Class 10 Science


Electricity NCERT In Text Questions

Part 1

Question 1: What does an electric current mean?

Answer: The flow of electric charge is known as electric current. Electric current is carried by moving electrons through a conductor. Electric current flows in opposite direction to the movement of electrons.

Question 2: Define the unit of current.

Answer: SI unit of electric current is ampere (A). Ampere is the flow of electric charges through a surface at the rate of one coulomb per second, i.e. if 1 coulomb of electric charge flows through a cross section for 1 second, it would be equal to 1 ampere.

Question 3: Calculate the number of electrons constituting one coulomb of charge.

Answer: We know that charge over 1 electron `= 1.6 xx 10^(-19)` coulomb

Thus, `1.6 xx 10^(-19)` C of charge = 1 electron

Therefore, 1 C of charge

`=(1)/(1.6xx10^(-19))\text(electrons)`

`=(10^19)/(1.6)\text(electrons)`

`=(10xx10^18)/(1.6)\text(electrons)`

`=6.25xx10^18\text(electrons)`

Question 4: Name a device that helps to maintain a potential difference across a conductor.

Answer: Battery or a cell

Question 5: What is meant by saying that the potential difference between two points is 1 V?

Answer: This means 1 joule of work is done to move a charge of 1 coulomb between two points.

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

Answer: Given, Charge `Q = 1C`, Potential difference, `V = 6V`

Therefore, Energy i.e. Work done, W =?

We know that,

`V=W/Q`

Therefore, `6V=(W)/(1C)`

Or, `6V xx1C=6J`

Thus, required energy = 6 J

Question 7: On what factors does the resistance of a conductor depend?

Answer: Resistance of a conductor depends upon:

  1. Nature of conductor
  2. Length of conductor
  3. Area of cross section of conductor

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?

Answer: Since, resistance is indirectly proportional to the area of cross section, thus current flows easily through a thick wire compared to a thin wire of the same material.

Question 9: Let the resistance of an electrical component remain 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?

Answer: We know, `R = V/I`

Therefore, if potential difference between two ends of the component will be halved, and resistance remains constant, then electric current would also be halved.

Question 10: Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?

Answer: Resistivity of alloy is generally higher than that of constituent metals. Alloys do not get oxidised (burnt) at high temperatures. Due to this, elements of heating appliances are made of alloys rather than a pure metal.

Use the data in Table 12.2 to answer the following

  1. Which among iron and mercury is a better conductor?

    Answer: Iron
  2. Which material is the best conductor?

    Answer: Silver

Question 11: Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω resistor, an 8 Ω resistor, and a 12 Ω resistor, and a plug key, all connected in series.

Answer:

circuit diagram

Question 12: Redraw the circuit of Question 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 Ω resistor. What would be the readings in the ammeter and the voltmeter?

Answer:

circuit diagram

The total resistance in the circuit = Sum of the resistances of all resistors

`= 5 Ω + 8 Ω + 12 Ω = 25 Ω`

We know,

`R=V/I`

Or, `25Ω=(6V)/(I)`

Or, `I=(6V)/(25Ω)=0.24A`

Since, resistances are connected in series, thus electric current remains the same through all resistors.

Here we have, electric current, I = 0.24 A, resistance, R = 12 Ω

Thus, potential difference, V through the resistor of `12 Ω = I xx R`

Or, `V = 0.24 A xx 12 Ω = 2.88 V`

Thus, reading of ammeter `= 0.24 A`

Reading of voltmeter through resistor of `12 Ω = 2.88 V`