Electricity Class 10 Science Chapter 11 Important Question Answer NCERT

Class 10th
Subject Science (NCERT)
Category Important Questions

Electricity Class 10 Science Chapter 11 Important Question Answer

Q1. The potential difference between the terminals of an electric heater is 80 V, when it draws a current of 5 A from the source. What current will the heater draw, if the potential difference is increased to 160 V?

Ans – Using Ohm’s law, V = IR

In first instance, V = 80 volts, I = 5 A
Therefore, R = V/I = 80/5= 16 Ω

In second instance, V = 160 volts, R (calculated above) = 16 Ω
Therefore, I = V/R = 160/16 = 10 A.

Thus, the heater will draw a current of 10 Ampere, if the potential difference is increased to 160 V.

Q2. An electric motor takes 4 A from a 220 V line. Find the power of the motor and energy consumed in 3 hours.

Ans – Given, V = 220 V, I = 4 A,

t = 3 hour = 3×60×60 sec = 10800 sec

Power of motor P = VI
= 220V × 4 A
= 880 watt.

Energy consumed by motor in 3 hours will be

E = P × t
= 880 × 10800 joule
= 9.504 × 106 j

Q3. Why are the coils of electric irons and toasters are made of an alloy rather than a pure metal ? Most Important

Ans – Because an alloy has a higher resistance than a pure metal. The alloys also do not melt easily at high temperature.

Q4. 150 Joule of heat is produced each second in a 6Ω resistance. Find the potential difference across the resistor.

Ans – Given, Heat energy H = 150 j, R = 6Ω, t = 1 sec

Using H = P × t

150 = P × 1

P = 150 watt

Now, using P = \displaystyle \frac{{{{V}^{2}}}}{R}

150 = \displaystyle \frac{{{{V}^{2}}}}{6}

V2 = 900

V = 30 V

Therefore, required potential difference is 30 V.

Q5. A 6 Ω resistance emits heat energy at the rate of 125 J / s. Find the potential difference across the resistor.

Ans – Given R = 6 Ω, P = 125 J/s, V = ?

Using, P = \displaystyle \frac{{{{V}^{2}}}}{R}

125 = \displaystyle \frac{{{{V}^{2}}}}{6}

V2 = 750

V = \displaystyle 5\sqrt{{30}} V = 27.39 V

Q6. How can three resistors of resistances 2Ω, 3Ω and 6Ω be connected to give a total resistance of 1Ω ? Most Important

Ans – If we connect all three resistors in parallel, we get

\displaystyle \frac{1}{{{{R}<em>{e}}}}=\frac{1}{{{{R}</em>{1}}}}+\frac{1}{{{{R}<em>{2}}}}+\frac{1}{{{{R}</em>{3}}}}

\displaystyle \frac{1}{{{{R}_{e}}}}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{{3+2+1}}{6}=\frac{6}{6}=1

Re = 1Ω

and we also need net resistance 1Ω

Therefore, we have to connect all given resistors in parallel.

Q7. A lamp of resistance 20 Ω and a conductor of resistance 8 Ω are connected with a battery of 8 V in series. Calculate the :
(a) total resistance in the circuit
(b) current flowing through the circuit

Ans – Given R1 = 20Ω , R2 = 8Ω, V = 8V

(a) Total resistance in series = R1 + R2 = 20Ω + 8Ω = 28Ω

(b) current flowing I = ?

Using V = IR

8 V = I × 28Ω

I = \displaystyle \frac{8}{{28}}=\frac{2}{7} A

Q8. An electric refrigerator of power 400 W is allowed to run 10 hrs. per day. What is the cost of energy to operate it for 30 days at Rs. 4.00 per kWh? Most Important

Ans – The total energy consumed by the refrigerator in 30 days would be

400 W × 8.0 hour/day × 30 days = 120000 Wh = 120 kWh

Thus, the cost of energy to operate the refrigerator for 30 days is

120 kwh × Rs. 4.00 per kWh = Rs. 480

Q9. On what factors does the resistance of a conductor depend? Most Important

Ans – (i) The resistance of a conductor depends directly on its length

(ii) Depends inversely on its area of cross-section,

(iii) Also on the nature of  material of the conductor

Q10. What do you mean by earthing? Why should electrical appliances be earthed ? Most Important

Ans – The third is the earth wire that has green insulation and this is connected to a metallic body deep inside earth. This process is called earthing.

It is used as a safety measure to ensure that any leakage of current to a metallic body does not give any severe shock to a user.

Q11. Why is tungsten used almost exclusively for filament of electric lamps?

Ans – Because it has a very high melting point due to which it does not melt even when it is heated to high temperatures due to the passage of electric current.

Q12. An electric current of 0.5 A flows through the filament of an electric bulb for 5 min. What will be the electric charge flowing through that wire ?

Ans – Given, I = 0.5 A, t = 5 min = 300s

Using, Q = It

= 0.5 A × 300s = 150 C

Q13. An electric bulb is connected to a 220 V generator. The current is 0.50 A. What is the power of the bulb?

Ans – Given V = 220V, I = 0.50 A

Using,  P = VI

P = 220 V × 0.50 A

P = 110 J/s = 110 W

Q14. When a 12 V battery is connected across an unknown resistor there is a current of 2.5mA in the circuit. Find the value of resistance of the resistor.

Ans – Given V = 12V, I = 2.5 mA = 2.5 × 10-3 A

Using V = IR

12 = 2.5 × 10-3 × R

R = 12/(2.5 × 10-3)

R = 4800 Ω

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