of the cell.
What is the function of an electric cell ?
The function of an electric cell is to provide potential difference in a circuit.
What do you mean by ‘open circuit’ and ‘closed circuit’ ?
An ‘open circuit’ is a circuit in which no current is flowing and ‘closed
circuit’ is circuit in which current is flowing.
SCHOOL SECTION
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6.
Ans.
7.
Ans.
8.
Ans.
9.
Ans.
10.
Ans.
11.
Ans.
12.
Ans.
13.
Ans.
14.
Ans.
15.
Ans.
16.
Ans.
17.
Ans.
SCIENCE & TECHNOLOGY
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What is ammeter ? How is it connected in a circuit ?
Ammeter is an instrument used to measure the electric current in the
circuit. It is connected in series with the cell in the circuit.
What is a voltmeter ? How is it connected in the circuit ?
Voltmeter is an instrument used to measure the potential difference across
the device. It is to be connected in parallel to the cell.
What is resistivity ?
Resistivity of a conductor is the resistance of a conductor of unit length
and unit area of cross section.
Write the formula for calculation of resistivity.
R
=
.

A
OR

=
A
.R

 = Resistivity of conductor
A = Area of the cross conductor
 = Length of the conductor
R = Resistance of the conductor
What is the effective resistance when resistances are connected in
1. Series
2.
Parallel ?
1. When resistances are connected in series the effective resistance is
greater than the individual resistances.
2. When resistances are connected in parallel the effective resistance is
lower than the individual resistances.
What is heating effect of electric current ?
The production of heat energy in a conductor by the electric current flowing
through it is called the heating effect of the current.
Give any four applications of heating effect of electric current.
Four applications of heating effect of electric current is
1. Electric bulb
2. Electric iron
3. Electric fuse
4. Electric heater.
What is the fuse wire made of ?
Fuse wire is made of an alloy of lead and tin which has low melting point.
From which material is the filament of bulb made of ?
The filament of an electric bulb is made from tungsten.
Name the gases that are filled in the bulbs and why ?
The bulbs are filled with chemically inactive nitrogen and argon gases to
prolong the life of the filament.
State Joule’s law mathematically.
V2t
VIt
I2Rt
H=
cal. or
cal or
cal.
4.18 R
4.18
4.18
What is the relation between calorie and joule ?
The relation between calorie and joule is : 1 cal = 4.18 J.
Q.VII Solve the following numericals :
Type A
Problem based on the formula : 1.
*1.
Ans.
Q = It
2.
W = VQ
Solved examples :
A current of 0.2 A is flowing through a bulb for 5 minutes. Find the
charge that is flowing through the circuit.
Given :
Current (I)
= 0.2 A
Time (t)
= 5 min.
= 300 sec.
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To find : Electric charge (Q) = ?
Formula : Q = It
 Q = 0.2 × 300
 Q = 60 C
The charge flowing through the circuit is 60 C.
*2.
Ans.
A charge of 420 C is flowing through a circuit for 10 minutes. Find the
current in the circuit.
Electric charge (Q) = 420 C
Given :
Time (t)
= 10 min.
= 600 sec.
To find : Electric current (I)
Formula : I
=
Solution : I
=
 I
=
Q
t
Q
t
420
600
7
10
 I
= 0.7 A
 The current in the circuit is 0.7 A.
 I
*3.
Ans.
=
Find the amount of work done if 3 C of charge is moved through a
potential difference of 9 V.
Electric charge (Q) = 3 C
Given :
P.D. (V)
= 9V
To find : Work done (W)
Formula : V
=
W
Q
Solution : V
=
W
Q




W = VQ
W = 9×3
W = 27 J
The work done is 27 joule.
HOME WORK ASSIGNMENT - A
1.
2.
3.
4.
90
A charge of 150 coulomb flows through a wire for 1 min 15 sec. Find the
(Ans. 2 A)
current flowing through the conductor.
If 100 J of work is done in moving a charge of 5 C from one point to another,
find the potential difference between the two points.
(Ans. 20 volt)
The potential difference between any two points in a circuit is 60 V. If a
charge of 24 C is transferred between these two points, find the work done
(Ans. 1440 J)
in joules.
Work of 250 J is done in moving a charge through a conductor having a
potential difference of 5 V in 2 seconds. Find the current flowing through
(Ans. 25 A)
the conductor.
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Type B
Problem based on the formula :1.
*1.
Ans.
*2.
Ans.
V = IR
2.
R=

A
Solved examples :
Find the resistance of a conductor if 0.24 A current is passing through
it and a potential difference of 24 V is applied across it.
Given :
Current (I) = 0.24 A
P.D. (V)
= 24 V
To find : Resistance (R)
Formula : V = IR
Solution : V = IR
V
 R =
I
24
 R =
0.24
 R = 100 
 The resistance of the conductor is 100 .
If the resistance of the filament of a bulb is 1000 . It is drawing a
current from a source of 230 V. How much current is flowing through it?
Given :
Resistance (R) = 1000
P.D. (V)
= 230 V
To find : Current (I)
= ?
V
=R
I
V
Solution :
=R
I
V
 I
=
R
230
 I
=
1000
Formula :
 I
= 0.23 A
The current flowing through the filament of bulb is 0.23 A.
*3.
Ans.
If a potential difference of 33 V is applied to a device whose resistance
is 110 . Find the current. If the same current is passed through a
device whose resistance is 500 , then how much potential difference
is to be applied ?
Given :
P.D. (V1)
= 33 V
Resistance (R 1) = 110
Resistance (R 2) = 500 
To find : 1. Current (I)
2. P.D. (V) for Resistance R2.
Formula : V = IR
Solution :
Case I :
V 1 = IR 1
V1
 I
=
R1
33
 I
=
110
 I
= 0.3 A
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Case II :
V 2 = IR 2
 V 2 = 0.3 × 500
 V 2 = 150 V
 The current is 0.3 A and potential difference to be applied is 150 V.
*4.
Ans.
If the length of the wire is 50 cm and radius is 0.5 mm. Find the
resistivity of wire if its resistance is 30 .
Given :
Length ()
= 50 cm
= 50 × 10–2m
Radius (r)
= 0.5 mm
= 0.5 × 10–3m
= 5 × 10–4 m
Resistance (R) = 30 
To find : Resistivity ()
Formula : R
=
Solution : R
=
 
=
 
=
 
=
 
=

A


A

RA
where = A = r2

30  3.14  (5  10 –4 )2
50  10 –2
30  3.14  25  10 –8
50  10 –2
30  3.14
× 10–6
2
 
= 15 × 3.14 × 10–6
 
= 47.1 × 10–6 m
The resistivity of the wire is 47.1 × 10–6 m.
5.
L
1.5L
A
2
(a)
Ans.
92
L/2
A
2
(b)
(c)
Three cylindrical copper conductors along with their cross - sectional
areas and lengths. Compare them according to the current through
them, if the same potential difference V is applied between their lengths.
Let V be the potential difference  is the resistivity which is a constant
for all three conductors.

R = 
A
V
 But V = IR  R =
I

V

= 
A
I
A
I

=

V
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Removing constants V and  we get
A
I
=

Case I :
A
I1 =
L
Case II :
A
 I2
=
2
1.5L
I2
=
A
3L
Case III :
A
 I3
=
L
 I3
=
A
L
2
2
Hence I1, I2, I3 are in the ratio
i.e. 1 :
*6.
Ans.
A
A
A
:
:
L 3L
L
1
: 1 respectively.
3
If the resistance of wire A is four times the resistance of wire B, find
the ratio of their cross sectional areas.
Given :
R 1 = 4R 2
To find : Ratio of A1 : A2

Formula : R = 
A

Solution : R 1 = 
A1

 A1
= 
R1
But R1 =
4R 2
=


4R2
 R2
=


A2
 A2
=


R2
 A1
Also,

4R 2


R2


A1
A2
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

A1
A2
A1
A2
=
=
EDUCARE LTD.
R2
4R2
1
4
A1
1
 The ratio of their cross sectional areas is A = .
4
2
HOME WORK ASSIGNMENT - B
1.
2.
3.
4.
Calculate the potential difference across a 7  resistor carrying a current
(Ans. 1.4 volt)
of 0.2 A.
A negligibly small current is passed through a wire of length 15 m and of
uniform cross-section 6.0 × 10 – 7m² and its resistance is measured to be
5.0. What is the resistivity of the material?
(Ans. 2.0 × 10–7 m)
A copper wire of length 2m and area of cross - section 1.7 × 10–6 m2 has a
resistance of 2 × 10–2 ohms. Calculate the resistivity of copper.
(Ans. 1.7 × 10–8 m)
What should be the potential difference across a conductor of resistance
15 in order to send a current of 5 A through it?
(Ans. 75 volt)
Type C
Problem based on the formula : 1.
2.
*1.
Ans.
*2.
Ans.
94
Rs = R1 + R2 + R3
1
1
1
1
=
+
+
R2
RP
R1
R3
Solved examples :
If three resistors 5, 6 and 8 each are connected in series, what is
the effective resistance in a circuit?
Given :
R1 = 5 
R2 = 6 
R3 = 8 
To find : Effective resistance in series (R S) = ?
Formula : R S = R1 + R2 + R3
 RS = 5 + 6 + 8
 R S = 19 
The effective resistance in a circuit is 19 .
Three resistances 15, 20 and 10 are connected in parallel. Find the
effective resistance of the circuit.
R 1 = 15 
Given :
R 2 = 20 
R 3 = 10 
To find : Effective resistance in parallel (R P)
1
1
1
1
+
+
Formula : R =
R
R
R
1
2
3
P
1
1
1
1
+
+
Solution : R =
R1 R 2 R 3
P
1
1
1
1
+
+
 R =
15
20
10
P
1
4+3+6
 R =
60
P
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
1
=
RP
 RP =
13
60
60
13
 R P = 4.615 
The effective resistance of the circuit is 4.615 .
3.
Resistors of 10 and 2.5 are connected in parallel combination and a
3 resistance in connected in series combination with them. Find the
combined resistance.
Ans.
R1 = 10
R3 = 3
R2 = 2.5
To find : Equivalent Resistance (R) = ?
Solution :
1. Resistance R1 and R2 are in parallel combination.
1
1
1
 R =
+ R
R
1
2
P
1
1
1
 R =
+
10
2.5
P
1
1+ 4
 R =
10
P
1
5
 R =
10
P
10
 RP =
5
 R P = 2
2.



*4.
Resistance RP and R3 are in series combination.
R = RP + R3
R = 2+3
R = 5
The effective resistance of the arrangement is 5.
Find the total resistance in the circuit.
8 
7.2 
12 
–
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–
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Ans.
= 8
= 12 
= 7.2 
Total resistance (R)
1
1
1
Formula : R =
+
R
R
2
P
1
R = RP + R3
1
1
1
Solution : R =
+
R
R
2
P
1
1
1
1
 R =
+
8
12
P
1
32
 R =
24
P
1
5
 R =
24
P
Given : R 1
R2
R3
To find :
 RP =




*5.
EDUCARE LTD.
24
5
R = 4.8 
R = RP + R 3
R = 4.8 + 7.2
R = 12 
The total resistance of the circuit is 12 .
Find the total resistance and current in the circuit.
5 
3 
–
Ans.
96
A +
2 
4V
+
–
Given : R 1
= 3
R2
= 2
R3
= 5
P.D. (V) = 4 V
To find : 1. Total resistance (R)
2. Current (I)
Formula : 1. R S = R 1 + R 2
1
1
1
2. R =
+
R
R
P
S
3
3. V = IR
Solution : R S = R 1 + R 2
 RS = 3 + 2
 RS = 5 
1
1
1
RP =
RS +
R3
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1
RP =
1
RP =
1
5
2
5
 RP =
5


 RP =
V =
+
1
5
2
2.5 
IR
 I
=
V
R
 I
=
4
2.5
 I
= 1.6 A
The total resistance is 2.5  and current is 1.6 A.
*6.
Ans.
If 6  and 4  resistors are connected in series. Find the current in the
circuit if 18 V potential difference is applied across it. Find also the
potential across each resistor.
Given : R 1 = 6
R 2 = 4
V = 18 V
To find : Current (I)
Potential across each resistance
i.e. V1, V2
Formula : R S = R1 + R2
I
=
Solution : R S =
 RS = 6 + 4
 R S = 10
 V = IRS
V
 I
=
RS
 I
=
V
R
R1 + R 2
18
10
 I
= 1.8 A
Also, V1 = IR 1
 V 1 = 1.8 × 6
 V 1 = 10.8 volt
V 2 = IR 2
 V 2 = 1.8 × 4
 V 2 = 7.2 volt
The current in the circuit is 1.8 A and potential across 6  resistor
is 10.8 volt and 4 resistor is 7.2 volt.
*7.
Ans.
If the resistors 5 , 10  and 30  are connected in parallel to battery
of 12 V. Find the effective resistance of a circuit. Calculate the total
current and current in each resistor.
Given : R 1 = 5 
R 2 = 10 
R 3 = 30 
V = 12 V
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To find : 1. Total current and current is each resistor i.e. I, I1, I2 and I3.
2. Effective resistance (R P).
1
1
1
1
+
+
Formula : 1. R =
R
R
R
P
1
2
3
2. V = IR
Solution :
V
1. I 1 =
R1
12
5
 I1
=
 I1
=
2.4 A
2. I 2
=
V
R2
 I2
=
 I2
=
12
10
1.2 A
3. I 3
=
V
R3
 I3
=
 I3
=
12
30
0.4 A
4. I
 I
 I
=
=
=
I1 + I2 + I3
2.4 + 1.2 + 0.4
4A
5.


1
=
RP
1
RP =
1
=
RP
 RP =
1
1
1
+
+
R1 R 2 R 3
1
1
1


5 10 30
6 + 3 +1
30
30
10
 R P = 3
1. The total current is 4 A and current in each resistors is 2.4 A,
1.2 A and 0.4 A.
2. The effective resistance in parallel is 3 .
*8.
Ans.
98
How will you connect three resistances of 4 ohms each to get 12 , 6 
and 1.33 , respectively?
Given : R 1 = 4 
R2 = 4 
R3 = 4 
To find : How to connect to obtain 12 , 6 , 1.33  ?
Formula : R S = R1 + R2 + R3
1
1
1
1
=
+
+
RP
R1
R2
R3
Solution :
Case I : To obtain 12 .
 R S = R1 + R 2 + R 3
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 RS = 4 + 4 + 4
 R S = 12 
To get a equivalent of 12  R1, R2, R3 should be connected in series.
Case II : To obtain 1.33 .

1
=
RP
1
1
1
+
+
R1
R2
R3

1
RP =
1
1
1
+
+
4
4
4
1
RP =
3
4
 R P = 1.33
To get a equivalent of 1.33  R1, R2, R3 should be connected in parallel.

Case III : To obtain 6 .

1
RP =
1
1
R1 + R 2

1
RP =
1
1
+
4
4





*9.
Ans.
1
2
RP =
4
RP = 2 
Connect Rp in series with R3
R = Rp + R 3
R = 2+4
R = 6
Hence to obtain 6 , R1 and R2 should be connected in parallel and
effective should be connected in series with R3.
If two resistors are connected in series the total resistance is 45  and
if the same resistors are connected in parallel the total resistance
becomes 10 . Find the individual resistors.
Let R1 and R2 be the two resistances.
 R1 + R2 = 45
 R2
= (45 – R1)
.......... (i)
Also,
1
RP =
1
1
+
R1
R2
1
RP =
R 2  R1
R1 R 2
 RP =
R1 R 2
R 2  R1

 10 =
R1  45  R1 
45
 450 = 45R1 – R12
 R12 – 45R1 + 450
 R12 – 30R1 – 15R1 + 450
SCHOOL SECTION
=
=
0
0
99
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R1 (R1 – 30) – 15 (R1 – 30)
= 0
 (R1 – 15) (R1 – 30)
= 0
 R1 = 15
OR
R 1 = 30
 Hence the values of the two resistances are 15  and 30 .
HOME WORK ASSIGNMENT - C
1.
2.
3.
4.
5.
Three resistances of 20 , 40  and 60  are connected
1. in series
2. in parallel.
Find their resultant resistance in each case. (Ans. 1. 120, 2. 10.9)
Three resistances of 50 , 100  and 150  are connected
1. in series
2. in parallel.
Find the equivalent resistance in each case. (Ans. 1. 300, 2. 27.27)
Two resistances each of 20  are connected in parallel. The combination
is connected in series with a resistance of 20 . Find the equivalent
resistance of the combination.
(Ans. 30)
Two resistances each of 50  are connected in parallel and the combination
is connected in series with a resistance of 20 . Find the equivalent
resistance of the combination.
(Ans. 45)
If two resistors are connected in series the total resistance is 9  and if
the same resistors are connected in parallel the total resistance becomes
(Ans. 6  and 3 )
2 . Find the individual resistors.
Type D
Problem based on the formula :
1. H =
*1.
Ans.
I2Rt
calorie
4.18
2.
H =
V 2t
calorie 3.
4.18R
P = VI = I2R =
V2
R
Solved examples :
Find the heat generated in calories if a current of 0.2 A is passed through
a coil of resistance 418  for 1 minute.
Given : Current (I)
= 0.2 A
Resistance (R) = 418 
Time (t)
= 1 min
= 60 sec
To find : Heat (H)
I2Rt
cal
4.18
I2Rt
Solution : H =
cal
4.18
(0.2)2  418  60
H =
4.18
 H = 240 cal
The heat generated is 240 cal.
Formula : H
*2.
Ans.
100
=
240 J heat is generated in a conductor of resistance 50 , when 0.2A
current is passed across it. Find for how long current will pass through
it.
Given :
Heat (H)
= 240 J
Resistance (R) = 50 
Current (I)
= 0.2 A
To find : Time (t)
Formula : H = I 2Rt
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Solution :
H = I 2Rt
240 = (0.2)2 × 50 × t
240 = 0.04 × 50 × t
240
 t
=
0.04  50
24  100
 t
=
4  5
 t
= 120 sec
t
= 2 min.
Time for which current is passed through the conductor is 2 min.
3.
Ans.
4.
Ans.
A potential difference of 250 volt is applied across a resistance of
1000  in an electric iron find :
1. The current and
2. Heat energy produced in joules in 12 sec.
= 250 V
Given : P.D. (V)
Resistance (R) = 1000 
Time (t)
= 12 sec.
To find : 1. Current (I)
2. Heat (H)
Formula : V = IR
H = VIt joule
Solution : V = IR
V
 I
=
R
250
 I
=
1000
 I
= 0.25 A
 H = VIt
 H = 250 × 0.25 × 12
 H = 750 joule
 The current is 0.25 A and heat in joules is 7500 joule.
Find resistance of 10W, 240V bulb.
= 10 W
Given : Power (P)
Potential difference (V) = 240 V
To find : Resistance (R)
V2
Formula : P =
R
V2
R
240  240
10 =
R
240 × 240
 R =
10
R = 5760 
The resistance of the filament is 5760
Solution : P
*5.
Ans.
=
An electric bulb is connected to 250 volts. The current is 0.27 A. What
is the power of the bulb.
Given :
P.D (V)
= 250 V
Current (I) = 0.27 A
To find : Power (P)
SCHOOL SECTION
101
Formula : P = VI
Solution : P = VI
 P = 250 × 0.27
 P = 67.5 W
The power of the bulb is 67.5 W.
*6.
Ans.
If a bulb of 60 W is connected across a source of 220 V, find the current
drawn by it.
Given : Power (P) = 60 W
P.D. (V)
= 220 V
To find : Current (I)
Formula : P = VI
Solution : P = VI
P
 I
=
V
60
 I
=
220
 I
= 0.2727 A
 The current flowing is 0.2727 A.
HOME WORK ASSIGNMENT - D
1.
2.
3.
4.
5.
6.
7.
Calculate the heat generated in calories when a current flows through a
wire of resistance 50 and P.D. 500V for 3 min 29 sec. (Ans. 250000 cal)
Calculate the heat in joules, generated in a coil of resistance 400 in 2
(Ans. 12000 J)
minutes when a P.D. of 200V is applied across it.
Calculate the heat generated in a coil of resistance 100 in 6 minutes
when P.D. of 10V is applied across it. (upto 2 decimals) (Ans. 86.12 cal)
Calculate the heat generated in calories, when a current of 0.3A flows
through a coil for 15 min. Resistance of the coil is 2.09. (Ans. 40.5 cal)
An electric current of 5A flows through a wire of resistance 41.8. How
long does the current flow through it to produce 3000 calories of heat?
(Ans. 12 sec)
Find resistance of 220W, 110V bulb.
(Ans. 55 )
The current flowing through a 20V Tungsten filament is 4A. Find the power
of the bulb.
(Ans. 80 W)
Type E
Problem based on the formula : Electric power =
1.
Ans.
102
Energy consumed
time in hours
Solved examples :
An electric iron rated 750W operates 2 hours/day. If the cost of unit is
Rs. 3.00 per kWh. Find the cost of energy to operate electric iron for
30 days.
Given :
Power (P) = 750 W
750
=
1000
Time (t)
= 2 × 30
= 60 hrs.
Unit cost
= 3 Rs.
To find : Cost of energy = ?
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Formula : Electric power =
SCIENCE & TECHNOLOGY
Energy consumed
Time in hours
Energy consumed
Time in hours
Energy consumed
= Electric power × time in hours
750
Energy consumed
=
× 60
1000
Energy consumed
= 45 kWh
Cost of energy
= 45 × 3
Cost of energy
= 135 Rs.
Hence cost of energy is 135 Rs.
Solution : Electric power =






2.
Ans.
3.
Ans.
A washing machine rated 300 W operates one hour/ day. If the cost of
unit is Rs. 3.00, find the cost of the energy to operate a washing machine
for the month of March.
Given : Power (P) = 300 W
300
=
= 0.3 kW
1000
Time (t)
= 31 hrs.
Unit cost
= Rs. 3
To find : Cost of energy
Energy consumed
Formula : Electric power =
Time in hours
Energy consumed
Solution : Electric power =
Time in hours
 Energy consumed
= Electric power × time in hours
= 0.3 × 31
= 9.3 kWh
 Cost of energy
= 9.3 × 3
= Rs. 27.9
 The cost of the energy to operate a washing machine for the month
of March is Rs. 27.9.
If a TV of rating 100W is operated for 6 hrs per day, find the number of
units consumed in any leap year.
Given : Power (P) = 100 W
100
=
= 0.1 kW
1000
Time (t)
= 6 × 366
= 2196
To find : Units consumed
Formula : Electric power =
Energy consumed
Time in hours
Energy consumed
Time in hours
 Energy consumed
= Electric power × time in hours
= 0.1 × 2196
= 219.6 kWh
 Hence number of units consumed is 219.6 kWh.
Solution : Electric power =
SCHOOL SECTION
103
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ACTIVITY BASED QUESTIONS
ACTIVITY : 4.1
Ans.
(For question refer to Text Book page No. 31)
1. The torch will not glow, because the cells are removed, as the current
is not flowing.
2. When cells are connected to the two ends the bulb glows.
3. When one end of wire is removed, circuit is incomplete, hence bulb will
once again stop glow.
ACTIVITY : 4.3
Ans.
(For question refer to Text Book page No. 34)
When the end ‘D’ is connected to A, B and C respectively, the brightness of
the bulb goes on increasing, as the number of cells are also increasing.
ACTIVITY : 4.4
Ans.
(For question refer to Text Book page No. 34)
The water flows from A to B i.e. from higher level to lower level.
ACTIVITY : 4.5
Ans.
(For question refer to Text Book page No. 36)
1. When copper wire is connected in the circuit the flow of current will be
more. Ammeter shows more deflection.
2. When aluminium wire is connected in the circuit, the flow of current
will be less than copper. Ammeter shows less deflection.
3. When glass rod is connected in the circuit flow of current will be stopped,
as glass is an insulator. Ammeter shows no deflection.
ACTIVITY : 4.8
Ans.
(For question refer to Text Book page No. 39)
In the figure (a) and figure (b) the intensity of the bulb will remain the
same as the current flowing through the bulb in each case is the same.
ACTIVITY : 4.10
(For question refer to Text Book page No. 42)
Sr. No. Number of cells P.D.
1
2
3
Ans.
104
2
2
3
4V
8V
12V
Current
9A
17A
25A
Temperature in ºC
T1ºC
T2ºC (T2 – T1)ºC
10ºC
10ºC
10ºC
23
41
53
13ºC
31ºC
43ºC
Note : The readings in the table are only for understanding purpose and
not based on practical calculations.
1. We can see that temperature increases as the number of cells increases.
2. If the current is passed for longer time the heat generated will be more.
SCHOOL SECTION
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HOTS QUESTIONS & ANSWERS
1.
Ramesh connected number of bulbs with a nichrome wire whereas
Suresh connected all the bulbs with copper wire of the same length and
thickness. In whose case will the bulbs be brighter ?
Ans. (a) The resistance offered by copper is less and so more current passes
through the wire.
(b) In nichrome the resistance is more and less current passes through
the wire.
(c) Thus the bulbs which Suresh connected will be brighter.
2.
If the length of a copper wire is doubled and its radius is halved, what is
the effect on its resistivity ?
Ans. As the resistivity depends only on the material used for the conductor, the
resistivity remains the same.
3. How are electrical appliances connected at home ? Why ?
Ans. (a) At home the electrical appliances are connected in parallel.
(b) Even if one appliance is switched off the others will continue to work in
such an arrangement.
4. A mica sheet is used in an electrical iron. Why ?
Ans. (a) The mica sheet is a bad conductor of electricity and good conductor of
heat.
(b) So this mica sheet prevents the leakage of current and thus protects
the user from getting an electric shock.
5. Why the commercial unit of power is different from its SI unit ?
Ans. (a) SI unit of power (energy) is watt second. Its value is very small to be
used commercially.
(b) The commercial unit of power (energy) is kilo watt hour is comparatively
larger and hence can be used commercially.
6. Heating appliances consume more electricity. Why ?
Ans. (a) Heating appliances contain coils of high resistance.
(b) High resistance coils require large current to get heated fast and reach
high temperature.
(c) Hence heating appliances consume more electricity.
7. Why tungsten wire is not used in electric fuse ?
Ans. (a) The fuse wire is made of an alloy of low melting point.
(b) Tungsten has a very high melting point so it cannot be used as a fuse
wire as a fuse wire requires low melting point.
8. Fuse is always connected in series. Why ?
Ans. (a) Fuse protects circuits and electrical appliances by stopping the flow of
excess of current.
(b) Fuse is connected in series. So that whatever current is passing through
appliances has to pass through fuse.

SCHOOL SECTION
105
S.S.C.
Marks : 30
CHAPTER 4 : ELECTRIC SPARK
Duration : 1 hr.
SCIENCE
Q.I
1.
2.
3.
4.
Q.I
1.
2.
3.
4.
[A] Choose the correct alternatives and rewrite the complete
sentences :
1 kilowatt hr = ....................... .
(a) 4.6 × 106 joule
(b) 3.6 × 106 joule
6
(d) 3.6 × 105 joule
(c) 30.6 × 10 joule
Electric current is measured with the help of a device called
....................... .
(a) an ammeter
(b) a galvanometer
(c) a volt meter
(d) a calorimeter
....................... is the electric discharge travelling from clouds at
high potential to the earth.
(a) Lightning
(b) Tornado
(c) Thunder
(d) Sparks
Fuse is made up of ....................... .
(a) iron and carbon
(b) lead and tin
(c) copper and zinc
(d) nichrome wire
4
[B] Match the columns :
Column A
Electric resistance
Electric current
Electric charge
Potential difference
4
(a)
(b)
(c)
(d)
(e )
Column B
It
Q/t
mgh
IR
V/I
Q.I
1.
2.
[C] State whether True or False :
Voltmeter is always connected in series with the device.
The SI unit of charge is volt.
2
Q.I
1.
2.
[D] Find the odd man out :
Ammeter, Ampere, Volt, Coulomb.
Electric iron, Electric bulb, Electric fan, Electric toaster.
2
... 2 ...
Q.II
1.
2.
3.
4.
5.
Define the following : (Any Four)
1 volt potential difference.
1 ampere.
1 Ohm.
Ohm’s Law.
Joule’s Law.
4
Q.III
1.
2.
3.
4.
Answer the following : (Any Three)
Distinguish between Resistance in series and parallel.
The melting point of filament of a bulb is very high. Why ?
The materials used for fuse has low melting point. Why ?
Connecting wires in a circuit is made of copper and aluminium.
Why ?
6
Q.IV Solve the following numericals :
1.
An electric current of 5A flows through a wire of resistance 41.8 W.
How long does the current flow through it to produce 3000 calories
of heat ?
2.
An electric iron rated 750 W operates 2 hours/day. If the cost of
unit is 3.00 per kwh. Find the cost of energy to operate electric iron
for 30 days.
4
Q.V
1.
2.
4
Answer the following questions in brief : (Any One)
Find the expression for resistivity of a material.
Find the expression for the resistance connected in series.
Best Of Luck 