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

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.

Potential difference = 1 V

Charge moved = 1C

Work done = Potential difference x Charge moved

                = 1 x 1 = 1 J

Volt

Given,

Potential difference = 12 V,  Charge moved = 2 C

We know that, 

Work done= p.d. x charge moved

                = 12 x 2

                = 24 joules

Coulomb

One coulomb of charge is that quantity of charge which exerts a force of 9 x 10⁹ Newton on an equal charge is placed at a distance of 1 m from it.

(a) volts; voltmeter; parallel

(b) conductor; insulator

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.

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

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

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.

(a)

      (b) V₁=220 V, V₂=230V, Charge moved=4C

           Thus, the potential difference= V₂- V₁

                                        =230-220

                                        =10V

           We know that,  

           Work done = Potential difference x Charge moved

           Work done = 40 joules

(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

(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,

(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

Work done = p.d. x charge moved

                = 2 x 1

Work done = 2 joules

Work done = p.d. x charge moved

                 = 6 x1

Work done = 6 joule

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.

Ampere

Electric Current

Electrons

Electrons

(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).

1A = 1C/s

Ampere

(a) 1 amp = 10³ milli amp

(b) 1 amp = 10⁶ micro amp

Ammeter is connected in series.

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

(i) Variable resistance

(ii) A closed plug key

Given, Q = 20 C, t=1s

I=?

We know that:

Given I=4amp, t=10s
Q=?

We know that:

Given, Q=20C, t=40s

I=?

We know that:

(a) electrons; closed

(b) amperes; ammeter; series

(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:-

(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.

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.

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

(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.

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.

We know that

Now,

When charge is coulombs, number of electrons = 1

When charge is 1 coulomb, number of electrons =

(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.

(a) Lamps are in series.

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

(c)

Ohm's law

Electric resistance.

Insulators

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

Potential difference, V = 20V

Resistance, R = 5ohms

Current, I = ?

We know that

V=IR

20 = I x 5

I = 20/5 = 4 A

 R = 20ohms

I = 2amp

We know that

V = IR

Thus,

V = 2 x 20

V = 40V

 I = 5amp

p.d., V = 3V

We know that

V=IR

Thus,

3 = 5 x R

R = 3/5 = 0.6 ohm

current

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.

Conductor :- mercury, aluminum, iron, metal coin

Resistor :- manganin, nichrome

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

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:

(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=4.8X10³ohm = 4800 ohm

(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)

(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

(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.

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

(b)

(c) Ohm's law

(d) Potential difference = Current x Resistane

We know that

V=IR

240 = 5 x R

R = 240/5 = 48 ohm

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

240 = I x 50

I =4.8 amp

Resistance.

(a) Ohm's law

(b)  Temperature

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

I = 200mA = 0.2 A

R = 4 x 10³ohm = 4000 ohm 

We know that

V=IR

V = 0.2 x 4000

V = 800 volt

The resistance decreases.

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

Iron

Nichrome

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.

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.

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.

(a) A long piece of nichrome wire.

(b) A thin piece of nichrome wire.

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

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

Ohms; increases; increases; decreases.

(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,

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

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

(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.

(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.

(a) Resistance will increase.

(b) Resistance will decrease.

(c) Resistance will increase.

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

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

(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 1m² 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.

(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 10¹⁵ 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 10³ ohm-m can be used for making solar cell because it is a semiconductor.

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

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

(c) Insulator= B (1 x 10¹⁰ ohm-m)

(d) Semiconductor= D (2.3 x 10³ ohm-m)

(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.

As per the law of combination of resistances in series,

R=R₁+ R₂+ R₃+ R₄+ R₅

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

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.

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

R₁=2ohm, R₂=6ohm

Case I: (Parallel combination)

1/R = 1/R₁+ 1/R₂

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

R = 6/4 = 1.5ohm

Case II: (Series combination)

R = R₁ + R₂ = 2+6 = 8ohm

(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

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.

Resistance of each part is R/2.

Resultant resistance R' is given by

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

R'=R/4

(a) R₁=500ohm, R₂=1000ohm

As per given figure,

R=R₁+R₂=500+1000=1500ohm.

(b) R₁=2ohm, R₂=2ohm

As per given figure,

1/R=1/R₁+1/R₂

1/R=1/2+1/2

R=1ohm

(c) R₁=4ohm, R₂=4ohm, R₃=3ohm

1/R=1/R₁+1/R₂

1/R=1/4+1/4

R=2ohm

Total resistance =R+R₃

=2+3=5ohm

R₁=6ohm, R₂=4ohm V=24V

The two resistances are connected in parallel.

Current across R₁=I₁=V/R₁=24/6=4amp

Current across R₂=I₂=V/R₂=24/4=6amp

(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.

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.

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

R₁=0.2ohm, R₂=0.4ohm, R₃=0.3ohm, R₄=0.5ohm, R₅=12ohm, V=9V

Resultant resistance=R₁+ R₂+ R₃+ R₄+ R₅

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.

(a) Total resistance of the circuit=R₁+R₂=20+4=24ohm

(b) We know that

V=IR

Therefore,

6=I x 24

I=6/24=0.25amp

(c) p.d. across bulb=IR₁=0.25X20=5V

(d) p.d. across resistance wire=IR₂=0.25X4=1V

      According to the diagram,

(i) Total current I=1amp is entering the parallel combination of R₁ and R₂. Let I₁ current flow through R₁ and I₂ current flow through R_2.

Then

(ii) p.d. across AB = IR₃ = 1 x 5 = 5V

Equivalent resisyance between B and C is

1/R' = 1/R₁ + 1/R₂ = 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 = R₃ + R' = 5 + 6 = 11 ohm

Given: Two resistors with resistances R₁=5ohm and R₂=10ohm, V=6volt

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

Total resistance = 5+10 = 15ohms

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

Total resistance R is given as

1/R=1/R₁+1/R₂

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

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

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

1/R' = 3/6

R' = 2ohms

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

I = 12/6 = 2amps

(iii) Potential difference across R₁=R₁ x I = 4 x 2 = 8V

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

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

V=12V

R₁, R₂ and R₃  are connected in parallel.

(a) Current through R₁ = V/R₁ = 12/5 = 2.4 A

     Current through R₂ = V/R₂ = 12/10 = 1.2 A

     Current through R₃ = V/R₂ = 12/30 = 0.4 A

(c)  Total resistance in the circuit=R

                
     1/R = 1/R₁+ 1/R₂ + 1/R₃

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

     1/R = 10/30

     R = 3 ohm

V = 4V, 

R₁ = 6 ohm, R₂ = 8 ohm (in series)

(a) Combined resistance, R = R₁ + R₂ = 6+2 = 8ohm

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

(c) p.d. across 6ohm resistor = I x R₁ = 0.5 x 6 = 3 V

V = 6V

R₁ = 3 ohm, R₂ = 6 ohm (in parallel)

(a) Combined resistance,

1/R = 1/R₁ + 1/R₂

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/R₁ = 6/3 = 2 A

Total current flowing through circuit, I = 6 A

R₁ = 3 ohm, R₂ = 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/R₁ = 12/3 = 4 A

(e) Current flowing through the 6 ohm resistor = V/R₂ = 12/6 = 2 A

      (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

V=6V, R₁=2ohms, R₂=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

Suppose total current flowing the circuit is I, then the current passing through resistance R₁ will be I₁ and current passing through resistance R₂ will be I_2.

Total current =I=I₁+I₂

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

I₁=V/R₁

I₂=V/R₂

Putting these eq in the above one, we get that

V/R=V/R₁+V/R₂

1/R=1/R₁+1/R₂

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

1/R=1/R₁+1/R₂

(b)

(i) Total reisitance=R

1/R=1/R₁+1/R₂

R₂=3+2=5ohms

R₁=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

Suppose total current flowing in the circuit is I, then the current passing through resistance R₁ will be I_1, current passing through resistance R₂ will be I₂ and current passing through resistance R₃ will be I_3.

Total current =I=I₁+I₂+I₃

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

I₁=V/R₁

I₂=V/R₂

I₃=V/R₃

Putting these eqs. in the above one, we get

V/R=V/R₁+V/R₂+ V/R₃

1/R=1/R₁+1/R₂+ 1/R₃

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

1/R=1/R₁+1/R₂+ 1/R₃

(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

(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.

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

Let the two resistances be R₁ and R₂.

If connected in series, then

9=R₁+R₂

R₁=9-R₂

If connected in parallel, then

1/2=1/R₁+1/R₂

From above equations we get that

1/2=(R₁+R₂)/R₁R₂

 1/2=9/(9-R₂) R₂

9R₂-R₂²=18

R₂²-9R₂+18=0

(R₂-6) (R₂-3)=0

R₂=6,3

So if R₂=6ohms, then R₁=9-6=3ohms.

If R₂=3ohms, then R₁=9-3=6ohms.

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/R₁+1/X

1/4.8=1/8+1/x

1/4.8 - 1/8 = 1/x

After solving we get that

X=12ohms

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

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

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

(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

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.

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

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

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 R₁ = I x R₁ = 3/R x R = 3V

Reading of voltmeter B= Voltage across R₂ = I x R₂ = 3/R x R = 3V

Reading of voltmeter C= Voltage across the series combination of R₃ and R₄ = I x (R₃+R₄) = 3/R x 2R = 6V

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.

The total current of 0.5 A flowing in the circuit distributes equally in the two arms having lamps (since the lamps have same resistances). So the current through each of these arms is 0.25 A. Hence A₂, A₃, A₄ and A₅, all will read 0.25 A.

No, they are wired in parallel.

All the other bulbs also stop glowing.

All the other bulbs keep glowing.

(a) Series

(b) Parallel

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

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.

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.

(a) Parallel circuit

(b) Parallel circuit

(c) Series circuit

(d) Series circuit.

(a) circuit (ii)

(b) circuit (iii)

(c) circuit (iii) 

Parallel arrangement because if one electrical bulb stops glowing due to some defect the other will keep glowing.

Parallel arrangement takes more current from the battery due to its lesser equivalent resistance.

(a) Parallel circuits - Because if one electrical appliance stops working due to some defect, then all other appliances in the circuit will keep working properly.

(b)  All the other lamps stop glowing.

(c) All lamps are connected in series.

(d)

(a)

(b) The brightness of the lamps can be changed by connecting the lamps in parallel.

(a) C will be the brightest. Voltage will be distributed equally between A and B, so they will have equal brightness but lesser than that of C.

(b) A gets the same voltage as before, so its brightness remains the same.

(c)  If B burns put, A will also stop glowing because it is connected in series with B. However, brightness of C remains the same.

The brightness of two lamps arranged in parallel is much more those arranged in series.

(a) In case of series connection, if filament of one lamp breaks, the other wil stop glowing.

(b) In case of parallel connection, if filament of one lamp breaks, the other will keep glowing.

(a) Turn the switch to right side so as the resistance decreases.

(b) Turn the switch to the left side so as the resistace increases.

Electrical energy consumed by an electrical appliance depends on:

1. Power rating of the appliance.

2. Time for which the appliance is used.

60 watt bulb, because power is inversely proportional to the resistance.

Kilowatt-hour is the commercial unit of electric energy.

V = 220 V, P = 100W

R=?

We know that

P = V²/R

Thus

R = V²/P = 220²/100 = 484ohm

(i) joule

(ii) watt

(i) Electric power

(ii) Electric energy

Electric power has the unit of watt.

kWh is the short form of kilowatt-hour, which is the commercial unit of electrical energy.

P=V²/R

R is fixed.

V becomes double.

Now, P = (2V)²/R = 4 V²/R

 So, the electric power becomes four times its previous value.

Other information is that it will consume energy at the rate of 36 J/s.

P=920W, V=230V, I=?

We know that

P = V x I,

920 = 230 x I

I = 920/230 = 4amp

When an electrical appliance consumes electrical energy at the rate of 1 joule per second, its power is said to be 1 watt.

1 watt = 1 volt x 1 ampere

One watt hour is the amount of electrical energy consumed when an electrical appliance of 1 watt power is used for 1 hour.

1 watt hour = 3600 joules

I=5amp, R=100ohms, t=2h

We know that

Electric energy consumed = P x t = I²Rt

                                   = 5² x 100 x 2

                                   = 5000 Wh

                                   = 5 kwh

Therefore, 5kwh = 5 x 3.6 x 10⁶ J = 18 x 10⁶ J

V=220V, I=0.5amp, P=?

We know that

P=VI=220X0.5

P=110 watt

(i) R = 300 ohm, I = 1 A, t = 1h

P = I²R = 1² x 300 = 300 W

E = P x t = 300 x 1 = 300 Wh

(ii) R = 100 ohm, I = 2 A, t = 1h

P = I²R = 2² x 100 = 400 W
E = P x t = 400 x 1 = 400 Wh

Hence, in case (ii), the electrical energy consumed per hour is more.

V=220V, P=2.2kW=2200W, t=3h

We know that

Electrical energy consumed = Pxt = 2.2x3 = 6.6 kWh

We have, P = V x I

2200 = 220 x I

I=10amp

V=250V, I=0.4amp

(i) We know that

Power=VI=250X0.4=100watt

      (ii) We have

      P=I²R

      100=0.4²XR

      R=625ohm

I=5amp, V=220volt, t=2h

P=?, E=?

P=VXI

  =220X5

  =1100watt

  =1.1kW

Energy consumed, E=PXt

                            =1.1X2

                            =2.2kWh

Case 1: TV set

P=250W=0.25kWh

t=1h

Energy consumed=PXt=0.25X1=0.25kWh

Case 2: Toaster

P=1200W=1.2kW, t=10min=10/60=1/6h

Energy consumed=PXt=1.2X(1/6)=0.2kWh

Thus, TV uses more energy.

Given 2 lamps:P₁=40W, P₂=60W

V=220V

(a)

(b) Voltage across both the bulbs is same and is equal to 220V.

Current through 40W lamp = I₁ = P₁/V = 40/220 A

Current through 60W lamp = I2 = P2/V = 60/220 A

Total current drawn from the electric supply = 40/220 + 60/220 = 0.45 A

(a) Energy consumed by 40 W lamp in 1 hr, E₁ = P₁ x t = 40 x 1 = 40Wh 

1Wh = 3.6 kJ

E₁ = 40 x 3.6 = 144 kJ

Energy consumed by 60W lamp in 1 hr, E₂ = P₂ x t = 60 x 1 = 60Wh = 216 kJ

Total energy consumed = 144 + 216 = 360 kJ

Given V=230V, I=10amp

(a) P=VI

P=230X10

P=2300watt = 2300 J/s

(b) Energy consumed in minute = P x t = 2300 J/s x 60s = 138000 J

For heater:

P=2kW, t=4h

E=Pxt=2x4=8kWh

For TV:

P=200W=0.2kW, t=4h

E=Pxt=0.2x4=0.8kWh

Lamps:

P=100W=0.1kW, t=4h, n=3

E=nxPxt=3x0.1x4=1.2kWh

Total energy consumed = 8+0.8+1.2 = 10kWh

Cost of 1kWh = Rs. 5.50

Cost of 10kWh = Rs. 5.50 x 10 = Rs. 55

I=13amp, V=230V

Power=VI

        =230X13

        =2990W

P=2.99kW

Given :- V=230V, I=0.4amp

Rate at which electric energy is transferred = Power

Power = VxI

          = 230 x 0.4

          = 92 W = 92 J/s

(a) The rate at which electrical work is done or the rate at which electrical energy is consumed, is known as electric power.

It is given by

P=VI=watt

(b) Given: V=3V, I=0.5amp

(i) R=?

3=0.5R

R=6ohms

(ii) Power of lamp=VI

                        =3x0.5

                        =1.5watt

(c) One kilowatt hour is the amount of electrical energy consumed when an electrical appliance having a power rating of 1 kilowatt is used for 1 hour.

1kWh=3.6x10⁶J

(d) Given P=500W=0.5kW, t=20hr

We know that

Energy consumed =Pxt=0.5X20

                         =10kwh

Total cost =10xcost per uint

Cost per unit=Rs. 3.9 per unit

Therefore, total cost=10x3.9=Rs 39

Given

P=4kw, V=220v

(a) I=?

Power=VI=250XI

4000=250I

I=16amp

(b) R=?

P=I²R

P=16²XR

R=4000/16²

R=15.25ohm

(c) Energy consumed in two hour=PXt

=4X2

=8kw-hr

(d) If 1kwh=Rs 4.6

total cost=8 x 4.6=Rs 36.8

By reducing the length of element the resistance will decrease.

Power is inversely proportional to resistance. So, this will result in more consumption of energy.

(a) Lamp; because least current is flowing through it.

(b) Large current drawn by the kettle; Earth connection needed.

(c) We know that

P=VI

V=240V, I=8.5A

P=240X8.5=2040W=2.04kW

(d) When connected to 240 V supply, P=2040W

R = V²/P = 240²/240

R=28.23ohm

Now, when V=120V, R=28.23ohm

I=V/R=120/28.23=4.25A

(a) 42919

(b) 42935

(c) 42935-42919=16 units

(d) 24 hours

(e) Cost of 1 unit = Rs. 5

Cost of 16 units = 16x5 = Rs. 80

P=10W, V=220V, I=5A

We know that

P=VI

=220X5

P=1100W

Power of one bulb=10W

Total no. of bulbs that can be connected=1100/10=110

Heat produced is directly proportional to the square of current.

If current I is doubled, heat H will be four times.

Two effects of produced by electric current are:

(a) Heating effect

(b) Magnectic effect

Heating effect

Heating effect

Electirc heater and electric fuse.

Argon and nitrogen.

Filament type electric bulbs are not power efficien because most of the electric power consumed by the filament of a bulb appears as heat and only a small amount of electric power is converted into light.

The connecting cord of the heater made of copper does not glow because negligible heat is produced in it by passing current (because of its extremely low resistance); but the heating element made of nichrome glows because it becomes red-hot due to the large amount of heat produced on passing current (because of its high resistance).

(a) Heat produced, H=I²Rt

(b) Given: R=20ohm, I=5amp, t=30s

We know that H=I²Rt

H=5²x20x30

H=15000 J

Heat produced by an electric current depends on the following factors:

(i) Heat produced is directly proportional to square of current.

(ii) Heat produced is directly proportional to resistance.

(iii) Heat produced is directly proportional to the time for which current flows.

(a) Joule's law of heating states that heat produced in joules when a current of I amperes flows in a wire of resistance R ohms for time t seconds is given by H = I²Rt.

Thus the heat produced in a wire is directly proportional to:

(i) Square of current

(ii) Resistance of wire

(iii) Time for which current is passed

We know that

Total resistance, R=40+60=100ohms

By Ohm's law,

V=IR

I=V/R

I=220/100=2.2amp

Putting the values of I, R and t in eq.  H=I²RT

H= 2.2²  X  100  X  30

H= 14520 J

If air is filled in an electric bulb, then the extremely hot tungsten filament would burn up quickly in the oxygen of air. So, the electric bulb is filled with a chemically unreactive gas like argon or nitrogen. Thes gases do not react with the hot tungsten filament and hence prolong the life of the filament of the bulb.

Tungsten is used for making the filaments of electric bulbs because it has a very high melting point. Due to its very high melting point, the tungsten filament can be kept white hot without melting away. Also, tungsten has  high flexibility and low rate of evaporation at high temperature.

The connecting wires of the heater get only slightly warm because they have extremely low resistance due to which negligible heat is produced in them by passing current.

Given: I=4amp, t=10min=10x60=600sec, H=2.88x10⁴J

(a) We have

H=I²RT

28800=4²xRx600

R=3ohms

We know that

P=I²xR

  =4²x3

P =48W

(b) V=?

We know that

V=IR

V=4x3

V=12V

Given: R=200ohms, I=2.5amp, t=1sec

We know that

H=I²RT

H=2.5²X200X1

H = 1250 J/s

Given: R=8ohms, I=15amp, t=1sec

We know that

H=I²RT

H=15²X8X1

H=1800J/s

Given: R=25ohms, V=12V, H=?, t=60sec

V=IR

12=25XI

I=0.48amp

We have

H=I²RT

H=0.48²X25X60

H=345.6J

Given: H=100J, t=1sec, R=4ohms,

We know that

H=I²RT

100=I²X4X1

100/4=I²

I=5amp

V=IR

V=5X4

  =20V

P=VI

I=P/V=12/12=1A

V=IR

R=V/I=12/1=12ohm

H=I²Rt 

H=1²x12x60

H=720J

(c)  The heat produced by the heater will become one-fourth because heat produced is directly proportional to the square of the current.

(d) When an electric current is passed through a high resistance wire, the wire becomes very hot and produces heat. This effect is knows as heating effect of current. This effect is used in room heaters and electric ovens.

 (e) Tungsten is used for making the filaments of an electric bulb.

(a) S; because it has high resistivity of 11X10^-7ohm-m (it is actually nichrome).

(b) Q; because it has very low resistivity of 1.7 X10^-8ohm-m (it is actually copper).

(c) R; because it has very very high resistivity of 1.0X10¹⁵ohm-m (it is actually rubber).

(a) The filament wire becomes white hot where as other wires in the circuit do not get heated much.

(b) High resistance of filament wire accounts for this difference.

In series, because total resistance in series connection is more than that in parallel connection.

Given: V=220V, P_min=360W, P_max=840W

For minimum heating case:

We know that

P_min=VI

360=220XI

I=1.63amp

R=V/I

R=220/1.63

R=134.96ohms

For maximum heating case:

We know that

P_max=VI

840=220XI

I=3.81amp

R=V/I

R=220/3.81

R=57.74ohms

Electric iron , electric oven, water heater, room heater.