NIOS Class 10 Science & Technology Chapter 16 Solution

NIOS Class 10 Science & Technology Chapter 16 Solution – Electrical Energy

NIOS Class 10 Science & Technology Solution Chapter 16 Electrical Energy. NIOS Class 10 Science & Technology Chapter 16 Question Answers Download PDF. NIOS Class 10 Science Notes.

NIOS Class 10 Science & Technology Chapter 16 Solution

Board NIOS
Class 10th (Secondary)
Subject Science and Technology (216)
Topic Question Answer, Solution, Notes






1) Define the units of (i) charge (ii) electric potential.

Ans: – (i) When 1C charge is placed at a distance of 1 m from an equal like charge then repeals with a of force of 9 × 109 N.

(ii) volt or electric potential is when 1C positive charge is brought from outside the field to a point against the field 1 J.


2) When a glass rod is rubbed with a piece of silk it acquires +10 micro coulomb of charge. How many electrons have been transferred from glass to silk?

Ans: –We know that number of electron= (coulomb charge/ 1.6×10^-19) = (10×10^-6 / 1.6×10^-19) = 6.25×10^13.


3) How will the force between two small electrified objects vary if the charge on each of the two particles is doubled and separation is halved?

Ans: –We know that electric force is inversely proportional to the distance^2 and directly proportional to charge. So, when the distance will be halved and charge is double then the force will be eight times.


4) How does the force between two small charged spheres change if their separation is doubled?

Ans: – If separation is double then the force will be 1/4th of the previous one.


5) A particle carrying a charge of 1 micro coulomb (μC) is placed at a distanceof 50 cm from a fixed charge where it has a potential energy of 10 J. Calculate

(i) the electric potential at the position of the particle

(ii) the value of the fixed charge.

Ans: – The electric potential is = (10/10^-6) =10^7.

The value of the fixed charge

= (10×0.5 / 9×10^-6×10^9).

= 5/9×10^-3C.


6) Two metallic spheres A and B mounted on two insulated stands as shown in the Fig. 16.4 are given some positive and negative charges respectively. If both the spheres are connected by a metallic wire, what will happen?

Ans: – The electron will flow from point B to point A. This flow will continue until the potential become equal.




1) Define the SI units of (i) current (ii) resistance.

Ans: – The SI unit of current is ampere and the resistance is ohm.


2) Name the instruments used to measure (i) current (ii) potential difference.

Ans: – The instrument is used for the measure of current is ammeter and for the potential difference is voltmeter.


3) Why is a conductor different from an insulator?

Ans: – A conductor passes the current from the one point to the other point whereas the insulator restricted the flow of current.


4) How is a volt related to an ohm and an ampere?

Ans: – We know volt or potential difference

= current × resistance.

Or, v=ir;


5) A number of bulbs are connected in a circuit. Decide whether the bulbs are connected in series or in parallel, when (i) the whole circuit goes off when one bulb is fused (ii) only the bulb that get fused goes off.

Ans: – In case one when the one bulb is goes off or fused then the all-other bulb in the circuit goes off this is an example of parallel connection.

And when the only one bulb is goes off then the connection is series connection.


6) When the potential difference across a wire is doubled, how will the following quantities be affected (i) resistance of the wire (ii) current flowing through the wire?

Ans: – We all know, voltage = current × resistance;

So, when the potential difference will be double then the resistance or current one of then get doubled.


7) What is the reading of ammeter in the circuit given below?

Ans: – 


8) How can three resistors of resistance 2Ω, 3Ω and 6Ω be connected to give a total resistance of (i) 11Ω (ii) 4.5Ω and (iii) 4Ω?

Ans: –  (i) When all are connected is series then the equivalent resistance is (2 +3 + 6)= 11ohm.

(ii) Resistors 2Ω and 6Ω in parallel, but 3Ω is in series to the combination of 2Ω and 6Ω.

(iii) Resistors 3Ω and 6Ω in parallel but 2Ω is in series with the of 3Ω and 6Ω.


9) State two advantages of connecting electrical devices in parallel with the battery instead of connecting them in series.

Ans: – The benefits of parallel connection-

  • Equivalent resistance decrease.
  • If one is affected then remaining all going on.



1) Which will produce more heat in 1 second – 1 ohm resistance on 10V or a 10-ohm resistance on the same voltage? Give reason for your answer.

Ans: – We know, Q/t = v^2 / R,

So, from the upper relation it is clear that the circuit which has more resistance will have low power.


2) How will the heat produced in a conductor change in each of the following cases ?

As heat (H)= I^2RT.

(i) The current flowing through the conductor is doubled.

Ans: – Heat will be four times from previous one.


(ii) Voltage across the conductor is doubled.

Ans: – Heat will be four times from the previous one.


(iii) Time for which current passed is doubled.

Ans: – Heat will be double.


3) 1 A current flows though a conductor of resistance 10 ohms for 1/2 minute. How much heat is produced in the conductor?

Ans: – Heat will be (H) = I^2× R×T,

Or, H= 1× 10× 1/2×60=300j.


4) Two electric bulbs of 40 W and 60 W are given. Which one of the bulbs will glow brighter if they are connected to the mains in (i) series and (ii) parallel?

Ans: – We know, power(p)= v^2/R;

So, from the upper relationship it is clear that the if the resistance is maximum will produce lowest brightness.


5) How is 1 kW h related with SI unit of energy?

Ans: – 1 kilowatt = 3.6 × 10^6j.


6) Name two household electric devices based on thermal effect of electric current.

Ans: –  Heater and electric cooker.




1) Which has a higher resistance, a 40W-220 V bulb, or a 1 kW-220V electric heater ?

Ans: – As we know, p = v ^ 2 / R;

So, 40w will have more resistance.


2) What is the maximum current that a 100W, 220 V lamp can withstand?

Ans: – As we know, I = p/v;

Here, I = ? P= power =100w, v = voltage = 220v,

So, I = 100/220 = 10/22.


3)How many units of electricity will be consumed by a 60 W lamp in 30 days ifthe bulb is lighted 4 hours daily?

Ans: –  Electricity consumed = power × time,

So, Q= 60×30×4= 7.2 KWH.


4)How many joules of electrical energy will a quarter horse power motor consume in one hour?

Ans: –  we know the hors power is 746w,

So, power consume by motor is (746×1/4× 3600) =6714000J.


5)An electric heater is used on 220 V supply and draws a current of 5 A. What is its electric power?

Ans: – Electric power= V×I,

Or, P= 220×5 = 1100W.


6)Which uses more energy, a television of 250 W in 60 minutes or a toaster of1.2 kW in (1/6) th of an hour?

Ans: – The energy used by television is more than the toaster.

E(Television)= 250×1/1000 = 0.25KW,

E(toaster)= 1.2×1/6 = 0.2KW.




1)Tick mark the most appropriate answer out of four given options at the end of each of the following statements:

(a) A charged conductor ‘A’ having charge Q is touched to an identical uncharged conductor ‘B’ and removed. Charge left on A after separation will be:

(i) Q (ii) Q/2 (iii) Zero (iv) 2Q.

Ans: – Option (ii).


(b) J C–1 is the unit of

(i) Current (ii) Charge (iii) Resistance (iv) Potential.

Ans: – Option (iv).


(c) Which of the following materials is an electrical insulator?

(i) Mica (ii) Copper (iii) Tungsten (iv) Iron.

Ans: – Option (i).


(d) The device which converts chemical energy into electrical energy is called

(i) Electric fan (ii) Electric generator

(iii) Electric cell (iv) Electric heater.

Ans: – Option (iii).


(e) The resistance of a conductor does not depend on its

(i) Temperature (ii) Length (iii) Thickness (iv) Shape.

Ans: – Option(iv).


(f) There are four resistors of 12 Ω each. Which of the following values ispossible by their combination (series and/or parallel)?

(i) 9 Ω (ii) 16 Ω (iii) 12 Ω (iv) 36Ω.

Ans: – Option (iv).


(g) In case of the circuit shown below in Fig. 16.12, which of the following statements is/are true:

(i) R1, R2, and R3 are in series

(ii) R2 and R3 are in series

(iii) R2 and R3 are in parallel

(iv) The equivalent resistance of the circuit is [R1+ (R2 R3/R2 + R3)].

Ans: – Option (iv).

(h) The equivalent resistance of two resistors of equal resistances connectedin parallel is —— the value of each resistor.

(i) Half (ii) Twice (iii) Same (iv) One fourth.

Ans: – Option(i).


2) Fill in the blanks.

(a) When current is passed through a conductor, its temperature ….. increase…………

(b) The amount of …….electric charge……… flowing past a point per unit ….are………… isdefined as electric current.

(c) A current carrying conductor carries an ……. magnetic……… field around it.

(d) One ampere equals one …… Columb………. per ……. second……….

(e) Unit of electric power is …..watt…………

(f) Of the two wires made of the same material and having same thickness,the longer one has ..more………….. resistance.


3)How many types of electric charge exist?

Ans: – There are two types of electric charges exist positive and negative charges.


4)In a nucleus there are several protons, all of which have positive charge. Why does the electrostatic repulsion fail to push the nucleus apart?

Ans: –In a nucleus there are several protons all of which have positive charge, this electrostatic repulsion fails to push the nucleus because of the strong nuclear force.


5)What does it mean to say that charge is conserved?

Ans: – Charges is conserved means the positive or the negative charges are placed in a certain places or store in certain area.


6)A point charge of +3.0 μC is 10 cm apart from a second point charge of–1.5 μC. Find the magnitude and direction of force on each charge.

Ans: – As we know force, F=kq1q2/r^2;

Here, k=9×10^9, q1=+3×10^-6; q2=-1.5×10^-6. R=10cm=0.1m.

So, F=9×10^9× (3×10^-6× -1.5×10^-6 / 0.1)

Or, F= 4.05N.


7) Name the quantity measured by the unit (a) VC (b) Cs–1.

Ans: – The unit of charge quantity is VC and ampere is measure by Cs^-1.


8) Give a one-word name for the unit (a) JC–1 (b) Cs–1.

Ans: – Potential difference is measure by JC^-1; and Cs^-1 is measure by current.


9) What is the potential difference between the terminals of a battery if 250 J of work is required to transfer 20 C of charge from one terminal of the batteryto the other?

Ans: – We know, Q=CV,

Here, Q=250J, C=20c, v=?

So, 250= 20 × V,

V= 12.5volt.


10) Give the symbols of (a) cell (b) battery (c) resistor (d) voltmeter.

Ans: – 


11) What is the conventional direction of flow of electric current? Do the charge carriers in the conductor flow in the same direction? Explain.

Ans: –  There are flow of electron in the current from one point to the other point in this way the negatively charged particles transform from to the positively charged particle.


12) Out of ammeter and voltmeter which is connected in series and which is connected in parallel in an electric circuit?

Ans: – The are connected in series whereas the voltmeter is connected in parallel in circuit of electrical system.


13)You are given two resistors of 3 Ω and 6 Ω, respectively. Combining these two resistors what other resistances can you obtain?

Ans: – The resistor which can be formed by these two resistors is 9(in series) and 2 (in parallel connection).


14)What is the current in SI unit if+100 coulombs of charge flow past a pointevery five seconds?

Ans: – As we know, current(i) = Q/t;

So, current= 100/5 = 20A.

Then 20A current will transfer through the circuit.


15) Reduce an expression for the electrical energy spent in flow of current through a conductor.

Ans: – We know that, V=J/Q,

Here j is joule or heat H,

So, V=H/Q,

Or, H=VQ,

Or, H=IR×It (as, V=IR, I=Q/t)

Then, H=I^2Rt.


16) Find the value of resistor X as shown in Fig. 16.13.

Ans: – 


17) In the circuit shown in Fig. 16.14, find

(i) Total resistance of the circuit.

(ii) Ammeter reading and (iii) Current flowing through 3 Ω resistor.



19) You are given three resistors of 1 Ω, 2 Ω and 3 Ω. Show by diagrams, how will you connect these resistors to get (a) 6/11 Ω (b) 6 Ω (c) 1.5 Ω?



20) A resistor of 8 Ω is connected in parallel with another resistor of X Ω. The resultant resistance of the combination is 4.8 ohm. What is the value of resistor X?

Ans: – The equation will be,

1/8 + 1/X = 1/4.8;

Or, (8+ X)/ 8X = 1/4.8;

Or, 3.2X= 38.4;

So, X= 38.4/3.2 = 12ohm.


21) In the circuit Fig. 16.16, find

(i) Total resistance of the circuit.

(ii) Total current flowing through the circuit.

(iii) The potential difference across 4 Ω resistor.



22) How many 132 Ω resistors should be connected in parallel to carry 5 A current in 220 V line?

Ans: – Here resistance, R= 220 / 5 = 44ohm,

Let the number of resistors is n.

By equation, 132/n =44;  or, n=3.

Updated: February 25, 2022 — 10:30 am

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