Current Electricity ICSE Class-10 Concise Physics Selina Solutions

Current Electricity ICSE Class-10 Concise Physics Selina Solutions Chapter-8 . We Provide Step by Step Answer of Exercise-8(A), MCQs-8(A), Numericals-8(A), Exercise-8(B), MCQ-8(B), Numericals-8(B) Exercise-8(C), MCQ-8(C) and Numericals -8(C)  Questions of Exercise-8 Current Electricity ICSE Class-10. Visit official Website CISCE  for detail information about ICSE Board Class-10.

Current Electricity ICSE Class-10 Concise Physics Selina Solutions Chapter-8


-: Select Exercise :-

Exe – 8(A), 

 MCQ – 8 (A), 

  Num – 8(A)

Exe – 8 (B)

MCQ – 8 (B) , 

Num – 8 (B)

Exe – 8 (C) ,  

MCQ – 8 (C) , 

Num – 8 (C)


How to Solve Numericals of Current Electricity ICSE Class-10

Current Electricity deals with ohm’s law, concepts of E.M.F., Potential Difference, resistance,resistances in series and parallel; internal resistance also deals in Current Electricity. Concepts of P.D. Current Electricity in Physics of Chapter 8 and resistance and charge. Simple Numerical Problems using the above relations in Current Electricity.


“Current Electricity” Selina Solutions ICSE Physics , Exercise – 8 (A)

Question 1

Define the term current and state its S.I. unit.

Answer 1

Current is defined as the rate of flow of charge.

I=Q/t

Its S.I. unit is Ampere.

Question 2

Define the term electric potential. State it’s S.I. unit.

Answer 2

Electric potential at a point is defined as the amount of work done in bringing a unit positive charge from infinity to that point. Its unit is the volt.

Question 3

How is the electric potential difference between the two points defined? State its S.I. unit.

Answer 3

The potential difference between two points is equal to the work done in moving a unit positive charge from one point to the other.

It’s S.I. unit is Volt.

Question 4

Explain the statement ‘the potential difference between two points is 1 volt’.

Answer 4

One volt is the potential difference between two points in an electric circuit when 1 joule of work is done to move charge of 1 coulomb from one point to other.

Question 5

(a) State whether the current is a scalar or vector? What does the direction of current convey?

(b) State whether the potential is a scalar or vector? What does the positive and negative sign of potential convey?

Answer 5

(a) Current is a scalar quantity. The direction of current conveys that the flow of electrons is opposite to the direction of flow of current.

(b) Potential is also a scalar quantity. The positive sign of potential conveys that work has to be done on the positive test charge against the repulsive force due to the positive charge in bringing it from infinity. The negative sign of potential conveys that work is done on the negative test charge by the attractive force.

Question 6

Define the term resistance. State it’s S.I. unit.

Answer 6

It is the property of a conductor to resist the flow of charges through it. It’s S.I. unit is Ohm.

Question 7

(a) Name the particles which are responsible for the flow of current in a metallic wire.

(b) Explain the flow of current in a metallic wire on the basis of movement of the particles named by you above in part

(c) What is the cause of resistance offered by the metallic wire in the flow of current through it?

Answer 7

(a) Particles responsible for the flow of current in a metallic wire are free electrons.

(b) In metals, free electrons are the moving charges that result in the conduction of electricity. If ‘n’ electrons pass through the metallic conductor in time ‘t’, then the total charge that has flown is given by Q (charge) = n x e (charge on an electron).

(c) A metal includes free electrons and fixed positive ions.

Positive ions give away their valence electrons and thus attain a positive charge.

Electrons are free for movement, but positive ions do not move; thus, when a potential difference is applied across the circuit, and when free electrons begin to move, they collide with these fixed ions.

This collision is the major cause of resistance offered by the metallic wire in the flow of current through it.

Question 8

State Ohm’s law and draw a neat labelled circuit diagram containing a battery, a key, a voltmeter, an ammeter, a rheostat and an unknown resistance to verify it.

Answer 8

It states that electric current flowing through a metallic wire is directly proportional to the potential difference V across its ends provided its temperature remains the same. This is called Ohm’s law.

V = IR

State Ohm's law and draw a neat labelled circuit diagram containing a battery

Question 9

(a) Name and state the law which relates the potential difference and current in a conductor.

(b) What is the necessary condition for a conductor to obey the law named above in part (a)?

Answer 9

(a)The law is called Ohm’s law. It states that the current flowing through the conductor is directly proportional to the potential difference across the ends of a conductor given the temperature remains constant.

(b) Temperature should remain constant.

Question 10

(a) Draw a V-I graph for a conductor obeying Ohm’s law.

(b) What does the slope of V-I graph for a conductor represent?

Answer 10

(a)

current 1

(b) Slope of VI graph represents the resistance.

Question 11

Draw an I – V graph for a linear resistor. What does its slope represent?

Answer 11

current 2

The slope of I-V graph (= ) is equal to the reciprocal of the resistance of the conductor, i.e.

Question 12

What is an ohmic resistor? Give one example of an ohmic resistor. Draw a graph to show its current – voltage relationship. How is the resistance of the resistor determined from this graph?

Answer 12

Ohmic Resistor: An ohmic resistor is a resistor that obeys Ohm’s law. For example: all metallic conductors (such as silver, aluminium, copper, iron etc.)

From above graph resistance is determined in the form of slope.

Question 13

What are non-ohmic resistors? Give one example and draw a graph to show its current-voltage relationship.

Answer 13

The conductors which do not obey Ohm’s Law are called non-ohmic resistors. Example: diode valve.

Question 14

Give two differences between an ohmic and non-ohmic resistor.

Answer 14

(1) Ohmic resistor obeys ohm’s law i.e., V/I is constant for all values of V or I; whereas Non-ohmic resistor does not obey ohm’s law i.e., V/I is not same for all values of V or I.

(2) In Ohmic resistor, V-I graph is linear in nature whereas in non-ohmic resistor, V-I graph is non-linear in nature.

Question 15 

Fig. 8.10 below shows the I-V curves for two resistors. Identify the ohmic and non-ohmic resistors. Give a reason for your answer.

Answer 15

Ohmic : (b), Non-Ohmic : (a)

Only for (a) the I-V graph is a straight line or linear while for (a), the graph is a curve.

Question 16

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?

Answer 16

In the above graph, TT2. 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 Tthan at low temperature T2. Thus, we can say that resistance of a conductor increases with the increase in temperature.

Question 17

(a) How does the resistance of a wire depend on its radius? Explain your answer.

(b) Two copper wires are of same length, but one is thicker than the other. Which will have more resistance?

Answer 17

(a) Resistance of a wire is inversely proportional to the area of cross-section of the wire.

This means if a wire of same length, but of double radius is taken, its resistance is found to be one-fourth.

Answer 17

(b) Resistance is directly proportional to the length and inversely proportional to the area of cross-section. The thicker wire has more area, and hence the resistance of the other wire will be more than that of the thicker wire.

Question 18

How does the resistance of a wire depend on its length? Give a reason of your answer.

Answer 18

Resistance of a wire is directly proportional to the length of the wire.

The resistance of a conductor depends on the number of collisions which the electrons suffer with the fixed positive ions while moving from one end to the other end of the conductor. Obviously the number of collisions will be more in a longer conductor as compared to a shorter conductor. Therefore, a longer conductor offers more resistance.

Question 19

How does the resistance of a metallic wire depend on its temperature? Explain with reason.

Answer 19

With the increase in temperature of conductor, both the random motion of electrons and the amplitude of vibration of fixed positive ions increase. As a result, the number of collisions increases. Hence, the resistance of a conductor increases with the increase in its temperature.

The resistance of filament of a bulb is more when it is glowing (i.e., when it is at a high temperature) as compared to when it is not glowing (i.e., when it is cold).

Question 20

Two wires, one of copper and other of iron, are of the same length and same radius. Which will have more resistance? Give reason.

Answer 20

Iron wire will have more resistance than copper wire of the same length and same radius because resistivity of iron is more than that of copper.

Question 21

Name three factors on which the resistance of a wire depends and state how it is affected by the factors stated by you?

Answer 21

(i) Resistance of a wire is directly proportional to the length of the wire means with the increase in length resistance also increases.

(ii) Resistance of a wire is inversely proportional to the area of cross-section of the wire. If area of cross-section of the wire is more, then resistance will be less and vice versa.

(iii) Resistance increases with the increase in temperature since with increase in temperature the number of collisions increases.

(iv) Resistance depends on the nature of conductor because different substances have different concentration of free electrons.Substances such as silver, copper etc. offer less resistance and are called good conductors; but substances such as rubber, glass etc. offer very high resistance and are called insulators.

Question 22

Define the term specific resistance and state its S.I. unit.

Answer 22

The resistivity of a material is the resistance of a wire of that material of unit length and unit area of cross-section. Its S.I. unit is ohm metre.

Question 23

Write an expression connecting the resistance of a wire and specific resistance of its material. State the meaning of symbols used.

Answer 23

Expression :

– resistivity

R – resistance

l – length of conductor

A – area of cross- section

Question 24

State the order of specific resistance of (i) a metal, (ii) a semiconductor and (iii) an insulator.

Answer 24

Metal < Semiconductor < Insulator

Question 25

(a) Name two factors on which the specific resistance of a wire depends?

(b) Two wires A and B are made of copper. The wire A is long and thin while the wire B is Short and thick. Which will have more specific resistance ?

Answer 25

(a) The specific resistance of a wire depends on the material of the substance and the temperature of the substance.

(b) Specific resistance depends on the material of the wire and not its dimensions. Hence, both the wires will have the same specific resistance .

Question 26

Name a substance of which the specific resistance remains almost unchanged by the increase in temperature.

Answer 26

Manganin

Question 27

How does specific resistance of a semi-conductor change with the increase in temperature?

Answer 27

Specific resistance of a semiconductor decreases with increase in temperature.

Question 28

How does (a) resistance, and  (b) specific resistance of a wire depend on its (i) length, and (ii) radius?

Answer 28

Resistance is directly proportional to the length and inversely proportional to the square of radius. Specific resistance is independent on the dimensions of a wire.

Question 29

(a) Name the material used for making the connection wires. Give reason for your answer.
(b) Why should a connection wire be thick?

Answer 29

(a) The materials used for making connection wires are copper or aluminium. These materials are chosen because they have small specific resistance.

(b) A connection wire should be thick so that it offers negligible resistance to the flow of current through the circuit.

Question 30

Name a material which is used for making the standard resistor. Give a reason for your answer.

Answer 30

Manganin is used for making the standard resistor because its resistivity is quite large and the effect of change in temperature on their resistance is negligible.

Question 31

Name the material used for making a fuse wire. Give a reason.

Answer 31

Generally fuse wire is made of an alloy of lead and tin because its resistivity is high and melting point is low.

Question 32

Name the material used for:

Filament of an electric bulb

Heating element of a room heater

Answer 32

(i) A wire made of tungsten is used for filament of electric bulb because it has a high melting point and high resistivity.

(ii) A nichrome wire is used as a heating element for a room heater because the resistivity of nichrome is high and increase in its value with increase in temperature is high.

Question 33

What is a superconductor? Give one example of it.

Answer 33

A superconductor is a substance of zero resistance at a very low temperature. Example: Mercury at 4.2 K.

Question 34

A substance has zero resistance below 1 K. What is such a substance called?

Answer 34

Superconductor


Chapter 8 (A) MULTIPLE CHOICE TYPE Current Electricity Selina Physics Solutions 

Question 1

Which of the following is an ohmic resistance ?

(a) LED

(b) junction diode

(c) filament of a bulb

(d) nichrome wire

Answer 1

(d) nichrome wire

Question 2

For which of the following substances, resistance decreases with the increase in the temperature?

(a) copper

(b) mercury

(c) carbon

(d) platinum

Answer 2

(c) carbon


 Numericals – 8 (A) Current Electricity” Selina Physics Solutions , 

Question 1

In a conductor 6.25 ×  electrons flow from its end A to B in 2 s. Find the current flowing

through the conductor (e = 1.6 ×  C)

Answer 1

Number of electrons flowing through the conductor,

N = 6.25 × 1016 electrons
Time taken, t = 2 s
Given, e = 1.6 × 10-19 c
Let I be the current flowing through the conductor.

Then, I = net

∴ I = (6.25×1016)(1.6×10-19)2 =5×10-jA

Or, I = 5 mA
Thus, 5 Ma current flows from B to A.

Question 2

A current of 1.6 mA flows through a conductor. If charge on an electron is – 1.6 × coulomb,
find the number of electrons that will pass each second through the cross section of that
conductor.

Answer 2

Current , I =1.6 mA = 1.6 ×  A

Charge, Q = – 1.6 × coulomb

t = 1 sec
I = Q/t
Q = I x t
Q = 1.6 × × 1
No. of electrons = 1.6 × /1.6 ×
=

Answer 3

Current (I) = 0.2 A

Resistance (R) = 20 ohm
Potential Difference (V) = ?
According to Ohm’s Law :
V = IR
V = 0.2 × 20 = 4 V


Excercise 8 (B) ICSE Class-10 Selina Solutions Physics Current Electricity

Question 1

Explain the meaning of the terms e.m.f., terminal voltage and internal resistance of cell.

Answer 1

e.m.f.: When no current is drawn from a cell, the potential difference between the terminals of the cell is called its electro-motive force (or e.m.f.).

Terminal voltage: When current is drawn from a cell, the potential difference between the electrodes of the cell is called its terminal voltage.

Internal Resistance: The resistance offered by the electrolyte inside the cell to the flow of electric current through it is called the internal resistance of the cell.

Question 2

State two differences between the e.m.f. and terminal voltage of a cell.

Answer 2

e.m.f. of cell Terminal voltage of cell
     1.It is measured by the amount of  work done in moving a unit positive charge in the complete circuit inside and outside the cell.      1. It is measured by the amount of work done in moving a unit positive charge in the circuit outside the cell.
     2.It is the characteristic of the cell i.e., it does not depend on the amount of current drawn from the cell 2. It depends on the amount of current drawn from the cell. More the current is drawn from the cell, less is the terminal voltage.
     3.It is equal to the terminal voltage when cell is not in use, while greater than the terminal voltage when cell is in use. 3. It is equal to the emf of cell when cell is not in use, while less than the emf when cell is in use.

 

Question 3

Name two factors on which the internal resistance of a cell depends and state how does it depend on the factors stated by you.

Answer 3

Internal resistance of a cell depends upon the following factors:

(i) The surface area of the electrodes: Larger the surface area of the electrodes, less is the internal resistance.

(ii) The distance between the electrodes: More the distance between the electrodes, greater is the internal resistance.

Question 4

A cell of emf and internal resistance r is used to send current to an external resistance R. Write expressions for (a) the total resistance of the circuit (b) the current drawn from the cell and (c) the p.d. across the cell (d) voltage drop inside the cell.

Answer 4

(a) Total resistance = R + r

(b) Current drawn from the circuit :

As we know that,

=V+v

= IR + Ir

=I(R+r)

I = e / (R + r)

(c) p.d. across the cell :

(d) voltage drop inside the cell:

Question 5

A cell is used to send current to an external circuit.

(a) How does the voltage across its terminal compare with its emf?

(b) Under what condition is the emf of a cell equal to its terminal voltage?

Answer 5

(a) Terminal voltage is less than the emf : Terminal Voltage < e.m.f.

(b) e.m.f. is equal to the terminal voltage when no current is drawn.

Question 6

Explain why the p.d. across the terminals of a cell is more in an open circuit and reduced in a closed circuit.

Answer 6

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.

Question 7

Write the expressions for the equivalent resistance R of three resistors R1, R2 and R3 joined in (a) parallel, (b) series.

Answer 7

(a) Total Resistance in series:

(b) Total Resistance in parallel:

Question 8

How would you connect two resistors in series? Draw a diagram. Calculate the total equivalent resistance.

Answer 8

If current I is drawn from the battery, the current through eac resistor will also be I.

On applying Ohm’s law to the two resistors separately, we further

have

V1 = I R1

V2 = I R2

V=V1 + V2

IR = I R1+ I R2

R= R1+ R2

Total Resistance in series R :

Question 9

Show by a diagram how two resistors R1 and R2 are joined in parallel. Obtain an expression for the total resistance of the combination.

Answer 9

 how two resistors R1 and R2 are joined in parallel. Obtain an expression for the total resistance of the combination

On applying Ohm’s law to the two resistors separately, we further

Have

I1 = V / R1

I2 = V / R2

I = I1 + I2

Question 10

State how are the two resistors are joined with a battery in each of the following cases when:

(a) same current flows in each resistor

(b) potential difference is same across each resistor.

(c) equivalent resistance is less than either of the two resistances.

(d) equivalent resistance is more than either of the two resistances.

Answer 10

(a) series

(b) parallel

(c) parallel

(d) series

Question 11

The V-I graph for a series combination and for a parallel combination of two resistors is shown in fig. Which of the two, A or B, represents the parallel combination? Give a reason for your answer.

Answer 11

For the same change in I, change in V is less for the straight line A than for the straight line B (i.e., the straight line A is less steeper than B), so the straight line A represents small resistance, while the straight line B represents more resistance. In parallel combination, the resistance decreases while in series combination, the resistance increases. So A represents the parallel combination.


MULTIPLE CHOICE TYPE – 8 (B) Selina Solutions Physics Current Electricity

Question 1

In series combination of resistances:
(a) p.d is same across each resistance
(b) total resistance is reduced
(c) current is same in each resistance
(d) all above are true

Answer 1

(c) In series combination of resistances, current is same in each resistance.

Question 2

In parallel combination of resistances:
(a) p.d is same across each resistance
(b) total resistance is increased
(c) current is same in each resistance
(d) all above are true

Answer 2 

(a) In parallel combination of resistances, P.D. is same across each resistance. 

Question 3

Which of the following combinations have the same equivalent resistance between X and Y?

Image result for Which of the following combinations have the same equivalent resistance between X and Y?

Answer 3 

(a) and (d)


NUMERICALS – 8 (B) Current Electricity Concise Physics Solution

Question 1

The diagram in Fig. 8.42 shows a cell of e.m.f. ε = 2 volt and internal resistance r = 1 ohm to an external resistance R = 4 ohm. The ammeter A measures the current in the circuit and the
voltmeter V measures the terminal voltage across the cell. What will be the readings of the ammeter and voltmeter when (i) the key K is open, (ii) the key K is closed.

Image result for The diagram below in Fig. 8.40 shows a cell of e.m.f. ε = 2 volt and internal resistance r = 1 ohm to an external resistance R = 4 ohm. The ammeter A measures the current in the circuit and the voltmeter V measures the terminal voltage across the cell. What will be the readings of the ammeter and voltmeter when (i) the key K is open, (ii) the key K is closed. 

Answer 1

(i) Ammeter reading = 0 because of no current
Voltage V = ϵ − Ir
V = 2 − 0 × 1 = 2 volt

(ii) Ammeter reading :
I =ε /(R + r)
I=2 / (4+1) = 2 / 5 = 0.4 amp
Voltage reading :
Voltage V = ϵ – Ir
V=2 – 0.4 x 1 = 2 – 0.4 = 1.6 V

Question 2

A battery of e.m.f 3.0 V supplies current through a circuit in which the resistance can be changed.
A high resistance voltmeter is connected across the battery. When the current is 1.5 A, the voltmeter reads 2.7 V. Find the internal resistance of the battery.

Answer 2

ε = 3 volt
I = 1.5 A
V = 2.7 V
V = ε − Ir
r = (e-V) / I
= (3 – 2.7) / 1.5 = 0.2 ohm

Question 3

A cell of e.m.f. and internal resistance 3 ohm is connected to two resistors of resistances 3 ohm and 6 ohm in series. Find:

(a) the current through the battery,

(b) the p.d. between the terminals of the battery.

Answer 3

Electricity numerical 8 (b) 3

Question 4

A battery of emf 15 V and internal resistance 3 ohm is connected to two resistors 3 ohm and 6 ohm connected in parallel. Find (a) the current through the battery (b) p.d. between the terminals of the battery.

(a)In parallel 1/R = 1/3 + 1/6 = 1/2

So R = 2 ohm

r = 3 W

=15 V

=I(R+r)

15=I(2+3)

I=15/5 = 1.25

(b)V=?

R=2 ohm

V=IR = 6 x 2 = 12 V (11.25 V)

Question 5

A Cell of e.m.f. ε and internal resistance ? sends current 1.0 A when it is connected to an external resistance 1.9Ω. But it sends current 0.5 A when it is connected to an external resistance 3.9 Ω.Calculate the values of ε and ?.

Answer 5

In first case
I = 1 A, R = 1.9 ohm
ε = I(R + r) = 1(1.9+r)
ε = 1.9 + r————(1)
In second case
I = 0.5 A, R = 3.9 ohm
ε = I(R + r) = 0.5 (3.9 + r)
ε = 1.95 + 0.5r —————-(2)
From eq. (1) and (2),
1.9 + r = 1.95 + 0.5r
r = 0.05/0.5 = 0.1 ohm
Substituting value of r
ε = 1.9 + r = 1.9 + 0.1 = 2 V

Question 6

Two resistors having resistance 4? and 6? are connected in parallel. Find their equivalent resistance.

Question 14

A uniform wire with a resistance of 27 ohm is divided into three equal pieces and then they are joined in parallel. Find the equivalent resistance of the parallel combination.

Answer 14

Wire cut into three pieces means new resistance = 27/3 = 9

Now three resistance connected in parallel :

Question 15

A circuit consists of a 1 ohm resistor in series with a parallel arrangement of 6 ohm and 3 ohm resistors. Calculate the total resistance of the circuit. Draw a diagram.

Answer 15

Question 16

Calculate the effective resistance between the points A and B in the network shown below in figure.

Answer 16

Question 17

Calculate the equivalent resistance between the points A and B in the adjacent diagram in the figure.

Answer 17

R1=3+2 = 5 ohm

R2=30 W

R3=6+4 = 10 ohm

R1 , R2 and R3 are connected in parallel

Question 18

In the network shown in adjacent figure 8.49, calculate the equivalent resistance between the points (a) A and B, and (b) A and C

Answer 18

(a) R1=2+2+2 = 6ohm

R2=2ohm

R1 and R2 are connected in parallel

(b) R1=2+2 = 4 ohm

R2=2+2=4 W

R1 and R2 are connected in parallel

Question 19

Five resistors, each 3 ohm, are connected as shown in figure 8.50. Calculate the resistance (a) between points P and Q (b) between points X and Y.

Answer 19

(a) R1=3+3=6 W

R2=3 W

R1 and R2 are connected in parallel

(b) As calculated above R=2 ohm

R3 = 3 ohm

R4 = 3 ohm

R’=R+R3+R4=2+3+3 = 8 ohm

Question 20

Two resistors of 2 ohm and 3 ohm are connected (a) in series, (b) in parallel, with a battery of 6.0 V and negligible internal resistance. For each case draw a circuit diagram and calculate the current through battery.

Answer 20

(a)

R1=2 ohm

R2=3 ohm

R = R1 + R2 = 2 + 3 = 5 ohm

V=6 V

(b)

R1 and R2 are connected in parallel

Question 21

A resistor of 6 ohm is connected in series with another resistor of 4 ohm. A potential difference of 20 V is applied across the combination. (a) Calculate the current in the circuit and (b) potential difference across the 6 ohm resistor.

Answer 21

(a) R1=6 ohm

   R2=4 ohm

   R = R1 + R2 = 6 + 4 = 10 ohm

   V=20 V

   I=V/R = 20/10 = 2 A

(b) R = 6 W

I = 2 A

V = ?

V = IR = 6 x 2 = 12 V

Question 22

Two resistors of resistance 4 Ω and 6 Ω are connected in parallel to a cell to draw 0.5 A current from the cell.

(a) Draw a labeled diagram of the arrangement

(b) Calculate current in each resistor.

Answer 22

(a) Circuit diagram

(b) Equivalent resistance of the circuit:

Hence, the emf of the cell is

Therefore, current through each resistor is

Question 23

Calculate current flowing through each of the resistors A and B in the circuit shown in figure 8.51?

Answer 23

For resistor A :

R=1 ohm

V=2 V

I=V/R = 2/1 = 2 A

For resistor B :

R = 2 ohm

V = 2 V

I=V/R = 2/2 = 1 A

Question 24

In figure 8.52, calculate :

(a) the total resistance of the circuit.

(b) the value of R, and

(c) the current flowing in R

Answer 24

(a)

V=4 V

I=0.4 A

Total Resistance R’=?

R’ = V/I = 0.4/4 = 10 ohm

(b)

R1= 20 ohm

R’ = 10 ohm

begin mathsize 12px style fraction numerator 1 over denominator straight R apostrophe end fraction equals 1 over straight R plus 1 over straight R subscript 1 1 over 10 equals 1 over straight R plus 1 over 20 1 over straight R equals 1 over 10 minus 1 over 20 equals 1 over 20 straight R equals 20 space straight capital omega end style

(c)

R=20 ohm

V=4 V

I=V/R = 4/20 = 0.2 A

Question 25

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.

Answer 25

(a) Resistance of 1m of wire = 3 ohm

Resistance of 1.5 m of wire = 3 x 1.5 = 4.5 W

(b) I=2 A

V=IR = 2 x 4.5 = 9 V

(c) R=3 ohm for 1 m

For 5 m : R=3 x 5 = 15 ohm

But Area A is double i.e. 2 A and Resistance is inversely proportional to area so Resistance will be half.

R=15/2 = 7.5 ohm

Question 26

A cell supplies a current of 1.2 A through two 2 ohm resistors connected in parallel. When resistors are connected in series, it supplies a current of 0.4 A. Calculate: (i) the internal resistance and (ii) e.m.f. of the cell.

Answer 26

In parallel R = ½ + ½ = 1 ohm

I = 1.2 A

=I(R+r) = 1.2(1+r)= 1.2 + 1.2 r

In series R = 2+2=4 ohm

I=0.4 A

=I(R+r) = 0.4(4+r) = 1.6 + 0.4 r

It means :

1.2 + 1.2 r = 1.6 + 0.4 r

0.8 r = 0.4

r = 0.4 / 0.8 = ½ = 0.5 ohm

(i) Internal resistance r = 0.5 ohm

(ii) =I(R+r) = 1.2(1+0.5) = 1.8 V

Question 27

A battery of emf 15 V and internal resistance 3 ohm is connected to two resistors 3 ohm and 6 ohm connected in parallel. Find (a) the current through the battery (b) p.d. between the terminals of the battery (c) the current in 3 ohm resistor (d) the current in 6 ohm resistor.

Answer 27

(a)In parallel 1/R = 1/3 + 1/6 = 1/2

So R = 2 ohm

r = 3 W

=15 V

=I(R+r)

15=I(2+3)

I=15/5 = 3 A

(b)V=?

R=2 ohm

V=IR = 3 x 2 = 6 V

(c)V=6 V

R = 3 ohm

I=V/R = 6/3 = 2 A

(d) R=6 ohm

V= 6 V

I=V/R = 6/6 = 1 A

Question 28

The circuit diagram in Fig. 8.53 shows three resistors  2 Ω, 4 Ω and R Ω connected to a battery of e.m.f. 2 V and internal resistance 3 ohm . If main current of 0.25 A flows through the circuit, Find :

(a)  the p.d. across the 4 ohm resistor?

(b)  the p.d. across the internal resistance of the cell,

(c) the p.d. across the R ohm or 2 ohm resistor, and

(d) the value of R.

Answer 28

(a)R = 4

I = 0.25 A

V=IR = 0.25 x 4 = 1 V

(b)Internal Resistance r=3 ohm

I = 0.25 A

V=IR = 0.25 x 3 = 0.75 V

(c) Effective resistance of parallel combination of two 2 ohm resistances = 1 ohm

V= I/R = 0.25/1 = 0.25 V

(d) I=0.25 A

=2 V, r = 3 ohm

=I(R’+r)

2=0.25(R’+3)

R’=5 W

Question 29

Three resistors of 6.0 ohm, 2.0 ohm and 4.0 ohm are joined to ammeter A and a cell of emf 6.0 V as shown in figure. Calculate :

(a) the effective resistance of the circuit.

(b) the reading of ammeter

Answer 29

(a) R1=6 W

R’=R2+R3=2+4=6 W

R1 and R’ in parallel :

(b) R= 3 ohm

V=6 V

I=?

I=V/R=6/3=2 A

Question 30

The diagram below in Fig. 8.55 shows the arrangement of five different resistances connected to a battery of e.m.f. 1.8 V. Calculate :

(a) The total resistance of the circuit, and

(b) The reading of ammeter A.

Answer 30

Question 31

A cell of e.m.f. 2 V and internal resistance 1.2 Ω is connected to an ammeter of resistance 0.8 Ω and two resistors of 4.5 Ω and 9 Ω as shown in fig.

Find:

(a) The reading of the ammeter,

(b) The potential difference across the terminals of the cells, and

(c) The potential difference across the 4.5 Ω resistor.

Answer 31

The total resistance of the circuit is

(a) Therefore, the current through the ammeter is

(b) The potential difference across the ends of the cells is

(c) The potential difference across the 4.5Ωresistor is


Chapter -8, “Current Electricity” Selina Physics Solution  Excercise – 8 (C)

Question 1

Write an expression for the electrical energy spent in flow of current through an electrical appliance in terms of current, resistance and time.

Answer 1

Question 2

Write an expression for the electrical power spent in flow of current through a conductor in terms of (a) resistance and potential difference, (b) current and resistance.

Answer 2

 

Electrical power P is given by the expression

(a)What do the symbols Q and V represent?

(b)Express the power P in terms of current and resistance explaining the meanings of symbols used there in.

Answer 3

 (a) Q represents Charge and V represents Voltage.

 (b) 

Question 4

Name the S.I. unit of electrical energy. How is it related to Wh ?

Answer 4

The S.I. unit of electrical energy is joule.

1 Wh = 3600 J

Question 5

Explain the meaning of the statement ‘the power of an appliance is 100 W’.

Answer 5

The power of an appliance is 100 W. It means that 100 J of electrical energy is consumed by the appliance in 1 second

Question 6

State the S.I. units of electrical power.

Answer 6

 The S.I. unit of electrical power is Watt.

Question 7

(i)State and define the household unit of electricity.

(ii)What is the voltage of the electricity that is generally supplied to a house?

(iii) What is consumed while using different electrical appliances, for which electricity bills are paid?

Answer 7

(i)The household unit of electricity is kilowatt-hour (kWh).

One kilowatt-hour (kWh) is the electrical energy consumed by an electrical appliance of power 1 kW when it is used for one hour.

(ii)The voltage of the electricity that is generally supplied to a house is 220 Volt.

(iii) electrical energy is consumed.

Question 8

Name the physical quantity which is measured in

(i) kW, (ii) kWh. (iii) Wh

Answer 8

(i)Electrical power is measured in kW and

(ii)Electrical energy is measured in kWh.

(iii)Energy consume is measured in Wh.

Question 9

Define the term kilowatt – hour and state its value in S.I. unit.

Answer 9

One kilowatt-hour (kWh) is the electrical energy consumed by an electrical appliance of power 1 kW when it is used for one hour.

Its value in SI unit is

Question 10

How do kilowatt and kilowatt-hour differ ?

Answer 10

Kilowatt is the unit of electrical power whereas kilowatt-hour is the unit of electrical energy.

Question 11

Complete the following:

(a)

(b)

1 kWh= ________ J

Answer 11

(a)

(b) 3.6 x 106 J

Question 12

What do you mean by power rating of an electrical appliance? How do you use it to calculate (a) the resistance of the appliance and (b) the safe limit of the current in it, while in use?

Answer 12

An electrical appliance such as electric bulb, geyser etc. is rated with power (P) and voltage (V) which is known as its power rating. For example: If an electric bulb is rated as 50W-220V, it means that when the bulb is lighted on a 220 V supply, it consumes 50 W electrical power.

(a) To calculate the resistance of the appliance, the expression is:

(b) The safe limit of current I is:

Question 13

An electric bulb is marked ‘100 W, 250 V’. What information does this convey?

Answer 13

It means that if the bulb is lighted on a 250 V supply, it consumes 100 W electrical power (which means 100 J of electrical energy is converted in the filament of bulb into the light and heat energy in 1 second).

Question 14

List the names of three electrical gadgets used in your house. Write their power, voltage rating and approximate time for which each one is used in a day. Hence find the electrical energy consumed by each in day.

Answer 14

Appliance Power

(in watt)

Voltage

(in volts)

Time

( hours)

Electrical energy
Fluorescent tube

Television set

Refrigerator

40

120

150

220

220

220

12

4

24

0.48 kWh

0.48 KWh

3.6 kWh

Question 15

Two lamps, one rated 220 V, 50 W and the other rated 220 V, 100 W, are connected in series with mains of 220 V. Explain why does the 50 W lamp consume more power.

Answer 15

   

Question 16

Name the factors on which the heat produced in a wire depends when current is passed in it, and state how does it depend on the factors stated by you.

Answer 16

When current is passed in a wire, the heat produced in it depends on the three factors:

(i) on the amount of current passing through the wire,

(ii) on the resistance of wire and (iii) on the time for which current is passed in the wire.

(iii)Dependence of heat produced on the current in wire: The amount of heat H produced in the wire is directly proportional to the square of current I passing through the wire.

(iv)Dependence of heat produced on the resistance of wire: The amount of heat H produced in the wire is directly proportional to the resistance R of the wire.

(v)Dependence of heat produced on the time: The amount of heat H produced in the wire is directly proportional to the time t for which current is passed in the wire.


Chapter :- 8, “Current Electricity” Selina Physics Solution MULTIPLE CHOICE TYPE – 8 (C)

Question 1.

When a current I flows through a resistance R for time t, the electrical energy spent is:

(a) IRt

(b) I²Rt

(c) IR²t

(d) I²R/t

Answer 1

(b) I²Rt

Question 2

An electrical appliances has a rating 100 W, 120 V. The resistance of element of appliance when in use is:

(a) 1.2 ohm

(b) 144 ohm

(c) 120 ohm

(d) 100 ohm

Answer 2

(b) 144 ohm


NUMERICALS – 8 (C) Chapter :- 8, “Current Electricity” Selina Physics Solution 

Question 1

An electric bulb of resistance 500 ohm draws current 0.4 A from the source. Calculate: (a) the power of bulb and (b) the potential difference at its end.

Answer 1

Resistance of electric bulb (R) = 500Ω

Current drawn from the source (I) = 0.4 A

Power of the bulb (P) = VI

V = I x R

V = 0.4 x 500 = 200 V

The potential difference at its end is 200 V.

Hence,

Power (P) = VI

P = 200 x 0.4 = 80 W

The power of the bulb is 80 Watt.

Question 2

A current of 2 A is passed through a coil of resistance 75 ohm for 2 minutes. (a) How much heat energy is produced ? (b) How much charge is passed through the resistance?

Question 3

Calculate the current through a 60 W lamp rated for 250 V. If the line voltage falls to 200 V, how is the power consumed by the lamp affected?

Answer 3

Question 4

An electric bulb is rated ‘100 W, 250 V’. How much current will the bulb draw if connected to a 250 V supply?

Answer 4

Question 5

An electric bulb is rated at 220 V, 100 W. (a) What is its resistance? (b) What safe current can be passed through it?

Answer 5

Question 6

A bulb of 40 W is used for 12.5 h each day for 30 days. Calculate the electrical energy consumed.

Answer 6

Question 7

An electric press is rated 750 W, 230 V. Calculate the electrical energy consumed by the iron in 16 hours.

Answer 7

Question 8

An electrical appliance having a resistance ofis operated at 220 V. calculate the energy consumed by the appliance in 5 minutes. (i) in joules (ii) in kWh.

Answer 8

Question 9

A bulb marked 12 V, 24 W operates on a 12 V battery for 20 minutes. Calculate:

(i)The current flowing through it, and

(ii)The energy consumed.

Answer 9

Question 10

A current of 0.2 A flows through a wire whose ends are at a potential difference of 15 V. Calculate:

(i)The resistance of the wire, and

(ii)The heat energy produced in 1 minute.

Answer 10

Question 11

What is the resistance, under normal working conditions, of an electric lamp rated at ‘240 v’, 60 W? If two such lamps are connected in series across a 240 V mains supply, explain why each one appears less bright.

Answer 11

When one lamp is connected across the mains, it draws 0.25 A current, while if two lamps are connected in series across the mains, current through each bulb becomes

(i.e., current is halved), hence heating (  ) in each bulb becomes one-fourth, so each bulb appears less bright.

Question 12

Two bulbs are rated 60 W, 220 V and 60 W, 110 V respectively. Calculate the ratio of their resistances.

Answer 12

Question 13

An electric bulb is rated 250 W, 230 V.

(i) the energy consumed in one hour, and

(ii) the time in which the bulb will consume 1.0 kWh energy when connected to 230 V mains?

Answer 13

Question 14

Three heaters each rated 250 W, 100 V are connected in parallel to a 100 V supply. Calculate:

(i)The total current taken from the supply,

(ii)The resistance of each heater, and

(iii)The energy supplied in kWh to the three heaters in 5 hours.

Answer 14

Question 15

A bulb is connected to a battery of p.d. 4 V and internal resistance . A steady current of 0.5 A flows through the circuit. Calculate:

(i)The total energy supplied by the battery in 10 minutes,

(ii)The resistance of the bulb, and

(iii)The energy dissipated in the bulb in 10 minutes.

Answer 15

Question 16

Two resistors A and B ofand  respectively are connected in parallel. The combination is connected across a 6 volt battery of negligible resistance. Calculate: (i) the power supplied by the battery, (ii) the power dissipated in each resistor.

Answer 16

Question 17

A battery of e.m.f. 15 V and internal resistance is connected to two resistors of 4 ohm and 6 ohm joined (a) in series. Find in each case the electrical energy spent per minute inresistor.

Answer 17

Question 18

Water in an electric kettle connected to a 220 V supply took 5 minutes to reach its boiling point. How long would it have taken if the supply had been of 200 V?

Answer 18

 

 

 

 

 

 

 

 

 

 

 

 

 Question 19

An electric toaster draws 8 A current in a 220 V circuit. It is used for 2 h. Find the cost of operating the toaster if the cost of electrical energy is Rs. 4.50 per kWh.

Answer 19

Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.

 

Question 20

An electric kettle is rated 2.5 kW, 250 V. Find the cost of running the kettle for two hours at Rs. 5.40 per unit.

Answer 20

Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.

 

Question 21

A geyser is rated 1500 W, 250 V. This geyser is connected to 250 V mains. Calculate:

(i)The current drawn,

(ii)The energy consumed in 50 hours, and

(iii)The cost of energy consumed at Rs 4.20 per kWh.

Answer 21

Return to Concise Selina ICSE Physics Class-10 

Thanks


 

Please Share with your friends

error: Content is protected !!