Ncert Science Class 10 Solutions Chapter 11 Electricity

Welcome to NCTB Solutions. Here with this post we are going to help 10th class students for the Solutions of NCERT Class 10 Science Book Chapter 11, Electricity. Here students can easily find step by step solutions of all the questions in Electricity. Also our Expert Science Teacher’s solved all the problems with easily understandable language with proper guidance so that all the students can understand easily. Here in this post students will get chapter 11 solutions. Here all Question Answer are based on NCERT latest syllabus.

Electricity Exercise question Solutions :

(1) 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

Answer :

The correct option is – (d) 25

As given R is resistance cut in to five equal parts,

So the each resistance Is R/5

Now if resistance is connected in parallel.

1/Rsum = 1/R1 + 1/R2 + 1/R3 + 1/R4 + 1/R5

Now putting each value of resistance,

1/Rsum = 5/R1 + 5/R2 + 5/R3 + 5/R4 + 5/R5

1/Rsum = 25/R

So, Rsum = R/25

And now ratio of R/Rsum = 25

(2) Which of the following terms does not represent electrical power in a circuit?

Answer :

The appropriate answer is option – (b) IR2

As we know P = VI……………..c

And From ohm’s law R = V/I

– V = RI

– I = V/R

P = (RI)I

P = RI2…………………………..a

And also,

P = VI

Putting value of I = V/R

P = V2/R…………………………d

(3) An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be

Answer :

The answer is alternative – (d) 25 W

Power consumed is given by

P = VI

P = V2/R

We Have Voltage and Power So we can use P = V2/R

R = V2/P

R = 220 × 220/100

R = 22 × 22

R = 284 ohm

Now when use 110 volt

P = V2/R

P = 110 × 110/484

P = 25 Watt

Therefore, the Answer is 25Watt

(4) Two 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

Answer :

The correct option is – (c) 1 : 4

Heat produced is given by

H = Pt

H = t × V2/R

Now heat produced in series

Hseries = t × V2/RS

Now heat produced in Parallel

Hparallel = t × V2/RP

Now removing Common,

Ratio = Hseries/Hparallel

= 1/RS/1/RP

= RP/RS ……eq 1

Now, Rs = R + R

Rs = 2R

and Rp = 1/1/(r+r)

RP = 1/1/2r

RP = R/2

Putting value in eq 1

Ratio = R/2/2R

Ratio = 1/4

There fore, the answer is 1/4

(5) How is a voltmeter connected in the circuit to measure the potential difference between two points?

Answer :

Voltmeter should be always connected in parallel between the two points whose potential difference has to measure.

(6) A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10–8 W m. What will be the length of this wire to make its resistance 10 W? How much does the resistance change if the diameter is doubled?

Answer :

We have formulae,

R = ρ 1/a

A is area of conductor,

A = πr^2

A = π〖*0.25〗^2

A = 0.1964 mm2

Now, l = R × A/ρ

l = 10 × 0.1964/1.6 × 10–8

l = 122.71 meter

Now, if diameter is doubled

A = πr^2

A = π〖*0.5〗^2

A = 0.7854 mm2

Now, R = ρ l/A

R = 1.6 × 10-8 × 122.71/0.7854

R = 2.5 Ohm

(7) The values of current (I) flowing through a given resistor of resistance (R), for the corresponding values of potential difference (V) across the resistor are as given below

V (volt)	0.5	1.0	1.5	2.0	2.5	3.0	4.0	5.0
I (amperes)	0.1	0.2	0.3	0.4	0.5	0.6	0.8	0.9

Plot a graph between current (I) and potential difference (V) and determine the resistance (R) of the resistor.

Answer :

(8) 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.

Answer :

As per the given question,

V = 12 v

I = 2.5 mA

R = ?

R = V/I

R = 12/2.5 × 10-3

R = 4.8 × 103 ohm

R = 4.8 KΩ

Therefore, the value of the resistance of the resistor will be 4.8 KΩ

(9) 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 would flow through the 12 Ω resistor?

Answer :

V = 9v

R = 0.2,0.3,0.4,0.5,12 Ω

Resister are connected in series,

Rsum = 0.2 + 0.3 + 0.4 + 0.5 + 12

= 13.4 Ω

R = V/I

I = V/R

I = 9/13.4

I = 0.671 Ampere.

Hence, 0.671 Ampere current will flow through the 12 Ω resistor.

(10) How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?

Answer :

V = 220 v

I = 5 A

R = 176 Ω

Actual resistance

Ra = V/I

= 220/5

= 44

Now these resister are connected in parallel so,

No of resister = R/Ra

= 176/44

= 4

Therefore, the number of resister will be 4.

(11) Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω, (ii) 4 Ω

Answer :

Lets take first 9 Ω,

If we connect resister in series it will be

6 + 6 + 6 = 18 Ω

And if we connect resister in parallel it will be,

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

R = 2

So we have to make combination of series and parallel connection,

one series and two parallel would give,

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

= 6 + 3

= 9 Ω

Now lets take second case,

Two series and one parallel resister would give,

Two series resister = 6 + 6 = 12

Now connect one parallel resister to it,

R = 1/1/12 + 1/6

= 72/18

= 4 Ω

(12) 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?

Answer :

As per the given question,

V = 220 v

P = 10 W

I = 5 A

R = V/I

R = 220/5

R = 44 Ω this is maximum resistance we can design,

Now resistance of each bulb

P = V2/R

R = 220 × 220/10

R = 4840 Ω

Now no of bulb we can connect parallel can be calculated be

1/R = 1/R1 + 1/R2 up to N numbers

1/R = 1/R × N

1/44 = 1/4840 × N

N = 4840/44

N = 110

Thus, we can connect maximum 110 bulbs.

(13) 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?

Answer :

As per the given data,

V = 220 v

R = 24 Ω

(1) When using separately,

I = V/R

I = 220/24

I = 9.166 A

(2) when using in series,

Resistance in series

R = 24 + 24 = 48

I = V/R

I = 220/48

I = 4.58 A

(3) when using in parallel,

Resistance in parallel,

1/R = 1/1/24 + 1/24

1/R = 1/12

R = 12 Ω

I = V/R

I = 220/12

I = 18.33 A

(14) Compare the power used in the 2 Ωresistor in each of the following circuits :

(i) a 6 V battery in series with 1 Ωand 2 Ωresistors,

(ii) a 4 V battery in parallel with 12 Ωand 2 Ω resistors

Answer :

Lets take first case,

V = 6 v

R = 1 + 2

R = 3 Ω

Now, I = V/R

I = 6/3

I = 2

The power across 2 Ω resister can be calculated as,

P = I2R

P = 22 × 2

P = 8 W

Lets take second case,

V = 4 v

And the resister are connected parallel so the voltage across two resister is same, So the power across 2 Ω resister can be calculated as,

P = V2/R

P = 42/2

P = 16/2

P = 8 W

(15) Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?

Answer :

According to the given data,

V = 220 v

P1 = 100 W

P2 = 60 W

Since both bulbs are connected parallelly

Now , P = V × I

I = P/V

I = 100/220 + 60/220

I = 160/220

I = 0.727 Ampere

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

Answer :

According to the question,

P1 = 250 W

T1 = 1 Hour = 3600 seconds

P2 = 1200 W

T2 = 10 Minute = 600 seconds

Now, H = P × t

H is the energy consumed by appliances.

H1 = 250 × 3600

H1 = 900 × 103 joule

H2 = 1200 × 600

H2 = 720 × 103 joule

From comparing both value the TV set uses more energy.

(17) 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.

Answer :

As per the question,

I = 15 A

R = 8 Ω

T = 2 Hour

Now Applying Formula,

P = I2R

P = (15)2 × 8

P = 1800 W or 1800 J/s

The rate of the heater is 1800 J/s.

(18) Explain the following.

Answer :

(a) Why is the tungsten used almost exclusively for filament of electric lamps?

= The tungsten does not burn or break by electricity because, It has high Melting point and resistance. That’s why tungsten is used almost exclusively.

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

= Pure metal have low resistance so other metal is used to form alloy of high resistance, Which can generate high heat when used as conductor.

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

= Because in series arrangement addition of resistance take place, so the voltage gets dropped in next device and device may burn due to handling of large current.

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

= Resistance is directly proportional to the cross sectional area of conductor, that’s why the resistance increases when the area of conductor increased.

(e) Why are copper and aluminium wires usually employed for electricity transmission?

= The conductivity of both metal is high and have low resistance so the copper and aluminum is usually employed for electricity transmission.

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