# Question-1

**Why is an ammeter likely to be burnt out of you connect it in parallel?**

**Solution:**

If an ammeter is connected in parallel, the resultant resistance of the circuit decreases and more current passes through the instrument. Hence, the ammeter is likely to be burnt out.

# Question-2

**State Ohm’s Law.**

**Solution:**

where I is the current through the conductor in units of amperes, V is the potential difference measured across the conductor in units of volts, and R is the resistance of the conductor in units of ohms. More specifically, Ohm's law states that the R in this relation is constant, independent of the current.

# Question-3

**What are the special features of a heating wire?**

**Solution:**

A heating wire must have high resistance and high melting point.

# Question-4

**What is the role of MnO**

_{2}in a dry cell?**Solution:**

MnO

_{2}acts as a depolariser in the dry cell to maintain the steady flow of current.

# Question-5

**What is the role of a fuse in electric circuits?**

**Solution:**

A fuse is a safety device. It is a short thin wire made of tin-lead alloy, which has a low melting point. As a result, this wire melts and breaks the circuit if the current exceeds a safe value. This saves costly electric appliances and buildings from damage.

# Question-6

**Name the device, which converts mechanical energy into electric energy.**

**Solution:**

Electrical Generator is used to convert electric energy into mechanical energy for doing work.

# Question-7

**What is the usual colour code followed for connecting live, neutral and earth wires? Why is it so important?**

**Solution:**

The following is the colour code:

Live wire - Red, Neutral wire - Black and Earth wire - Green.

This colour code of electric wires must be followed, so that even a layman will be able to identify the correct connecting wire.

# Question-8

**Explain what is electric potential and potential difference.**

**Solution:**

Electric potential or electrostatic potential at a point in an electric field is defined as the work done in bringing a unit positive charge from infinity to that point. The S.I. unit of electric potential is volt.

Potential difference between two points A and B in an electrostatic field is the amount of work done in carrying unit positive charge from A to B along any path between the two points.

# Question-9

**A current of 0.5A flows through a conductor having resistance of 10 ohms for ½ hr. Find the amount of electric energy consumed by this conductor.**

**Solution:**

Current I = 0.5A; Resistance R = 10 ohm

T = (1/2) hr = 30 x 60 = 1800 seconds

Electric energy consumed = i^{2}Rt = (0.5)^{2} x 10 x 1800 = 4500 Joule.

# Question-10

**What is the resistance of an ideal ammeter?**

**Solution:**

The resistance of an ideal ammeter is zero.

# Question-11

**Which one has more resistance –a 100 watt bulb or 60 watt bulb?**

**Solution:**

The resistance of a 60 watt bulb is greater than the resistance of a 100 watt bulb.

# Question-12

**Define the principle of conservation of electric charge. Explain it with an example.**

**Solution:**

The total electric charge in an isolated system is conserved. Electric charge can neither be created nor destroyed.

If a metal conductor kept on an isolated stand has a charge of 12 units, and it is brought into contact with an identical uncharged insulated metal conductor and then removed, the charge on each metal conductor will be 6 units, so that the total charge is 6 + 6 = 12 units, which remains constant or conserved.

# Question-13

**How does the current flow in a wire?**

**Solution:**

Electric current is the flow of electrons in a metal wire when a cell or battery is applied across its ends. A metal wire has a plenty of free electrons in it.

(i) When the metal wire has not been connected to a source of electricity like a cell or a battery, the electrons present in it move at random in all directions between the atoms of the metal wire.

(ii) When a source of electricity like a cell or a battery is connected between the ends of the metal wire, the electric force acts on the electrons present in the wire. Since the electrons are negatively charged, they start moving from negative end to the positive end of the wire. This flow of electrons constitutes the electric current in the wire.

# Question-14

**What is the resistance of a conductor? On what factors does it depend?**

**Solution:**

The property of a conductor due to which, it opposes the flow of current through it, is called resistance. The resistance of a conductor is numerically equal to the ratio of potential difference across its ends to the current flowing through it, i.e

Resistance = Potential difference / Current

R = V/I

The resistance of a conductor depends on the length, thickness, nature of material and temperature of the conductor. A long wire has more resistance and a short wire has less resistance. A thick wire has less resistance and a thin wire has more resistance. The rise in temperature of a wire increases its resistance.

# Question-15

**Derive the expression for the resistivity of a substance and give the meaning of each symbol which occurs in it.**

**Solution:**

The resistance of a given conductor is directly proportional to its length, i.e.,

R ∝ l ----(1)

The resistance of a given conductor is inversely proportional to its area of cross-section, i.e.,

R ∝ ----(2)

From equation (1) and (2), we get

R ∝

R = ----(3)

where p is the constant known as resistivity of the material of the conductor.

Rearranging equation (3), we get

p =

where

R = Resistance of the conductor

A = Area of cross-section of the conductor

l = length of the conductor

# Question-16

**(a) What happens to the electrical resistance when mercury is cooled to 4.12 K?**

(b) What name is given to this phenomenon?

(b) What name is given to this phenomenon?

**Solution:**

(a) When mercury is cooled to 4.12 K, the electrical resistance of mercury disappears completely and becomes zero.

(b) This phenomenon of loss of electrical resistance of a substance on cooling it to an extremely low temperature is known as superconductivity.

# Question-17

**Derive the formula for electrical power which is used only when current I and resistance R are known to us.**

**Solution:**

We know that P = V × I ----(1)

From Ohm’s law = R

or V = I × R ----(2)

Substituting the value of V in equation (1), we get

P = I × R × I

= I

^{2}× R

# Question-18

**State Coulomb’s law and give the unit of force constant.**

**Solution:**

The electric force between two charges depends directly on the product of charges and inversely as the square of the distance between the charges,

F ∝ q_{1}q_{2} and F ∝ ⇒ F = k

Where q_{1}, q_{2} are the charges, r is the distance between them and k is the constant of proportionality.

# Question-19

**Derive the formula for electrical power, which is used only when voltage V and resistance R are known to us.**

**Solution:**

We know that P = V × I

From Ohm’s law, we have

= R ----(1)

or V = I × R

or I = ----(2)

Substituting the value of I in equation (1), we get

P = V ×

or Power, P =

# Question-20

**State the three factors on which the heat produced by an electric current depends.**

**Solution:**

Heat produced in a wire depends on

(i) the square of the current (I

^{2})

(ii) resistance of wire (R)

(iii) time (t) for which current is passed

Since the heat produced is directly proportional to the square of current,

H ∝ I^{2},

if the current is doubled, the heat produced will become four times and if the current is halved, the heat generated will become one-fourth.

Since the heat produced in a wire is directly proportional to the resistance,

H ∝ R,

if the resistance is doubled, then the heat produced will also get doubled and if the resistance ishalved, the heat produced will also be halved. This means that a given current will produce more heat in a high resistance wire than a low resistance wire.

Since the heat produced in a wire is directly proportional to the time for which current flows

H ∝ t,

if the current is passed through a wire for double the time, then the heat produced is doubled and if the time is halved, the heat produced is also halved.

# Question-21

**What is resistance? How is resistance caused. What are the factors it depends? How does it vary with temperature?**

**Solution:**

The resistance of a conductor is the ratio of potential difference between the ends of a conductor to the current flowing through it.

Resistance =

⇒ R =

S. I. unit of resistance is ohm.

1 ohm = .

As electrons inside a conductor are not entirely free to move, their motion is restrained by the attraction to the atoms. This retards the motion of electrons through the conductor, and is called resistance.

The resistance of a conductor depends on its length, cross sectional area and resistivity.

R = ρ (Where l is length, A is cross sectional area, ρ is resistivity).

As resistivity changes with temperature, resistance also depends on temperature.

Resistance of a conductor increases with increase in temperature.

# Question-22

**Define electric Power. What is its unit?**

**Solution:**

The rate at which electrical energy is consumed or dissipated is known as electric power.

Electric power = = = I^{2} R. Its unit is watt.

# Question-23

**An electric bulb is rated at 220V, 200W. What is its resistance?**

**Solution:**

Power, P = ⇒ 200 =

⇒ R = = 242 ohms.

# Question-24

**A torch bulb is rated at 2.5V, 500 A. Find the power and resistance.**

**Solution:**

R = = = (200 mA = 0.2 ampere)

Or R = 12 ohms.

# Question-25

**What is the S.I. unit of Power?**

**Solution:**

The S. I. of Power is Watt, which is denoted by the letter W. The power of 1 watt is a rate of working of 1 joule per second. That is,

1 watt = .

# Question-26

**A heater coil connected to a 220 V ha a resistance of 150Ω . How long will it take for this coil to heat 1 kg of water from 20**

^{o}C to 60^{o}C, assuming that all the heat is taken up by the water?**Solution:**

Heat produced by coil = . t = . t (where t = time period)

Heat required by water = m. s. θ

Where m = mass of water, s = specific heat of water = 4180 joules/kg. and θ is increase in temperature = 60 – 20 = 40.

Equating the heat produced and heat required.

. t = 1 × 4180 × 40 ⇒ = 518 seconds. (Time taken to heat water up to 60° C)

# Question-27

**How much current will an electric heater draw from a 220 volt line if the resistance of the heater is 44 ohms?**

**Solution:**

V = 220 volt, R = 44 ohms,

By Ohm’s law, V = IR

Or I = = = 5 amperes.

# Question-28

**Potential difference across the end of a wire is 0.6V which carries a current of 8 amperes. Find the resistance of the wire.**

**Solution:**

By Ohm’s law, the resistance of a wire, R = = = 0.075 ohm.

The resistance of the wire is 0.075 ohms.

# Question-29

**In the given circuit, calculate the current shown by the ammeter and total resistance of the circuit.**

**Solution:**

In the circuit, two resistances of 2 Ω each are connected in series and then connected to 4Ω in parallel.

The equivalent resistance of resistances in series = 2+2 = 4Ω .

The equivalent resistance of 4Ω and the resultant of two 2 ohms (4Ω )

= + = + = =

or R = 2 ohms.

As total equivalent resistance is 2 ohms,

Current in circuit will be = = 2 amperes.

The ammeter will show a current of 2 amps, and the total resistance of the circuit is 2 ohms.

# Question-30

**An electric bulb of 40W is connected to a source of 2 20v. What will be the current drawn and the resistance of the bulb?**

**Solution:**

In this case we have been given power P and voltage V, so the formula to be used for calculating the current will be:

P = V × I

Here, Power, P = 40 watts

Voltage, V = 220 volts

And, Current, I =? (To be calculated)

Now, putting these values in the above formula, we get:

40 = 220 × I

I = =

Thus, Current, I = 0.18 amperes.

# Question-31

**Read the following statements. Write true or false against each:**

(i) The quantity of charge flowing past a point multiplied by the time gives the current.

(ii) The flow of charge through a conducting wire, connected to a cell, is due to the chemical reaction inside the cell.

(iii) The resistivity of all pure metals increases with rise in temperature.

(iv) Ohm’s law is a relation between the power used, the current and the potential difference in a circuit.

(v) A series circuit has only one conducting path for the electrons that move through it; a parallel circuit has multiple conducting paths.

(vi) A conducting wire offers resistance to the flow of electrons because electrons repel each other in the wire.

(i) The quantity of charge flowing past a point multiplied by the time gives the current.

(ii) The flow of charge through a conducting wire, connected to a cell, is due to the chemical reaction inside the cell.

(iii) The resistivity of all pure metals increases with rise in temperature.

(iv) Ohm’s law is a relation between the power used, the current and the potential difference in a circuit.

(v) A series circuit has only one conducting path for the electrons that move through it; a parallel circuit has multiple conducting paths.

(vi) A conducting wire offers resistance to the flow of electrons because electrons repel each other in the wire.

**Solution:**

(i) False. The current is defined as the amount of charge flowing per second. I = Q / t.

(ii) True.

(iii) True.

(iv) False, Ohm’s law is a relation between the current in a wire and the potential difference between its ends (V = IR)

(v) True.

(vi) False, the resistance is because of attraction of the atoms among which they move

# Question-32

**Explain the following:**

(i) Why is tungsten used almost exclusively as filament for incandescent lamps?

(ii) Why are the conductors of electric heating devices, such as toasters and electric irons, made of an alloy rather than a pure metal?

(iii) Why is the series arrangement not used for domestic circuits?

(iv) How does the resistance of a wire vary with its cross-sectional area?

(i) Why is tungsten used almost exclusively as filament for incandescent lamps?

(ii) Why are the conductors of electric heating devices, such as toasters and electric irons, made of an alloy rather than a pure metal?

(iii) Why is the series arrangement not used for domestic circuits?

(iv) How does the resistance of a wire vary with its cross-sectional area?

**Solution:**

(i) A thin wire of tungsten has high resistance and does not get oxidized easily at high temperatures. Hence a tungsten filament is used for making filaments of electric bulbs.

(ii) The resistance of all pure metals increases on raising the temperature and decreases on lowering the temperature. But the resistance of alloys like manganin, constantan and nichrome is almost unaffected by temperature.

(iii) If we connect all the electrical appliances like bulbs, fans and sockets, etc., in a series, then if one appliance is switched off or gets fused then all the appliances will also stop working because their electricity supply will be cut-off.

(iv) When the area of a cross-section of a wire is doubled, its resistance gets halved; and if the area of a cross-section of wire is halved, then its resistance will get doubled. Hence a thick wire which has a greater cross-section has less resistance and a thin wire which has a smaller area of cross-section has high resistance.

# Question-33

**A piece of wire having resistance, R is cut into four equal parts.**

(i) How will the resistance of each part compare with the original resistance?

(ii) If the four parts are placed in parallel, how will the resistance of the combination compare with the resistance of the original wire?

(i) How will the resistance of each part compare with the original resistance?

(ii) If the four parts are placed in parallel, how will the resistance of the combination compare with the resistance of the original wire?

**Solution:**

(i) Since the resistance R is cut into 4 equal parts each part has 1/4

^{th}of the total length. Thus the resistance is 1/4

^{th}the original resistance i.e. each has a resistance of R/4. This is because R is proportional to the length of the wire.

(ii) When the 4 resistances of resistance R/4 each are connected in parallel the equivalent resistance of the circuit is R/16, i.e. it is 1/16

^{th}of the original resistance R.

# Question-34

**A piece of wire is redrawn by pulling it until its length is doubled. Compare the new resistance with the original value?**

**Solution:**

Let the original dimensions of the wire be of length l, area of cross section A and resistivity ρ .

Therefore resistance R =

Since the total volume of the cylindrical wire is constant, the length of the wire is doubled by pulling it and hence the cross sectional area of the wire is halved.

Therefore the new resistance R′ = = 4R

Thus the resistance is 4 times its original value.

# Question-35

**A copper wire has a diameter of 0.5 mm and resistivity of 1.6 ohm cm. How much of this wire would be required to make a 10 ohm coil?**

**Solution:**

Diameter of copper wire, D = 0.5mm

Therefore Area of cross section A = = 0.1963 mm^{2 }

Resistivity of the copper wire ρ = 1.6 ohm cm = 16 ohm mm

Let the required length be l

Then l = R ×

=10 ×

= 0.12269 mm

= 122. 7 × 10^{-6} m

# Question-36

**Draw a schematic diagram of a circuit consisting of a battery of four 2 V cells, a 5 ohm resistor, an 8 ohm resistor and a 12 ohm resistor and a plug key all connected in series.**

**Solution:**

# Question-37

**Redraw the circuit of problem 6, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the voltage across the 12 ohm resistor. What would be the readings in the ammeter and the voltmeter.**

**Solution:**

Total voltage applied V = 4 × 2 = 8 volts

Total resistance R = 5 + 8 + 12 = 25 Ω

Therefore reading of the ammeter I = = 0.32 A

Reading of voltmeter V = IR = 0.32 × 12 = 3.84 V.

# Question-38

**Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to a 220 V supply. What current is drawn from the supply line?**

**Solution:**

Let the current drawn from the supply be I and the individual currents drawn by each lamp be I

_{1}and I

_{2}respectively. Since they are connected in parallel the voltage drop ‘V’ across each lamp is the same, which is also equal to the supply voltage 220 V.

Then current in lamp 1

I_{1} = = 0.4545 A

Similarly current in lamp 2

I_{2} = = 0.2727 A

Therefore the net current drawn I = I_{1} + I_{2} = 0.7272 A

# Question-39

**A heater coil connected to a 220 V has a resistance of 150 ohm. How long will it take for this coil to heat 1 kg of water from 20**

^{0}C to 60^{0}C, assuming that all heat is taken up by water?**Solution:**

Mass of the water, m = 1 kg

Specific heat capacity of water, C = 4200 J / kg /

^{0}C

Change in temperature, (t

_{2 }- t

_{1}) = 60

^{0}C – 20

^{0}C = 40

^{0}C

Therefore the heat required to raise the temperature of 1 kg of water from 20

^{0}C to 60

^{0}C

Q = mC(t_{2}-t_{1})

= 1 × 4200 × 40

= 168000 joules

Heat produced Q = joules

i.e. 168000 =

Therefore time taken t = = 520.66 s

# Question-40

**In the circuit shown in figure below, calculate (i) the current flowing through the arms AB, AC and CDE; (ii) the potential difference across AB, CD and DE.**

**Solution:**

(i) Current through AB = = 0.125 A

Current through AC is given by the resistors 6Ω and 4Ω , hence

= 0.15 A

Current through CDE = = 0.15 A

(ii) Potential difference across AB = IR = 0.125 × 12

= 1.5 V

Potential difference across CD = 0.15 × 6 = 0.9 V

Potential difference across DE = 0.15 × 4 = 0.6 V

# Question-41

**A 40 watt lamp requires 0.182 A of current at 220 volts, while a 60 watt lamp requires 0.272 A of current at 220 volts. If a 40 watt lamp and a 60 watt lamp are connected in series with a 220 volt line, how many amperes of current will flow through each lamp?**

**Solution:**

Let the current flowing through the two lamps, which are connected in series, be I.

Let their individual resistances be R

_{1}and R

_{2}and their power be P

_{1}and P

_{2}

Then for the lamp1 R

_{1}= = 1210 Ω

Similarly for the lamp2 R

_{2}= = 806.67 Ω

Therefore when they are connected in series the net resistance of the two lamps is the sum of their individual resistances

R = R_{1} + R_{2} = 2016.67 Ω

Therefore the current I = = 0.109 A.

# Question-42

**Five dry cells each of 1.5 volt have internal resistance of 0.2, 0.3, 0.4, 0.5 and 12 ohms. When connected in series, what current will leave these five cells furnish through 10 ohm resistance?**

**Solution:**

Total voltage produced by the batteries V = 5 × 1.5 = 7.5 V

Total resistance R = Internal resistance + external resistance

= (0.2 + 0.3 + 0.4 + 0.5 + 1.2) + 10 = 12.6 Ω

Therefore current I = = 0.595 A.

# Question-43

**The value of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below:**

I (amperes) 0.5 1.0 2.0 3.0 4.0

V (volt) 0.5 2.5 6.75 11.0 15.0.

I (amperes) 0.5 1.0 2.0 3.0 4.0

V (volt) 0.5 2.5 6.75 11.0 15.0.

**Solution:**

Since V is an independent variable it is taken along x-axis. The dependent I variable is taken along y-axis.

Resistance of the resistor R = = = 3.22 Ω.

# Question-44

**Calculate the resistance of one metre of copper wire that has a cross–sectional area of about 2 x 10**

^{-2}cm^{2}(using the value of resistivity of copper to be 1.2 × 10^{-8 }ohm m) Compare the value of this resistance with that of a flashlight bulb, which has a power rating of 1 W and operates at 3 V. What does this comparison tell you ?**Solution:**

Length of the wire l = 1 m

Area of cross section of the wire A = 2 × 10

^{-2}cm

^{2}

= 2 × 10

^{-6}m

^{2}

Resistivity of copper ρ = 1.2 × 10

^{-8 }ohm m

Therefore R = = = 0.6 × 10

^{-2}Ω

The resistance of a flash bulb of power rating 1W and voltage 3V has a resistance

R′ = = 9 ΩThis clearly shows that the resistance of a flashbulb is much greater than the resistance of the copper wire of 1m length.