NCERT Solutions Of Electricity With Most Important Questions | Asterisk ClassesNCERT Solutions Of Electricity With Most Important Questions | Asterisk Classes
Electricity Chapter Class 10th NCERT Solutions With Examples, Intextual Question and Excercise that is important for various board exams.
Class 10th Electricity Chapter Q&A
Class 10th Electricity Chapter – Questions and Answers
3.1 A current of 0.5 A is drawn by a filament of an electric bulb for 10 minutes. Find the amount of electric charge that flows through the circuit.
Electric charge (Q) = current (I) × time (t)
Given:
I = 0.5 A
t = 10 minutes = 600 seconds
Q = 0.5 A × 600 s = 300 C
3.2 How much work is done in moving a charge of 2 C across two points having a potential difference 12V?
Work done (W) = charge (Q) × potential difference (V)
W = 2 C × 12 V = 24 J
Question 1. Name a device that helps to maintain a potential difference across a conductor.
A battery or a cell maintains a potential difference across a conductor.
Question 2. What is meant by saying that the potential difference between two points is 1 V?
It means that 1 joule of work is done to move 1 coulomb of charge between those two points.
Question 3. How much energy is given to each coulomb of charge passing through a 6 V battery?
Energy per coulomb = Potential difference = 6 V
Thus, 6 J of energy is given to each coulomb of charge.
3.3 (a) How much current will an electric bulb draw from a 220 V source, if the resistance of the bulb filament is 1200 Ω?
Using Ohm’s law: I = V / R
I = 220 V / 1200 Ω = 0.183 A
3.3 (b) How much current will an electric heater coil draw from a 220 V source, if the resistance of the heater coil is 100 Ω?
I = 220 V / 100 Ω = 2.2 A
3.4 The potential difference between the terminals of an electric heater is 60 V when it draws a current of 4 A from the source. What current will the heater draw if the potential difference is increased to 120 V?
Assuming resistance remains constant:
Resistance (R) = V / I = 60 V / 4 A = 15 Ω
At V = 120 V, I = V / R = 120 V / 15 Ω = 8 A
3.5 Resistance of a metal wire of length 1 m is 26 Ω at 20°C. If the diameter of the wire is 0.3 mm, what will be the resistivity of the metal at that temperature? Using Table 3.2, predict the material of the wire.
Resistivity (ρ) = R × A / L = 26 × 7.0686 × 10-8 / 1 ≈ 1.837 × 10-6 Ωm
Using Table 3.2, the resistivity suggests that the wire is made of an alloy.
Example 3.6
A 4Ω resistance wire is doubled on it. Calculate the new resistance of the wire.
Solution
We are given, \( R = 4 \, \Omega \)
When a wire is doubled on it, its length would become half and the area of cross-section would double.
That is, a wire of length \( l \) and area of cross-section \( A \) becomes of length \( \frac{l}{2} \)
and area of cross-section \( 2A \). From Eq. (3.10), we have:
Factors on which the resistance of a conductor depends:
Length (L)
Cross-sectional area (A)
Material’s resistivity (ρ)
2. Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?
Current flows more easily through a thick wire because a thicker wire has a larger cross-sectional area, resulting in lower resistance.
3. 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?
Using Ohm’s law, I = V / R. If V is halved and R remains constant, I will be halved.
4. Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
Alloys have higher resistivity and better mechanical properties, making them efficient for heating and more durable under high temperatures.
5. Use the data in Table 3.2 to answer the following:
a. Which among iron and mercury is a better conductor? Mercury is a better conductor than iron because it has lower resistivity.
b. Which material is the best conductor? Among common materials, silver is the best conductor.
3.7 An electric lamp, whose resistance is 20 Ω, and a conductor of 4 Ω resistance are connected to a 6 V battery. Calculate:
(c) The potential difference across the electric lamp and conductor:
V_lamp = I × R = 0.25 A × 20 Ω = 5 V
V_conductor = I × R = 0.25 A × 4 Ω = 1 V
Drawing a schematic diagram:
[Since visual diagrams cannot be displayed, here is a textual description]
A battery of three cells of 2 V each (total 6 V) connected in series.
Connected in series with a 5 Ω resistor, an 8 Ω resistor, a 12 Ω resistor, and a plug key (switch).
2. 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?
Assuming total voltage = 6 V and total resistance =5 +8 +12=25 Ω
Current, I = V / R = 6 V /25 Ω = 0.24 A
Potential difference across 12 Ω resistor:
V = I × R = 0.24 A ×12 Ω = 2.88 V
3.8 In the circuit diagram given in Fig. 3.10, suppose the resistors R1, R2 and R3 have the values 5 Ω, 10 Ω, 30 Ω respectively, which have been connected to a battery of 12 V. Calculate:
(a) The current through each resistor:
Assuming series connection: Same current flows through each resistor.
I = V / (R1 + R2 + R3) = 12 V / (5 +10 +30) Ω = 12 /45 ≈ 0.267 A
(b) The total current in the circuit:0.267 A
(c) The total circuit resistance:45 Ω
3.9 In Fig. 3.12, R1 = 10 Ω, R2 = 40 Ω, R3 = 30 Ω, R4 = 20 Ω, R5 = 60 Ω and a 12 V battery is connected to the arrangement. Calculate:
Judge the equivalent resistance when the following are connected in parallel:
a. 1 Ω and 106 Ω
1/R_eq =1/1 +1/106 ≈1
R_eq ≈1 Ω
b. 1 Ω and 103 Ω and 106 Ω
1/R_eq =1 +0.001 +0.000001 ≈1.001001
R_eq ≈0.999 Ω
2. An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?
First, calculate the total current:
I_lamp = 220 V /100 Ω = 2.2 A
I_toaster = 220 V /50 Ω = 4.4 A
I_filter =220 V /500 Ω = 0.44 A
Total current, I_total =2.2 +4.4 +0.44 = 7.04 A
Resistance of electric iron, R = V / I =220 V /7.04 A ≈ 31.25 Ω
Current through electric iron is 7.04 A
3. What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
Each device receives the full voltage of the battery.
Devices operate independently; failure of one does not affect others.
Allows different currents to flow through each device as per their resistance.
4. How the three resistors of resistance 2 Ω, 3 Ω, and 6 Ω be connected to give a total resistance of:
a) 4 Ω
Connect 2 Ω and 6 Ω in parallel, then in series with 3 Ω.
R_eq = (2 ×6)/(2+6) +3 =12/8 +3 =1.5 +3 = 4.5 Ω
b) 1 Ω
Connect all three resistors in parallel.
1/R_eq =1/2 +1/3 +1/6 =1 +0.333 +0.166 ≈1.5
R_eq =1 /1.5 ≈ 0.667 Ω
5. What is:
a) The highest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω: Connect all in series. R_total =4 +8 +12 +24 =48 Ω
b) The lowest total resistance: Connect all in parallel.
1/R_eq =1/4 +1/8 +1/12 +1/24 =0.25 +0.125 +0.0833 +0.0417 ≈0.5
R_eq =1/0.5 = 2 Ω
3.10 An electric iron consumes energy at a rate of 840 W when heating is at the maximum rate and 360 W when the heating is at the minimum. The voltage is 220 V. What are the current and the resistance in each case?
At maximum:
Power (P) = 840 W
Voltage (V) = 220 V
Current, I = P / V =840 /220 ≈ 3.818 A
Resistance, R = V / I =220 /3.818 ≈ 57.5 Ω
At minimum:
P = 360 W
I =360 /220 ≈ 1.636 A
R =220 /1.636 ≈ 134.6 Ω
Why does the cord of an electric heater not glow while the heating element does?
The cord has a lower resistance and larger cross-sectional area, resulting in less heat generation, while the heating element has a higher resistance to generate heat.
2. Compare the heat generated while transferring 96,000 coulomb of charge in one hour through a potential difference of 50 V.
Heat generated, H = V × Q =50 V ×96,000 C =4,800,000 J
3. An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 seconds.
Heat developed, H = I2 × R × t =52 ×20 ×30 =25 ×20 ×30 =15,000 J
An electric bulb is connected to a 220 V generator. The current is 0.50 A. What is the power of the bulb?
Power, P = V × I =220 ×0.50 =110 W
3.13 An electric refrigerator rated 400 W operated 8 hours/day. What is the cost of the energy to operate it for 30 days at Rs 3.00 per kW h?
Energy consumed per day = Power × time =400 W ×8 h =3,200 Wh =3.2 kWh
Energy consumed in 30 days =3.2 ×30 =96 kWh
Cost =96 kWh ×3.00 Rs/kWh =Rs 288
What determines the rate at which energy is delivered by a current?
The rate at which energy is delivered by a current is determined by the product of current and potential difference (P = V × I).
2. An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 hours.
Power, P = V × I =220 ×5 =1100 W
Energy consumed, E = P × t =1100 W ×2 h =2200 Wh =2.2 kWh
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 \(\frac{R}{R’}\) is —
Answer: Each part has a resistance of \( \frac{R}{5} \). Since they are connected in parallel, the equivalent resistance is \( \frac{R’}{R} = 25 \). Thus, \( R’ = \frac{R}{25} \) and the ratio \( \frac{R}{R’} \) is 25:1.
2. Which of the following terms does not represent electrical power in a circuit?
3. An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be —
(a) 100 W
(b) 75 W
(c) 50 W
(d) 25 W
Answer: (d) 25 W
Calculation: \( P = \frac{V^2}{R} \), so \( P’ = \frac{110^2 \times 100}{220^2} = 25 W \).
4. Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then in parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be —
(a) 1:2
(b) 2:1
(c) 1:4
(d) 4:1
Answer: (c) 1:4
5. How is a voltmeter connected in the circuit to measure the potential difference between two points?
Answer: A voltmeter is connected in parallel with the component across which the potential difference is to be measured.
6. A copper wire has a diameter of 0.5 mm and resistivity of \(1.6 \times 10^{-8}\) Ωm. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?
Answer:
The length \( L \) can be calculated using the formula \( R = \rho \frac{L}{A} \), where \( A = \pi \left(\frac{d}{2}\right)^2 \).
For \( d = 0.5 \, \text{mm} \), \( A = \pi \times \left(\frac{0.25 \times 10^{-3}}{2}\right)^2 \)
\( L = \frac{R \times A}{\rho} = \frac{10 \times \pi \times \left(0.25 \times 10^{-3}\right)^2}{1.6 \times 10^{-8}} \approx 49.2 \, \text{m} \)
If the diameter is doubled, the resistance decreases by a factor of 4 (since \( R \propto \frac{1}{d^2} \)).
7. The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below. Plot a graph between V and I and calculate the resistance of that resistor.
I (amperes)
0.5
1.0
2.0
3.0
4.0
V (volts)
1.6
3.4
6.7
10.2
13.2
The resistance \( R = \frac{V}{I} \) calculated for each pair of values gives an average of approximately 3.3 Ω.
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: Using Ohm’s Law \( R = \frac{V}{I} \), \( R = \frac{12 \, \text{V}}{2.5 \times 10^{-3} \, \text{A}} = 4800 \, \Omega \)
9. A battery of 9 V is connected in series with resistors of 0.2Ω, 1.2Ω, 0.3Ω, 0.4Ω, 0.5Ω and 12Ω, respectively. How much current would flow through the 12Ω resistor?
Answer: The total resistance \( R_t = 0.2 + 1.2 + 0.3 + 0.4 + 0.5 + 12 = 14.6 \, \Omega \). The current \( I = \frac{9 \, \text{V}}{14.6 \, \Omega} \approx 0.616 \, \text{A} \).
10. How many 176Ω resistors (in parallel) are required to carry 5 A on a 220 V line?
Answer: Total resistance required \( R = \frac{V}{I} = \frac{220 \, \text{V}}{5 \, \text{A}} = 44 \, \Omega \). For resistors in parallel, \( \frac{1}{R_{eq}} = \frac{1}{176} + \frac{1}{176} + \ldots \). Thus, number of resistors \( n = \frac{176}{44} = 4 \). So, 4 resistors are required.
11. Show how you would connect three resistors, each of resistance 6Ω, so that the combination has a resistance of (i) 9Ω (ii) 4 Ω.
(i) To get 9Ω: Connect two resistors in parallel and one in series.
(ii) To get 4Ω: Connect all three resistors in parallel.
12. Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10W. 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: Power \( P = VI \). For each bulb, \( I = \frac{P}{V} = \frac{10}{220} = 0.0455 \, \text{A} \). Maximum number of bulbs = \( \frac{5 \, \text{A}}{0.0455 \, \text{A}} \approx 110 \) bulbs.
13. A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24Ω resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?
Case 1: Separate use \( I = \frac{V}{R} = \frac{220 \, \text{V}}{24 \, \Omega} \approx 9.17 \, \text{A} \).
Case 2: Series combination \( R = 48 \, \Omega \), \( I = \frac{220 \, \text{V}}{48 \, \Omega} \approx 4.58 \, \text{A} \).
Case 3: Parallel combination \( R = \frac{24}{2} = 12 \, \Omega \), \( I = \frac{220 \, \text{V}}{12 \, \Omega} \approx 18.33 \, \text{A} \).
14. Compare the power used in the 2Ω resistor in each of the following circuits:
(i) A 6 V battery in series with 1Ω and 2Ω resistors: \( I = \frac{6}{3} = 2 \, \text{A} \), Power \( P = I^2 R = 2^2 \times 2 = 8 \, \text{W} \).
(ii) A 4 V battery in parallel with 12Ω and 2Ω resistors: \( V = 4 \, \text{V} \), \( P = \frac{V^2}{R} = \frac{4^2}{2} = 8 \, \text{W} \).
15. Two lamps, one rated 100 W at 220 V and the other 60W 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: Total current \( I = \frac{100}{220} + \frac{60}{220} = \frac{160}{220} \approx 0.727 \, \text{A} \).
16. Which uses more energy, a 250W TV set in 1 hr, or a 1200W toaster in 10 minutes?
Answer: Energy used by TV \( = 250 \, \text{W} \times 1 \, \text{hr} = 250 \, \text{Wh} \). Energy used by toaster \( = 1200 \, \text{W} \times \frac{10}{60} \, \text{hr} = 200 \, \text{Wh} \). TV uses more energy.
17. An electric heater of resistance 8 Ω draws 15A current from the service mains in 2 hours. Calculate the rate at which heat is developed in the heater.
Answer: Power \( P = I^2 R = 15^2 \times 8 = 1800 \, \text{W} \). Heat \( H = P \times t = 1800 \times 2 \times 3600 \, \text{J} \approx 12.96 \times 10^6 \, \text{J} \).
18. Explain the following:
(a) Why is tungsten used almost exclusively for filament of electric lamps?
Answer: Tungsten has a very high melting point and emits white light at high temperatures, making it ideal for lamp filaments.
(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?
Answer: Alloys have higher resistivity and do not oxidize (burn) easily at high temperatures, making them suitable for heating elements.
(c) Why is the series arrangement not used for domestic circuits?
Answer: In a series arrangement, if one appliance fails, the entire circuit is broken. Also, different appliances require different currents, which cannot be provided in a series circuit.
(d) How does the resistance of a wire vary with its area of cross-section?
Answer: Resistance \( R \) is inversely proportional to the cross-sectional area \( A \), i.e., \( R \propto \frac{1}{A} \).
(e) Why are copper and aluminium wires usually employed for electricity transmission?
Answer: Copper and aluminium have low resistivity, which reduces energy losses due to heat during electricity transmission.
