Important Questions for Class 10 Science Chapter 12 – Electricity

# Important Questions for Class 10 Science Chapter 12 – Electricity

Important questions for class 10 science chapter 12 – Electricity encompasses all about Electricity. Electrons move through a conductor to form an electric current. Generally, current flow is the opposite of electron flow. Electric current is measured in amperes in the SI system. Batteries or cells initiate electron motion in electrical circuits. The terminals of a cell have a potential difference. Conductor resistance is its ability to resist electron flow. Controls the magnitude of the current. Conductor resistance is directly proportional to its length, inversely proportional to its cross-sectional area, and material-dependent. The sum of the individual resistances of a series of resistors is the equivalent resistance.

While solving important questions for class 10 science chapter 12 – Electricity, students are required to pay attention dealing with the following topics:

• Current and circuit in electricity
• The prospectives of electricity and the differences in their potential
• Conductor resistance factors
• A system of resistors’ resistance
• Effect of electric current on heating

Below are some of the important questions for class 10 science chapter 12 – Electricity from an examination point of view.

## Class 10 Science Important Questions Chapter 12 – Electricity

Q1. Assertion (A) : The metals and alloys are good conductors of electricity.

Reason (R) : Bronze is an alloy of copper and tin and it is not a good conductor of electricity.

(a) Both (A) and (R) are true and (R) is the correct explanation of the assertion (A).

(b) Both (A) and (R) are true, but (R) is not the correct explanation of the assertion (A).

(c) (A) is true, but (R) is false.

(d) (A) is false, but (R) is true.

Solution:

(c) : Metals and alloys are good conductors of electricity. Bronze is an alloy of copper and tin which are metals and thus is a good conductor of electricity.

Q2. Assertion (A) : Alloys are commonly used in electrical heating devices like electric iron and heater.

Reason (R): Resistivity of an alloy is generally higher than that of its constituent metals but the alloys have low melting points then their constituent metals.

(a) Both (A) and (R) are true and (R) is the correct explanation of the assertion (A).

(b) Both (A) and (R) are true, but (R) is not the correct explanation of the assertion (A).

(c) (A) is true, but (R) is false.

(d) (A) is false, but (R) is true.

Solution:

(a) Both (A) and (R) are true and (R) is the correct explanation of the assertion (A).

Q3. (i) List three factors on which the resistance of a conductor depends.

(ii) Write the SI unit of resistivity.

Solution:

(i) Resistance of a conductor depends upon the following factors:

(1) Length of the conductor : (Treater the length (I) of the conductor more will be the resistance (R).

R ∝ I

(2) Area ol cross section of the conductor: (Ireater the cross-sectional area of the conductor, less will be the resistance.

R ∝ 1A

(3) Nature of conductor.

(ii) SI unit of resistivity is Ω m.

Q4. If the radius of a current carrying conductor is halved, how does current through it change?

Solution:

If the radius of conductor is halved, the area of cross-section reduced to (14) of its previous value.

Since, R ∝ 1A, resistance will become four times

From Ohm’s law, V = IR

For given V, I ∝ 1R

So, current will reduce to one-fourth of its previous value.

Q5. If a person has five resistors each of value 1/5 Ω, then the maximum resistance he can obtain by connecting them is

(a) 1 Ω

(b) 5 Ω

(c) 10 Ω

(d) 25 Ω

Solution:

(a) The maximum resistance can be obtained from a group of resistors by connecting them in series. Thus,

Rs = 1/5+1/5+1/5+1/5+1/5 1 Ω

Q6. The maximum resistance which can be made using four resistors each of 2 Ω is

(a) 2 Ω

(b) 4 Ω

(c) 8 Ω

(d) 16 Ω

Solution:

(c) : A group of resistors can produce maximum resistance when they all are connected in series.

∴ Rs = 2 Ω + 2 Ω + 2 Ω + 2 Ω = 8 Ω

Q7. List the advantages of connecting electrical devices in parallel with an electrical source instead of connecting them is series.

Solution:

(a) When a number of electrical devices are connected in parallel, each device gets the same potential difference as provided by the battery and it keeps on working even if other devices fail. This is not so in case the devices are connected in series because when one device fails, the circuit is broken and all devices stop working.

(b) Parallel circuit is helpful when each device has different resistance and requires different current for its operation as in this case the current divides itself through different devices. This is not so in series circuit where same current flows through all the devices, irrespective of their resistances.

Q8. (a) Write the mathematical expression for Joules law of heating.

(b) Compute the heat generated while transferring 96000 coulomb of charge in two hours through a potential difference of 40 V.

Solution:

(a) The Joule’s law of healing implies that heat produced in a resistor is

(i) directly proportional to the square of current lor a given resistance,

(ii) directly proportional to resistance for a given current, and

(iii) directly proportional to the time for which the current flows through the resistor.

i.e., H = I² Rt

(b) Given, charge q = 96000 C, time t = 2 h = 7200 s and potential difference V = 40 V

We know, H = I²Rt = Q^2/t^2×V/Q × t × t = VQ

= 40 × 96000 = 3.84 × 10^6 J = 3.84 MJ

Q9. (a) Why is tungsten used for making bulb filaments of incandescent lamps?

(b) Name any two electric devices based on heating effect of electric current.

Solution:

(a) (i) Tungsten is a strong metal and has high melting point (3380°C).

(ii) It emits light at high temperatures (about 2500°C).

(b) Electric laundry iron and electric heater are based on heating effect of electric current.

Q10. (a) Define power and state its SI unit.

(b) A torch bulb is rated 5 V and 500 mA. Calculate its

(i) power

(ii) resistance

(iii) energy consumed when it is lighted for 2.5 hours.

Solution:

(a) Power is defined as the rate at which electric energy is dissipated or consumed in an electric circuit.

P = VI = I²R = V²/R

The SI unit of electric power is watt (W). It is the power consumed by a device that carries 1 A of current when operated at a potential difference of IV.

1 W = 1 volt × 1 ampere = 1 V A

(b) Given, V = 5 V and I = 500 mA = 0.5 A

(i) Power, P = V × 7 = 5 × 0.5 = 2.5 W

(ii) As, P = V^2/R⇒R=V^2/P=25/2.5 = 10 Ω

(iii) Given, time t = 2.5 hrs = 9000 s

∴ The energy consumed, E = P × t

= 2.5 × 9000 = 2.25 × 10^4 J

= 6.25 Watt hour