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Q 1. A particular resistance wire has a resistance of 3 ohm per
meter. Find : (a) The total resistance of three lengths of this wire each 1.5 m
long, in parallel. (b) The potential difference of the battery which gives a
current of 2 A in each of the 1.5 m length when connected in the parallel to
the battery (assume that resistance of the battery is negligible). (c) The
resistance of 5 m length of a wire of the same material, but with twice the
area of cross section.
Solution
Resistance of 1m of wire
= 3 ohmResistance of 1.5 m of wire = 3 x 1.5 = 4.5 W
I = 2 AV = IR = 2 x 4.5 = 9 V R = 3 ohm for 1
mFor 5 m : R=3 x 5 = 15 ohmBut Area A is double i.e. 2A and resistance is
inversely proportional to area so the resistance will be half.→ R = 15 / 2 =
7.5 ohm
Q 2. Name the physical quantity whose unit is volt/ampere.
Solution
Resistance or electrical
resistance.
Q 3. What is meant by the statement that the potential difference
between two points is 1 volt?
Solution
It means that one joule
of work is done to move a charge of one coulomb from one point to another.
Q 4. How will you connect the resistances of 4 ohms each to get 12
Ω, 6 Ω and Ω, respectively.
Solution
(1) Series combination RS =
4Ω + 4Ω + 4Ω = 12 Ω (2) For this, two resistances are connected in parallel and
third resistance is connected in series with it as shown in fig. 
(3) Parallel
combination 
(3) Parallel
combination
Q 5. A piece of resistance wire has a resistance 16 ohm.Its diameter
is doubled.what will be the new" R"
Solution
Q 6. Give reason: Connecting wires in a circuit are made of copper
and aluminum.
Solution
For conduction of
electricity, good conducting material is required. Copper and aluminum are good
conductors as they provide conducting path for motion of electron through the
circuit. They have free electrons which can be accelerated by a potential
difference. Therefore, connecting wires in a circuit are made of copper and
aluminum.
Q 7. Distinguish between a closed circuit and an open circuit, with
the use of suitable labelled diagrams.
Solution
A circuit is said to be
closed when every part of it is made of a conductor and on plugging in the key
or on being complete, current flows through the circuit. A circuit is said to
be open when no current flows through it. It can happen when the key is not
plugged in or when any one of its components is not made of a conductor or when
the circuit is broken.
Q 8. 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.
Solution
(1) Wires P and Q
have the same resistivity. As they are made of the same metal, (2) Wires
P and Q have different resistances. Resistance, R = ϱ (l/A)
where l is length and A is the area of cross section of a conductor. Here
ϱ and l are same for P and Q, but A is not hence the wires have
different resistances.
Q 9. What can you say in general about the equivalent resistance
when the following are connected in parallel?1)1 ohm and 105 ohm
2)1 ohm ,102ohm, 103 ohm
Solution
When resistors are
connected in parallel, the equivalent resistance would be less than the
smallest resistance. Using this property, the equivalent resistance in both the
circuits would be less than 1 ohm.
Q 10. Explain why the p.d. across the terminals of a cell is more in
an open circuit and reduced in a closed circuit.
Solution
When the electric cell is
in a closed circuit the current flows through the circuit. There is a fall of
potential across the internal resistance of the cell. So, the p.d. across the terminals
in a closed circuit is less than the p.d. across the terminals in an open
circuit by an amount equal to the potential drop across the internal resistance
of the cell.
Q 11. What happens to equivalent resistance and current in a
parallel circuit when more and more resistances are added?
Solution
Increasing the number of
resistors in a parallel circuit decreases the total resistance of the circuit.
And with decrease in total resistance, the total current should increase as
current and resistance are inversely proportional for a constant battery
voltage.
Q 12. In cold places it is difficult to start the car. Why?
Solution
It is difficult to start
the car in a cold place as the engine oil becomes viscous and needs higher
amount of current to warm up the oil.
Q 13. Explain the analogy between the flow of charge (or
current) in a conductor under a potential difference with the free fall of a
body under gravity.
Solution
If a body is free to
fall, on releasing it from a height, it falls downwards towards the earth's
surface. For, this one end has to be at higher level and other at lower level,
so that gravity could effect on this difference and body could freely fall. Same
way to make flow of the charge through a conductor, the gravity of course has
no role of play; there should be difference of electric potential. This
difference gives the flow of charge in a conductor.
Q 14. What is meant by potential difference between two points ?
Solution
Potential difference
between two points in an electric field is the amount of work done to move a
unit charge from one point to other
Q 15. What happens to the resistance of a conductor when its area of
cross-section is increased?
Solution
On increasing the area of
cross-section, resistance decreases. This is because resistance is inversely
proportional to area.
Q 16. Find out the quantity
of heat produced in a resistance 10Ω; when 2A current is supplied to it for 8
minutes.
Solution
The heat produced , H H =
I2Rt H = 2 x 2 x 10 x 8 x 60 H = 19200 Joules.
Q 17. Draw a V – I graph for a conductor at two different
temperatures. What conclusion do you draw from your graph for the variation of
resistance of conductor with temperature?
Solution
In the given graph T1 > T2.
The straight line A is steeper than the line B which leads us to conclude that
the resistance of conductor is more at high temperature T1 than
at low temperature T2. Thus, we can say that resistance of a
conductor increases with the increase in temperature.
Q 18. Differentiate between: Resistance and resistivity.
Solution
Resistance
Resistivity
The property of the conductor due to which it opposes flow of current is called
resistance. The resistivity of a conductor is the resistance of a conductor of
unit length and unit area of cross section. The resistance of a conductor
depends on its length and area of cross section. The resistivity of conductor
does not depend on its length and area of cross section. R = V/I 
where ϱ is the
resistivity and is a constant. Its SI unit is ohm. Its SI unit is ohm-metre.
where ϱ is the
resistivity and is a constant. Its SI unit is ohm. Its SI unit is ohm-metre.
Q 19. Explain electrical energy and derive its formula.
Solution
The energy consumed by an
electric circuit to flow current through it is referred as electrical energy.
This energy is provided by the battery to every electric charge. The work
required to keep the charge Q in motion by the battery of voltage V is W = VQ
From the definition of electric current, Q = It Therefore, W = VIt According to
Ohm’s law, V = I R Therefore, W = (IR) I t Or, W = I2Rt Thus, the
current flowing through the resistor R for time t is I, the electrical energy
consumed to is W which is converted into heat energy.
Q 20. Determine the p.d. (voltage), which must be applied to a 2kΩ
resistor in order that a current of 10mA may flow.
Solution
Resistance R = 2 kΩ =
2 × 103 Ω = 2000 Ω Current I = 10mA = 10 × 10-3 A
= 0.01 A From Ohm’s law, potential difference is V = IR = (0.01) (2000) = 20 V
Q 21. (i) Define the term current and state its S.I. unit.
(ii) A current of 1.6 mA flows through a conductor. If charge of an
electron is -1.6 x 10-19 coulomb, find the number of electrons
that will pass each second through the cross section of that conductor.
Solution
(i) Current is
defined as the rate of flow of charge. It is given as I=Q/t. Its
S.I. unit is Ampere. (ii) Current , I = 1.6 mA = 1.6 x 10-3 A Charge, Q =
-1.6 x 10-19 coulomb t=1 sec I = Q/t Q = I × t Q = 1.6 x 10-3 x
1 = 1.6 x 10-3 C Q = ne →n = Q/e No. of electrons = 1.6 x
10-3/1.6 x 10-19
=1016
Q 22. Name the material used for making the connection wires. Give a
reason for your answer. Why should a connection wire be thick?
Solution
'Copper or Aluminium’ is
used as a material for making connection wires because the resistivity of these
materials is very small and thus wires made of these materials possess
negligible resistance. The connection wires are made thick so that their
resistance can be considered as negligible. Resistance of conductor pf length
'l' and area of cross section 'A' is Therefore, greater the area
of cross-section, lesser shall be the resistance.
Therefore, greater the area
of cross-section, lesser shall be the resistance.
Q 23. Which type of circuit (series/parallel) would you use to
decorate the Christmas tree and for light fittings in house. Write reasons for
your choice also.
Solution
For decorating Christmas
tree it would be better to connect bulbs in series combination because If the
bulbs are connected in series then all the bulbs can be controlled by a single
switch. It is safer because the amount of current in it is smaller.
Joining the bulbs in series has one disadvantage that, if any one bulb gets
fused due to some reason, then the whole circuit would break and all bulbs
would be turned off. For light connection in house it would be better to
connect bulbs in parallel combination because There is separate switch for each
light or appliance. Thus, lights or appliances could be individually turned ON
or OFF. Each light or appliance would get the same voltage as that of power
supply. Due to this, all lights would glow as brightly as compared to when
connected in series. The overall resistance of the household circuit is
reduced, due to which the current from the power supply is high. Therefore,
every appliance can draw the required amount of current for its working.
Q 24. An electric heater consumes 4.4 kW power when connected with a
220 V line voltage then, (i) Calculate the current passing through the heater.
(ii) Calculate resistance of a heater. (ii) Calculate the energy consumed in 2
hours.
Solution
Q 25. Which material is used to make heating coils in electric home
appliances and why?
Solution
Heating coils in electric
home appliances are made of nichrome, an alloy of nickel, chromium, manganese
and iron. The reasons for using nichrome in a heating coil are
i) Nichrome has a high melting point.
ii) The nichrome coil can remain in red-hot condition for a
long time. iii) It has high resistance.
Q 26. An electric geyser works in a 230V supply which draws a
current of 6A for 5 minutes. Find out the quantity of heat produced in the
heater.
Solution
V= 230V I = 6A t = 5
minute = 5 x 60 seconds = 300 seconds H = I²Rt = VIt
= 230 x 6 A x 300 s = 414000J
Q 27. The resistance of a wire of uniform diameter "d" and
length "l"is R. What will be the resistance of same material
but of diameter "2d" and length 4l ?
Solution
Q 28. A nine-volt battery supplies power to a cordless curling iron
with a resistance of 18 ohms. How much current is flowing through the curling
iron?
Solution
From ohm’s law, we have
V = IR
I = V/R Given: R = 18 Ω; V = 9V
I = 9/18
I = 0.5 A
Q 29. If two resistors are connected in series, the total resistance
is 45 Ω and if the same resistors are connected in parallel the total
resistance becomes 10 Ω. Find the value of individual resistors.
Solution
Given: RS =
45 Ω, RP = 10 Ω, R1 =? , R2=? In
series combination, →
RS =
R1+ R2 =45 Ω
-------(1)
Parallel combination: 
-----
(2) Combining equations (1) and (2) and without considering the unit of
resistance, we get 
→ R1 = 30 Ω and R2 = 15 Ω or
R1 = 15 Ω and R2 = 30 Ω
-----
(2) Combining equations (1) and (2) and without considering the unit of
resistance, we get
→ R1 = 30 Ω and R2 = 15 Ω or
R1 = 15 Ω and R2 = 30 Ω
Q 30. Differentiate between: Resistances in series and parallel.
Solution
Resistances
in series Resistances in parallel 1. In a series combination of resistances,
the resistances are connected one after another so that the current
through each of them is the same. In a parallel combination of resistances, the
resistances are connected between two common points so that the potential
difference across each of them is the same. 2. When n resistors of resistances,
R1, R2, ---- Rn, are connected in series, the effective resistance
Rs of the combination is given by Rs = R1+ R2+
---- Rn. When n resistors of resistances, R1, R2,---- Rn, are
connected in parallel, the effective resistance RP of the combination is
given by 
3. The effective resistance
of the combination is greater than any of the resistances in the combination. The
effective resistance of the combination is less than any of the resistances in
the combination. 4. This combination is used to increase the effective
resistance. This combination is used to decrease the effective resistance. 5.
This combination decreases the current in the circuit. This combination
increases the total current in the circuit.
3. The effective resistance
of the combination is greater than any of the resistances in the combination. The
effective resistance of the combination is less than any of the resistances in
the combination. 4. This combination is used to increase the effective
resistance. This combination is used to decrease the effective resistance. 5.
This combination decreases the current in the circuit. This combination
increases the total current in the circuit.
Q 31. Name the physical quantity which is same and different, in all
the bulbs when three bulbs of: (a) same wattage are connected in series (b)
same wattage are connected in parallel (c) different wattage are connected in
series (d) different wattage are connected in parallel
Solution
(a) Same wattage are
connected in series: (i) Current and voltage are same
(ii) Nothing is different (b) Same wattage are connected in parallel:
(i) Current and voltage are same (ii) Nothing is
different (c) Different wattage are connected in series: (i)
Current is same (ii) Voltage is different (d) Different wattage
are connected in parallel: (i) Voltage is same (ii)
Current is different
Q 32. Why do the wires connecting an electric heater to the mains
not glow while its heating element does?
Solution
The wires of the
connecting cord of electric heater are made of copper. Copper has extremely low
resistance, due to which there is negligible heat produced. The heating element
of an electric heater is made of nichrome wire. It glows because large amount of
heat is produced due to its high resistance.
Q 33. What is the commercial unit of energy?
Solution
Kilowatt hour or kWh
Q 34. Are there any applications of heating effects of current?
Solution
Yes there are
applications of heating effect of current. For example: electric iron, heater,
electric oven and kettle. In a device the heat released due to electric energy
is used to heat either the coils or plates and then used for practical
purposes.
Q 35. Name the instrument used for measuring: (i) Potential difference
(ii) Current
Solution
(i) Volmeter (ii) Ammeter
Q 36. How is the current flowing in a conductor changed if the
resistance of conductor is doubled keeping the potential difference across it
the same?
Solution
V = IR or, I =
V/R….(i) If R is doubled, Then, I' = V/2R = I/2……………(ii) From (i) and (ii), it
is clear that current will be halved.
Q 37. A coil has a current of 50 mA flowing through it when the
voltage is 12 V. What is the resistance of the coil?
Solution
Given: Current I = 50 mA
= 0.05 A; Voltage, V = 12 V We know that according to ohm’s law
,V=IR R = V/I R = 12V / 0.05 A R = 240 Ω Hence the resistance of the
coil = 240 Ω.
Q 38. Define 1 kWh. How is this unit of energy related to 1 joule ?
Solution
1 kWh is the energy
consumed when one kilowatt power is used for 1 hour 1 kWh = 3.6 X 106 joule
Q 39. Are the headlights of a car connected in series or parallel?
Why?
Solution
Headlights of a car are
connected in parallel. In parallel connection, each headlight is exposed to the
full potential difference supplied by the car's electrical system, giving
maximum brightness. Another advantage is that if one headlight burns out, the
other one keeps shining.
Q 40. Define: (i) 1 volt PD (ii) 1 ampere
Solution
(i)
The potential difference between two points is said to be 1 volt
if 1 joule of work is done in moving 1 coulomb of electric charge from one
point to another.
(ii)
If one coulomb of charge is passing through any cross section of a
conductor in one second, the amount of current flowing through it is called one
ampere. 1 A = 1 C /
s
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