# Question-1

**What does an electric circuit means?**

**Solution:**

The continuous and closed path of an electric current is called an electric circuit. If the circuit is broken anywhere the current stops flowing.

# Question-2

**Define the unit of current.**

**Solution:**

The unit of current is ampere(A). 1 ampere is constituted by the flow of 1 coulomb of charge per second, .

# Question-3

**Calculate the number of electrons constituting one coulomb of charge.**

**Solution:**

The S.I unit of electric charge is coulomb (C), which is equivalent to the charge contained in nearly 6x10

^{18}electrons.

# Question-4

**Name a device that helps to maintain a potential difference across a conductor.**

**Solution:**

Voltmeter is a device that helps to maintain a potential difference across a conductor.

# Question-5

**What is meant by saying that the potential difference between the two points is 1 V?**

**Solution:**

1 V is the potential difference between the two points in a current carrying conductor when 1 joule of work is done to move a charge of 1 coulomb from one point to other.

Therefore,

1 V=1 JC

^{-1 }

# Question-6

**How much energy is given to each coulomb of charge passing through a 6 V battery?**

**Solution:**

Energy, W = VQ = 6 V × 1.6 × 10

^{-19 }C = 9.6 × 10

^{-19}C

Therefore, 9.6 × 10

^{-19}C of energy is given to each coulomb of charge passing through a 6 V battery.

# Question-7

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

**Solution:**

The resistance of a conductor depends on its Length, Area of cross-section and on the nature of its material.

# Question-8

**Will current flow more easily through a thick wire or a thin wire of the same material, When connected to the same source? Why?**

**Solution:**

The current flow more easily through a thick wire because are Area µ thickness

Resistance ∝

Current ∝

If thickness more area is more, area is more, resistance is less and hence current flow is more.

# Question-9

**Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?**

**Solution:**

2I

_{2}=I

_{1}

therefore,

So that the current decreases to half of its former value.

# Question-10

**Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?**

**Solution:**

The resistivity of an alloy is generally higher than that of its constituent metals. Alloys do not oxidise readily at high temperatures. For this reason alloy is commonly used in electric toasters and electric irons.

# Question-11

**Use the data in the table to answer the following:**

**(a) Which among iron and mercury is a better conductor?**

**(b) Which material is the best conductor?**

**Solution:**

(a) Iron is the better conductor than mercury.

(b) Silver is the best conductor.

# Question-12

**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-13

**Redraw the circuit of the previous problem, 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-14

**Judge the equivalent resistance when the following are connected in parallel****(a) 1 Ω and 10 ^{6} Ω **

**(b) 1 Ω and 10 ^{3 }Ω^{ }and 10^{6 }Ω **

**Solution:**

(a) R

_{1}=1 Ω , R

_{2}=10

^{6}Ω

_{}

R

_{T}=1 Ω.

(b) R1=1 Ω , R2=10

^{3 }Ω

^{ }, R

_{3}=10

^{6 }Ω

^{ }

=1

Therefore, R

_{T }=1 Ω.

# Question-15

**An electric lamp of 100 Ω , a toaster of resistance 50 Ω , and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all the three appliances, and What is the current through it?****Solution:**

R

_{1 }= 100 Ω , R

_{2 }= 50 Ω , and R

_{3 }= 500 Ω

= 0.032 Ω

We know that, V=220 V

Total current,

= 220 V x 0.032 Ω

= 7.04 V

resistance of electric iron =

=

= 31.25 Ω.

# Question-16

**What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?**

**Solution:**

A parallel circuit divides the current through the electrical gadgets. The total resistance in a parallel circuit is decreased. This is helpful particularly when each gadget has different resistance and requires different to operate properly.

# Question-17

**How can three resistors of resistances 2Ω , 3Ω and 6Ω be connected to give a total resistance of (a) 4 Ω (b) 1 Ω**

**Solution:**

(a) Resistance 3 Ω and 6 Ω are connected in parallel and 2 Ω is connected in series with this combination.

(b) All the three resistance are connected in parallel.

# Question-18

**What is (a) the highest, (b) the lowest total resistance that can be secured by combination of four coils of resistance 4 Ω , 8 Ω , 12 Ω , 24 Ω ?****Solution:**

(a) Highest resistance =4 + 8 + 12 + 24 = 48 Ω

All resistance should be connected in series.

(b) If all the resistance are connected in parallel the lowest resistance

= 0.5

R_{T} = = 2 Ω.

# Question-19

**Why does the cord of an electric heater not glow while the heating element does?**

**Solution:**

# Question-20

**Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.**

**Solution:**

Give charge(Q) = 96000 C

Time (t)= 1hr= 60*60= 3600s

Potential difference V = 50 Volts

Now we Know that H= V/t

So we have to calculate first

Ad I = Q/t

I = 96000/3600

=80/3 A

Now we can calculate the heat generated

H= 50* 80*3600/3 J

= 50*80*1200 J

=4800000J

= 4.8*10

^{6}J.# Question-21

**An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.****Solution:**

R = 20 Ω , I = 5 A, t = 30 s.

H = I

^{2}Rt

= 25 × 20 × 30

= 15000 J

The heat developed in 30 s = 15000 J.

# Question-22

**What determines the rate at which energy is delivered by a current?**

**Solution:**

The rate at which electric energy is consumed in an electric circuit is termed as electric power. The power P is given by

P=VI

Or, P=I

^{2}R =

# Question-23

**An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 hr.**

**Solution:**

V = 220 V, I = 5 A

Power, P = VI = 220 V x 5 A = 1100 W

Energy=Pt = 1100 W x 2 hr = 2200 W hr.

# Question-24

**A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R’, then the ratio R/R’ is(a) 1/25 (b) 1/5 (c) 5 (d) 25**

**Solution:**

(d) 25

# Question-25

**Which of these following terms does not represent electrical power in a circuit?**

**(a)**

**I**

^{2}R (b) IR^{2}(c) VI (d)**Solution:**

(b) IR

^{2 }

# Question-26

**An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be (a) 100 W (b) 75 W (c) 50 W (d) 25 W**

^{ }

**Solution:**

(c) 50 W

# Question-27

**Tow conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be (a) 1:2 (b) 2:1 (c) 1:4 (d) 4:1**

**Solution:**

(c) 1:4

# Question-28

**How is a Voltmeter connected in the circuit to measure the potential difference between two points?**

**Solution:**

A Voltmeter should be connected between the two points in the circuit to measure the potential difference.

# Question-29

**A copper wire has diameter 0.5 mm and resistivity of 1.6x10**^{-8 }Ω^{ }m. What will be the length of this wire to make its resistance 10 Ω ? How much does the resistance change if the diameter is doubled?**Solution:**

R=10 Ω , r=

^{ }

l=1.23

The length of this wire is 1.23

^{ }

If d

_{1 }=2d, r

_{1 }=2r

Therefore the resistance reduces by quarter of the original value.

# Question-30

**The values 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 (Volts) |
1.6 |
3.4 |
6.7 |
10.2 |
13.2 |

** **

**Solution:**

=2 Ω.

# Question-31

**When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.**

**Solution:**

V=12 V, I=2.5x10

^{-3 }mA

R=4.8x10

^{-3}Ω

# Question-32

**A battery of 9 V is connected in series with resistors of 0.2 Ω , 0.3 Ω , 0.4 Ω , 0.5 Ω and 12 Ω , respectively. How much current will flow through the 12 Ω resistor?****Solution:**

Current through 12 Ω resistor= = = 0.67 A.

# Question-33

**How many 176 Ω resistors are required to carry 5 A on a 220 V line?****Solution:**

V = 220 V, I = 5 A

= = 44 Ω

= 4

Therefore no of resistors = 4.

# Question-34

**Several electric bulbs designed to be used on a 220 V electric supply line , are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?**

**Solution:**

V=220 V

P=10 W

=4840 Ω

R_{T}=__220 __

5

=44

No of lamps=

=110.

# Question-35

**A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistance, which may be used separately, In series, or in parallel. What are the currents in the three cases?****Solution:**

V=220 V, R=24 Ω

When the resistance coils are used separately, current

=9.17 Ω

When the resistance are connected in series,

=4.58 Ω

When the resistance are connected in parallel, I=

__ __

=18.33 Ω

# Question-36

**Compare the power used in the 2 Ω resistor in each of the following circuits**

**(i) A 6 V battery in series with 1 Ω and 2 Ω resistors, and **

**(ii) A 4 V battery in parallel with 12 Ω and 2 Ω resistors. **

**Solution:**

(i)The voltage across 2 Ω resistor=6x

=4 V

Power used in the 2 Ω resistor=

^{ }

=

=8 W

(ii) The voltage across 2 Ω resistor=4 V

Power used in the 2 Ω resistor=

^{ }

=

=8 W.

# Question-37

**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-38

**Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?**

**Solution:**

Energy=Pt

Energy in TV set =250x1

=250 Whr

Energy in toaster =1200x

=200 Whr

Therefore TV set consumes more energy than toaster.

# Question-39

**An electric heater of resistance 8 Ω draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater**.**Solution:**

I=15A, R=8 Ω

=28.125

The rate at which heat is developed in the heater is 28.125

# Question-40

**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.