# Class 10 LAKHMIR SINGH AND MANJIT KAUR Solutions Physics Chapter 1 - Electricity

## Electricity Exercise 5

### Solution 1

### Solution 2

### Solution 3

(b) Voltmeter is used to measure p.d.

### Solution 4

Electric potential at a point is 1 volt means 1 joule of work is done in moving 1 unit positive charge from infinity to that point.

### Solution 5

Potential difference = 1 V

Charge moved = 1C

Work done = Potential difference x Charge moved

= 1 x 1 = 1 J

### Solution 6

Volt

### Solution 7

Given,

Potential difference = 12 V, Charge moved = 2 C

We know that,

Work done= p.d. x charge moved

= 12 x 2

= 24 joules

### Solution 8

Coulomb

### Solution 9

One coulomb of charge is that quantity of charge which exerts a force of 9 x 10^{9} Newton on an equal charge is placed at a distance of 1 m from it.

### Solution 10

(a) volts; voltmeter; parallel

(b) conductor; insulator

### Solution 11

Conductors:- Those substances through which electricity can flow are known as conductors. E.g., Copper, silver etc.

Insulators:- Those substances through which electricity cannot flow are known as insulators. E.g., Plastic, cotton etc.

### Solution 12

Conductor:- Silver, Copper, Aluminum, Nichrome, Graphite, Mercury, Manganin

Insulators:- Sulphur, Cotton, Air, Paper, Porcelain, Mica, Bakelite, Polythene

### Solution 13

The electric potential (or potential) at a point in an electric field is defined as the work done in moving a unit positive charge from infinity to that point.

Unit of electric potential is volt.

### Solution 14

(a)

(b) V_{1}=220 V, V_{2}=230V, Charge moved=4C

Thus, the potential difference= V_{2}- V_{1}

=230-220

=10V

We know that,

Work done = Potential difference x Charge moved

= 10 x 4Work done = 40 joules

### Solution 15

(a) Voltmeter

(b) Given : Potential difference=12V, Charge moved=1C

We know that,

Work done = Potential difference x charge moved

= 12 x 1 = 12 joules

Since work done on each coulomb of charge is 12 joules, the energy given to each coulomb of charge is also 12 joules.

## Electricity Exercise 6

### Solution 16

(a) Potential difference between two points in an electric circuit is defined as the amount of work done in moving a unit charge from one

point to the other point.

(b) The potential difference between two points is 1 volt means 1 joule of work is done in moving 1 coulomb of electric charge from one point to the other.

(c) Given: Work done = 250J, Charge moved = 20C

We know that,

(d) A voltmeter is a device which is used to measure the potential difference between two points in an electric circuit. Voltmeter is always connected in parallel across the two points where the potential difference is to be measured.

### Solution 22

(a) If three cells of 2 volt each are connected in series to make a battery, then the total potential difference between terminals of the battery will be 6V.

(b) (i) Given: p.d. = 2V, Charge moved = 1C

We know that

Work done = p.d. x charge moved

= 2 x 1

Work done = 2 joules

(ii) Given: p.d. = 6V, Charge moved = 1C

Work done = p.d. x charge moved

= 6 x1

Work done = 6 joule

### Solution 23

Copper has free electrons that are loosely held by the nuclei of the atoms. These free electrons result in conduction of electricity.

The electrons present in rubber are strongly held by the nuclei of its atoms. So, rubber does not have free electrons to conduct electricity.

## Electricity Exercise 11

### Solution 1

Ampere

### Solution 2

Electric Current

### Solution 3

Electrons

### Solution 4

Electrons

### Solution 5

(a) Conventional current flows from positive terminal of a battery to the negative terminal, through the outer circuit.

(b) Electrons flow from negative terminal to positive terminal of the battery (opposite to the direction of conventional current).

### Solution 6

1A = 1C/s

### Solution 7

Ampere

### Solution 8

(a) 1 amp = 10^{3} milli amp

(b) 1 amp = 10^{6} micro amp

### Solution 9

Ammeter is connected in series.

### Solution 10

Ammeter is connected in series in a circuit whereas voltmeter is connected in parallel.

### Solution 11

(i) Variable resistance

(ii) A closed plug key

### Solution 12

Given, Q = 20 C, t=1s

I=?

We know that:

### Solution 13

Given I=4amp, t=10s

Q=?

We know that:

### Solution 14

Given, Q=20C, t=40s

I=?

We know that:

### Solution 15

(a) electrons; closed

(b) amperes; ammeter; series

### Solution 16

(a) Cell or battery helps to maintain potential difference across a conductor.

(b) Given: p.d. = 10 V, I = 2amp, t = 1 min = 60s

We know that:-

### Solution 17

(a) An electric currrent is a flow of electric charges (electrons) through a conductor.

Potential difference between the ends of the wire makes electric current to flow in the wire.

(b) When 1 coulomb of charge flows through any cross-section of a conductor in 1 second, the electric current flowing through it is said to be 1 ampere.

### Solution 18

Ammeter is a device used for the measurement of electric current. It is always connected in series with the circuit in which the current is to be measured.

### Solution 19

(b) I = 0.36 A, t = 15 min = 900 s

### Solution 20

(a) The resistance of an ammeter should be very small so that it may not change the value of the current flowing in the circuit.

(b) The resistance of a voltmeter should be very large so that it takes a negligible current from the current.

### Solution 21

### Solution 22

A diagram which indicates how different components in a circuit have been connected by using the electrical symbols for the components is called a circuit diagram.

A voltmeter has a large resistance.

### Solution 23

We know that

Now,

When charge is coulombs, number of electrons = 1

When charge is 1 coulomb, number of electrons =

### Solution 24

## Electricity Exercise 12

### Solution 25

### Solution 26

(a) Electric current is the flow of electric charges (electrons) in a conductor such as a metal wire.

SI unit of electric current is ampere.

(b) 1 ampere.

(c) An ammeter is used to measure electric current. It should be connected in series with the circuit.

(d) Conventional direction of flow of electric current is from positive terminal of a battery to the negative terminal, through the outer circuit. The direction of flow of electrons is opposite to the direction of conventional current, i.e. from negative terminal to positive terminal.

### Solution 32

(a) Lamps are in series.

(b) Student has connected ammeter in parallel with lamps. It should be connected in series.

(c)

### Solution 33

### Solution 34

### Solution 35

## Electricity Exercise 18

### Solution 1

Ohm's law

### Solution 2

### Solution 3

Electric resistance.

### Solution 4

Insulators

### Solution 5

### Solution 6

Strength of electric current flowing in a given conductor depends on

(i) potential difference across the ends of the conductor

(ii) resistance of the conductor

### Solution 7

### Solution 8

### Solution 9

Potential difference, V = 20V

Resistance, R = 5ohms

Current, I = ?

We know that

V=IR

20 = I x 5

I = 20/5 = 4 A

### Solution 10

R = 20ohms

I = 2amp

We know that

V = IR

Thus,

V = 2 x 20

V = 40V

### Solution 11

I = 5amp

p.d., V = 3V

We know that

V=IR

Thus,

3 = 5 x R

R = 3/5 = 0.6 ohm

### Solution 12

current

### Solution 13

Those substances which have very low electrical resistance are called as good conductors. E.g., copper and aluminium.

Those substances which have comparatively high resistance than conductors are known as resistors. E.g., nichrome and manganin.

Those substances which have infinitely high electrical resistance are called insulators. E.g., rubber and wood.

### Solution 14

Conductor :- mercury, aluminum, iron, metal coin

Resistor :- manganin, nichrome

Insulator :- rubber, polythene, wood, bakelite, paper, thermocol

## Electricity Exercise 19

### Solution 15

Ohm's law gives a relationship between current (I) and potential difference (V). According to ohm's law: At constant temperature, the current flowing through a conductor is directly proportional to the potential difference across its ends.

If I is the current flowing through a conductor and V is the p.d. across its ends, then according to the ohm's law:

### Solution 16

(a) The property of a conductor due to which it opposes the flow of current through it is called resistance of the conductor.

(b) V = 12volt, I=2.5 x 10^{-3} A

We know that

V=IR

R=V/I

R=12/(2.5X10^{-3})

R=4.8X10^{3}ohm = 4800 ohm

### Solution 17

(a) 1 ohm is the resistance of a conductor such that when a potential difference of 1 volt is applied to its ends, a current of 1ampere flows through it.

(b) Its resistance will increase.

(c)

### Solution 18

(a) Electricians wear rubber hand gloves while working with electricity because rubber is an insulator and protects them from electric shocks.

(b) I=6amp, R=40ohm

We know that

V=IR

V = 6 x 40 = 240 V

### Solution 19

(i)

(ii) Since the graph is a straight line passing through the origin, so current is directly proportional to the potential difference.

Hence, the ration remains constant.

From graph, when V=1.5 volt, I=0.6 amp

So, =

For p.d. 0.8V, 1.2v and 1.6V, the value of ratio remains the same i.e., 2.5 ohm.

(iii) The resistance of the wire is equal to the ratio of potential difference applied and the current passing through it.

### Solution 20

(a) The ratio of potential difference and current is known as resistance.

(b)

(c) Ohm's law

(d) Potential difference = Current x Resistane

(e) V = 240 volt, I = 5ampWe know that

V=IR

240 = 5 x R

R = 240/5 = 48 ohm

## Electricity Exercise 20

### Solution 30

In first case,

I = 2.4 amp, V = 120 volt

V=IR

120 = 2.4 x R

R = 120/2.4 = 50 ohm

In second case,

V = 240 volt, R = 50 ohm

V = IR 240 = I_{ }x 50

I =4.8 amp

### Solution 31

Resistance.

### Solution 32

(a) Ohm's law

(b) Temperature

### Solution 33

In first case,

I = 0.02 amp, V = 10 volt

V=IR

10 = 0.02 x R

R = 10/0.02 = 500 ohm

In second case,

I = 250 x 10^{-3} amp, R = 500 ohm

V = 250 x 10^{-3}_{ x} 500

V = 125 volt

### Solution 34

I = 200mA = 0.2 A

R = 4 x 10^{3}ohm = 4000 ohm

We know that

V=IR

V = 0.2 x 4000

V = 800 volt

## Electricity Exercise 26

### Solution 1

The resistance decreases.

### Solution 2

### Solution 3

Resistance of a conductor depends on the following factors:-

Length of the conductor, area of cross section of the conductor, nature of material of the conductor and temperature of the conductor

### Solution 4

### Solution 5

Iron

### Solution 6

### Solution 7

Nichrome

### Solution 8

Nichrome is an alloy of nickel, chromium, manganese ad iron having a resistivity of about 60 times more than that of copper. It is used for making the heating elements of electrical heating appliances.

### Solution 9

Nichrome alloy is used for making the heating elements of electrical appliances because:

(i) nichrome has very high resistivity

(ii) nichrome does not undergo oxidation (or burn) easily even at high temperature.

### Solution 10

Because

(i) resistivity of an alloy is much higher than that of a pure metal

(ii) an alloy does not undergo oxidation (or burn) easily even at high temperature.

### Solution 11

(a) A long piece of nichrome wire.

(b) A thin piece of nichrome wire.

### Solution 12

(a) On decreasing the temperature, the resistance decreases.

(b) Presence of impurities in a metal increases the resistance.

### Solution 13

Ohms; increases; increases; decreases.

### Solution 14

(a) Resistivity is the characteristic property of a substance which depends on the nature of the substance and its temperature. It is numerically equal to the resistance between the opposite faces of a 1 m cube of the substance.

(b) l = 1m

r = d/2 = 0.2/2 mm = 0.1 mm = 0.0001m,

R = 10 ohm

We know that,

### Solution 15

### Solution 16

(a) Silver and copper are good conductors of electricity because they have free electrons available for conduction.

### Solution 17

Current will flow more easily through thick wire because the resistance of the thick wire will be lesser than that of thin wire.

### Solution 18

(a) Resistance of a conductor increases (or decreases) with increase (or decrease) in the length of the conductor.

(b) Resistance of a conductor decreases (increases) with increase (decrease) in the area of cross-section of the conductor.

(c) Resistance of a conductor increases on raising the temperature and decreases on lowering the temperature.

### Solution 19

(a) If we take two similar wires of same length and same diameter, one of copper metal and other of nichrome alloy, we will find that the resistance of nichrome wire is about 60 times more than that of the copper wire. This shows that the resistance depends on the nature of material of the conductor.

### Solution 20

(a) Resistance will increase.

(b) Resistance will decrease.

(c) Resistance will increase.

### Solution 21

(a) By increasing the area of cross section, the resistance will decrease.

(b) By increasing the diameter, the resistance will decrease.

## Electricity Exercise 27

### Solution 22

### Solution 23

### Solution 24

(b) Ohm-meter

(c) 1. Resistance is the property of the conductor, while resistivity is the property of the material of the conductor.

2. Resistance of a conductor is the opposition to the flow of electric current through it. Resistivity of a substance is the opposition to the flow of electric current by a rod of that substance which is 1m long and 1m^{2} in cross section.

3. Resistance of a conductor depends on length, thickness, nature of material and temperature of the conductor; while resistivity of a substance depends on the nature of the substance and temperature.

(d) Resistivity of a substance depends on the nature of the substance and its temperature. It does not depend on the length or thickness of the conductor.

### Solution 33

### Solution 34

(a) Material Q with resistivity 2.63 X 10^{-8} ohm-m can be used for making electric wires because it has very low resistivity.

(b) Material R with resistivity 1.0 X 10^{15} ohm-m can be used for making handle of soldering iron because it has very high resistivity.

(c) Material P with resistivity 2.3 X 10^{3} ohm-m can be used for making solar cell because it is a semiconductor.

### Solution 35

(a) Good conductor = C (10 x 10^{-8}ohm-m)

(b) Resistor = A (110 x 10^{-8} ohm-m)

(c) Insulator= B (1 x 10^{10} ohm-m)

(d) Semiconductor= D (2.3 x 10^{3} ohm-m)

### Solution 36

(a) E is best conductor of electricity due to its least electrical resistivity.

(b) C, because its resistivity is lesser than that of A.

(c) B, because it has the highest electrical resistivity.

(d) C and E, because of their low electrical resistivities.

## Electricity Exercise 37

### Solution 1

### Solution 2

As per the law of combination of resistances in series,

R=R_{1}+ R_{2}+ R_{3}+ R_{4}+ R_{5}

R=0.2+0.2+0.2+0.2+0.2=1ohm

### Solution 3

According to the law of combination of resistance in parallel, the reciprocal of the combined resistance of a number of resistances connected in parallel is equal to the sum of the reciprocals of all the individual resistances.

### Solution 4

### Solution 5

Since the resultant resistance is less than the individual resistances, so the resistances should be connected in parallel.

### Solution 6

In case of parallel combination, the resultant resistance will be less than either of the individual resistances.

### Solution 7

R_{1}=2ohm, R_{2}=6ohm

Case I: (Parallel combination)

1/R = 1/R_{1}+ 1/R_{2}

1/R = 1/2 + 1/6 = 4/6

R = 6/4 = 1.5ohm

Case II: (Series combination)

R = R_{1} + R_{2} = 2+6 = 8ohm

## Electricity Exercise 38

### Solution 17

### Solution 8

(a) By connecting in parallel: Since equivalent resistance will be

1/ R = 1/4 + 1/4 = 2/4 = 1/2

Therefore, R = 2 ohm

(b) By connecting in series : Since equivalent resistance will be

R = 4 ohm + 4 ohm = 8 ohm

### Solution 9

Resistance of arrangement A is 10 ohm.

Combined resistance of arrangement B is caculated as follows:

1/R = 1/10 + 1/1000 = (100+1)/1000

R = 1000/101 = 9.9 ohm

Therefore, arrangement B has lower combined resistance.

### Solution 10

Resistance of each part is R/2.

Resultant resistance R' is given by

1/R' = 2/R+2/R

R'=R/4

### Solution 11

(a) R_{1}=500ohm, R_{2}=1000ohm

As per given figure,

R=R_{1}+R_{2}=500+1000=1500ohm.

(b) R_{1}=2ohm, R_{2}=2ohm

As per given figure,

1/R=1/R_{1}+1/R_{2}

1/R=1/2+1/2

R=1ohm

(c) R_{1}=4ohm, R_{2}=4ohm, R_{3}=3ohm

1/R=1/R_{1}+1/R_{2}

1/R=1/4+1/4

R=2ohm

Total resistance =R+R_{3}

=2+3=5ohm

### Solution 12

R_{1}=6ohm, R_{2}=4ohm V=24V

The two resistances are connected in parallel.

Current across R_{1}=I_{1}=V/R_{1}=24/6=4amp

Current across R_{2}=I_{2}=V/R_{2}=24/4=6amp

### Solution 13

(i) Series combination

When two or more resistances are connected end to end consecutively, they are said to be connected in series combination. The combined resistance of any number of resistances connected in series in equal to the sum of the individual resistances.

R=R_{1}+R

_{2}+......

The resultant resistance is more than either of the individual resistances.

(ii) Parallel combination

When two or more resistances are connected between the same two points, they are said to be connected in parallel combination. The reciprocal of the combined resistance of a number of resistances connected in parallel is equal to the sum of the reciprocals of all the individual resistances.

1/R=1/R

_{1}+1/R

_{2}+........

The resultant resistance is less than either of the individual resistances.

### Solution 14

R_{1}=0.2ohm, R_{2}=0.4ohm, R_{3}=0.3ohm, R_{4}=0.5ohm, R_{5}=12ohm, V=9V

Resultant resistance=R_{1}+ R_{2}+ R_{3}+ R_{4}+ R_{5}

R=0.2+0.4+0.3+0.5+12=13.4ohm

Thus the current flow through 12ohm resistance will be=V/R

I=9/13.4

I=0.67amp.

### Solution 15

(a) Total resistance of the circuit=R_{1}+R_{2}=20+4=24ohm

(b) We know that

V=IR

Therefore,

6=I x 24

I=6/24=0.25amp

(c) p.d. across bulb=IR_{1}=0.25X20=5V

(d) p.d. across resistance wire=IR_{2}=0.25X4=1V

### Solution 16

According to the diagram,

(i) Total current I=1amp is entering the parallel combination of R_{1} and R_{2}. Let I_{1} current flow through R_{1} and I_{2} current flow through R_{2. }

Then

(ii) p.d. across AB = IR_{3 }= 1 x 5 = 5V

Equivalent resisyance between B and C is

1/R' = 1/R_{1} + 1/R_{2} = 1/10 + 1/15

1/R' = 5/30

R' = 6 ohm

Total resistance between A and C is R = 5+6 = 11 ohm

p.d. across AC = IR = 1x11 = 11V

(iii) Total resistance = R_{3} + R' = 5 + 6 = 11 ohm

## Electricity Exercise 39

### Solution 18

Given: Two resistors with resistances R_{1}=5ohm and R_{2}=10ohm, V=6volt

(a) For minimum current these two should be connected in series. For maximum current these two should be connected in parallel.

(b)In series,Total resistance = 5+10 = 15ohms

Therefore total current drawn = V/R = 6/15 = 0.4amps

In parallel,Total resistance R is given as

1/R=1/R_{1}+1/R_{2}

1/R=1/5+1/10

1/R=3/10

R=10/3ohm

Therefore total current drawn by the circuit = V/R = 6/(10/3) =1.8amps

### Solution 19

(i) Total resistance of two resistors that are connected in parallel is

1/R' = 1/3+1/6

1/R' = 3/6

R' = 2ohms

Total resistance of the circuit = 2+4ohms = 6ohms

(ii) Total current flowing through the circuit=V/total resistance

I = 12/6 = 2amps

(iii) Potential difference across R_{1}=R_{1} x I = 4 x 2 = 8V

### Solution 20

Given :-

1 amp current is flowing through 5ohm resistor.

We know that in case of parallel connection, the p.d. across each resistor is same and is equal to the voltage applied.

Therefore, applied voltage, V = IR = 1 x 5 = 5V

So,

Current through 4 ohm resistor = V/R = 5/4 = 1.25 A

Current through 10 ohm resistor = V/R = 5/10 = 0.5 A

### Solution 21

### Solution 22

Given V=220V

R_{A} = R_{B} = 24 ohm

(a) Current drawn when only coil A is used:

I = V/R_{A} = 220/24

=9.16amps

(b) Current drawn when coils A and B are used in series:

Total resistance, R = R_{A} + R_{B} = 24+24 = 48ohms

I = V/R = 220/48

=4.58amps

(c) Current drawn when coils A and B are used in parallel:

Total resistance, 1/R = 1/R_{A} + 1/R_{B }= 1/24 + 1/24 = 2/24 = 1/12

R=12ohms

I = V/R = 220/12

=18.33amps

### Solution 23

## Electricity Exercise 40

### Solution 24

V=12V

R_{1}, R_{2} and R_{3} are connected in parallel.

(a) Current through R_{1 }= V/R_{1} = 12/5 = 2.4 A

Current through R_{2} = V/R_{2} = 12/10 = 1.2 A

Current through R_{3} = V/R_{2} = 12/30 = 0.4 A

(b) Total current in the circuit = 2.4 + 1.2 + 0.4 = 4 A

(c) Total resistance in the circuit=R

1/R = 1/R_{1}+ 1/R_{2} + 1/R_{3}

1/R = 1/5 + 1/10 + 1/30

1/R = 10/30

R = 3 ohm

### Solution 25

V = 4V,

R_{1} = 6 ohm, R_{2} = 8 ohm (in series)

(a) Combined resistance, R = R_{1} + R_{2} = 6+2 = 8ohm

(b) Current flowing, I = V/R = 4/8=0.5amp

(c) p.d. across 6ohm resistor = I x R_{1} = 0.5 x 6 = 3 V

### Solution 26

V = 6V

R_{1} = 3 ohm, R_{2} = 6 ohm (in parallel)

(a) Combined resistance,

1/R = 1/R_{1} + 1/R_{2}

1/R = 1/3 + 1/6 = 3/6 = 1/2

R = 2 ohm

(b) Current flowing in the main circuit, I = V/R = 6/2 = 3 A

(c) Current flowing in 3 ohm resistor = V/R_{1} = 6/3 = 2 A

### Solution 28

Total current flowing through circuit, I = 6 A

R_{1} = 3 ohm, R_{2} = 6 ohm

(a) Combined resistance R is

1/R=1/3+1/6

1/R=3/6

R=2ohms

(b) p.d. across the combined resistance = IR = 6 x 2 = 12 V

(c) p.d. across the 3 ohm resistor = p.d. across the combined resistance = 12 V

(d) Current flowing through the 3 ohm resistor = V/R_{1} = 12/3 = 4 A

(e) Current flowing through the 6 ohm resistor = V/R_{2} = 12/6 = 2 A

### Solution 27

## Electricity Exercise 41

### Solution 29

(a)

(b) Effective resistance = 20 + 20 = 40 ohms

(c) Current flowing through the circuit = I = V/R = 5/40 = 0.125 amps

(d) p.d. across each resistance = I x R = 0.125 x 20 = 2.5 V

### Solution 30

V=6V, R_{1}=2ohms, R_{2}=3ohms

(a) Resistors are connected in parallel

(b) p.d. across each resistor is same and is equal to 6V.

(c) 2 ohms resistance have bigger share of current because of its lower resistance.

(d) Effective resistance=R

1/R=1/2+1/3

1/R=5/6

R=1.2ohms

(e) Current flowing through battery, I=V/R=6/1.2=5amps

### Solution 33

### Solution 31

### Solution 32

## Electricity Exercise 42

### Solution 34

Suppose total current flowing the circuit is I, then the current passing through resistance R_{1} will be I_{1} and current passing through resistance R_{2} will be I_{2.}

Total current =I=I_{1}+I_{2}

Let resultant resistance of this parallel combination is R. By applying the Ohm's law to the whole circuit, we get that I=V/R

Since the potential difference across the both the resistances is same, so applying the Ohm's law to each resistance we get that

I_{1}=V/R_{1}

I_{2}=V/R_{2}

Putting these eq in the above one, we get that

V/R=V/R_{1}+V/R_{2}

1/R=1/R_{1}+1/R_{2}

If two resistance are connected in parallel than, the resultant resistance will be

1/R=1/R_{1}+1/R_{2}

(b)

(i) Total reisitance=R

1/R=1/R_{1}+1/R_{2}

R_{2}=3+2=5ohms

R_{1}=5ohms

1/R=1/5+1/5

1/R=2/5

R=2.5ohms

(ii) Current flowing through the circuit

I=V/R=4/(2.5)

=1.6amps

### Solution 35

Suppose total current flowing in the circuit is I, then the current passing through resistance R_{1} will be I_{1,} current passing through resistance R_{2} will be I_{2} and current passing through resistance R_{3} will be I_{3.}

Total current =I=I_{1}+I_{2}+I_{3}

Let resultant resistance of this parallel combination is R. By applying the Ohm's law to the whole circuit, we get that

I=V/R

Since the potential difference across all the resistances is same, so applying the Ohm's law to each resistance we get that

I_{1}=V/R_{1}

I_{2}=V/R_{2}

I_{3}=V/R_{3}

Putting these eqs. in the above one, we get

V/R=V/R_{1}+V/R_{2}+ V/R_{3}

1/R=1/R_{1}+1/R_{2}+ 1/R_{3}

If two resistance are connected in parallel, then the resultant resistance will be

1/R=1/R_{1}+1/R_{2}+ 1/R_{3}

(b) If switch is open, then only upper two resistances (connected in parallel) are in the circuit.

Effective resistance is 1/R_{eq}=1/R+1/R=2/R

R_{eq} = R/2

So the current=I=V/(R/2)=0.6A (given)

V/R = 0.3 A

When the switch closes, the third resistance also comes in the circuit. The effective resistance of the circuit becomes R/3

Hence, Current I = V/(R/3) = 3 (V/R) = 3 x 0.3 = 0.9 A## Electricity Exercise 43

### Solution 43

(i)

Resultant resistance for parallel circuit=R

1/R=1/6+1/6

1/R=2/6

R=3

Effective resistance=6+3=9ohms

(ii)

Resultant resistance for each parallel circuit=R

1/R=1/6+1/6+1/6

1/R=3/6

R=2

Therefore effective resistance=2+2=4ohms.

### Solution 44

Two resistances when connected in parallel, resultant value is 2ohms.

Let the two resistances be R_{1} and R_{2}.

If connected in series, then

9=R_{1}+R_{2}

R_{1}=9-R_{2}

If connected in parallel, then

1/2=1/R_{1}+1/R_{2}

From above equations we get that

1/2=(R_{1}+R_{2})/R_{1}R_{2}

1/2=9/(9-R_{2}) R_{2}

9R_{2}-R_{2}^{2}=18

R_{2}^{2}-9R_{2}+18=0

(R_{2}-6) (R_{2}-3)=0

R_{2}=6,3

So if R_{2}=6ohms, then R_{1}=9-6=3ohms.

If R_{2}=3ohms, then R_{1}=9-3=6ohms.

### Solution 45

Given:

A resistor of 8ohm is connected in parallel with a resistor of X.

And resultant is 4.8.

Then X=?

We know that for parallel case

1/R=1/R_{1}+1/X

1/4.8=1/8+1/x

1/4.8 - 1/8 = 1/x

After solving we get that

X=12ohms

### Solution 46

## Electricity Exercise 44

### Solution 47

Given: Three resistances of 2ohms, 3ohms, 5ohms.

Their resultant, R=2.5ohms

Resistance of first line = 2+3 = 5 ohm

So, 1/R = 1/5 + 1/5

On solving we get that

R=2.5ohms

### Solution 48

(a) Connect 2ohms resistor in series with a parallel combination of 3ohms and 6ohms.

(b) Connect 2ohms, 3ohms, and 6ohms in parallel.

### Solution 49

(a) For obtaining the highest resistance by combining the given resistances, we must connect them in series.

We get,

R=4+8+12+24=48ohms

(b) For obtaining the lowest resistance by combining the given resistances, we must connect them in parallel.

We get,

1/R=1/4+1/8+1/12+1/24

On solving we get, R=2ohms

### Solution 50

So, the resultant of these three resistances = 20+20+10 = 50ohms.

This 50ohms is in parallel with 30ohms. So resultant of these two will be

1/R=1/30+1/50

1/R=80/1500

R=18.75ohms

Now, the resistances 10 ohms, 18.75 ohms and 10 ohms are in series.

Therefore, resultant resistance = 18.75+10+10 = 38.75ohms.

### Solution 51

Given: n=100, R=1 ohm

For obtaining the smallest resistance, these resistances are connected in parallel:

Equivalent resistance = 1/1 + 1/1 + 1/1.....100 times = 100/1

R_{eq} = 1/100 = 0.01 ohm

For obtaining the largest resistance, these resistances are connected in series:

Equivalent resistance = 1 + 1 + 1.....100 times = 100

R_{eq} = 100 ohm

### Solution 52

For obtaining 250ohms, connect two 100ohms in series with a parallel combination of two 100ohms.

### Solution 53

R_{eq} = R+R+R+R = 4R ohm

Total current in the circuit, I = V/R = 12/4R = 3/R

Reading of voltmeter A = Voltage across R_{1} = I x R_{1} = 3/R x R = 3V

Reading of voltmeter B= Voltage across R_{2} = I x R_{2} = 3/R x R = 3V

Reading of voltmeter C= Voltage across the series combination of R_{3} and R_{4} = I x (R_{3}+R_{4}) = 3/R x 2R = 6V

### Solution 54

1/R = 1/16 + 1/16 + 1/16 + 1/16 = 4/16

R = 4 ohm

Four such combinations are connected in series, so total resistance = 4+4+4+4 = 16 ohm.

## Exercise

### Solution

## Electricity Exercise 47

### Solution 6

A series arrangement is not used for connecting domestic electrical appliances in a circuit because if one electrical appliance stops working due to some defect, then all other appliance also stop working as the whole circuit is broken.

### Solution 7

Different electrical appliances in a domestic circuit are connected in parallel because of the following advantages:

(i) If one electrical appliance stops working due to some defect, then all other appliances keep working properly.

(ii) Each electrical appliance has its own switch due to which it can be turned on or turned off independently, without affecting other appliances.

(iii) Each electrical appliance gets the same voltage as that of the power supply line.

### Solution 8

(a) Parallel circuit

(b) Parallel circuit

(c) Series circuit

(d) Series circuit.

### Solution 9

(a) circuit (ii)

(b) circuit (iii)

(c) circuit (iii)

### Solution 1

No, they are wired in parallel.

### Solution 2

All the other bulbs also stop glowing.

### Solution 3

All the other bulbs keep glowing.

### Solution 4

(a) Series

(b) Parallel

### Solution 5

The two lamps (of 4V each) should be arranged in parallel with the two 2V cells.