CBSE Class 10 Science Electricity

Electricity is the flow of charge through a conductor, and it powers almost every device we use. This CBSE Class 10 Science chapter on Electricity covers the mind map of electric current, potential difference, Ohm's law, resistance and resistivity, resistances in series and parallel, electric power and energy, and the heating effect of current, with the formulae, circuit diagrams and crossword, followed by the question bank with answers.

ELECTRICITY: MIND MAP

Electricity is of two kinds. Static electricity is due to charges at rest. Current electricity is due to charges in motion.

Electric current
Electric current is the flow of charge per unit time. It is given by I = Q / T. Its SI unit is the ampere (A), and it is measured with an ammeter.
1 ampere = 1 coulomb / 1 second.

Electric potential difference
Electric potential difference is the work done in moving a unit charge from one point to another. It is given by V = W / Q. Its SI unit is the volt (V), and it is measured with a voltmeter.
1 volt = 1 joule / 1 coulomb.

Ohm's law
Ohm's law states the relation between electric current and potential difference. The current through a conductor is directly proportional to the potential difference across its ends, that is, V ∝ I, or V = IR. Here R is a constant called resistance. Its SI unit is the ohm (Ω).

Factors on which resistance depends
1. Length of the conductor: R ∝ L (resistance is directly proportional to length).
2. Area of cross-section: R ∝ 1/A (resistance is inversely proportional to area).
Combining these, R ∝ L / A, or R = ρ(L / A), where ρ is a constant known as resistivity or specific resistance.

Resistivity
Resistivity is given by ρ = RA / L. Its SI unit is the ohm metre (Ω m). A material with low resistivity is a conductor, and a material with high resistivity is an insulator.

Resistances in series
1. The same current passes through each resistor.
2. The potential difference across different resistors is different.
3. The total potential difference is V = V₁ + V₂ + V₃.
4. The equivalent resistance is R = R₁ + R₂ + R₃.

Resistances in parallel
1. The voltage across each resistor is the same.
2. The current is divided in each branch.
3. The total current is I = I₁ + I₂ + I₃.
4. The equivalent resistance is \( \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \).

[Figure: Figure 12.4, resistors R₁, R₂ and R₃ connected in series in one line between points X and Y, with a voltmeter V across the combination, an ammeter A in the circuit, a cell battery and a plug key K]

[Figure: Figure 12.5, resistors R₁, R₂ and R₃ connected in parallel between points A and B, with a voltmeter V across the parallel group, a plug key K and an ammeter A in the main circuit, and a cell battery]

[Figure: Figure 12.3, the electric circuit for studying Ohm's law, with a resistor R between points X and Y, a voltmeter V connected across R, an ammeter A and a plug key K in the circuit, and a rheostat with its sliding contact connected to a cell]

ELECTRIC POWER AND ENERGY

Electric power is the rate at which electric energy is consumed. Its SI unit is the watt (W).

P = W / T, or Power = electrical energy / time
P = VI = I²R = V²/R

Electrical energy = power × time. Its SI unit is the watt hour (Wh), and its commercial unit is the kilowatt hour (kWh), also called one unit of electricity.

1 kWh = 1 unit
1 kWh = 3.6 × 10⁶ joule

[Figure: Mind map with Electric Power at the centre, stating that it is the rate at which electric energy is consumed and that its SI unit is the watt, linked to the formulae P = V²/R, P = I²R, P = VI and P = W/T]

HEATING EFFECT OF CURRENT

Joule's law of heating: The heat produced in a conductor is H = I²RT, where I is the current, R is the resistance and T is the time.

Practical applications of the heating effect
1. Devices which work on this effect include the toaster, the oven, the heating iron and the heater.
2. An electric bulb produces light due to this effect, because the tungsten filament is heated. Tungsten has a high melting point, and it is covered by a glass bulb. An inert atmosphere is provided by filling the bulb with nitrogen or argon, which are inert gases.
3. A fuse, used in electric circuits, works on this effect.

FORMULAE

CIRCUIT DIAGRAMS

[Figure: Figure 12.1, a schematic diagram of an electric circuit comprising a cell, an electric bulb, an ammeter and a plug key, all connected in series in a single closed loop]

[Figure: Table 12.2, symbols of commonly used components in circuit diagrams: an electric cell (one long and one short line); a battery or a combination of cells (several cells in a row); a plug key or switch (open) and (closed); a wire joint (a dot on the wire); wires crossing without joining (one wire with a hop over the other); an electric bulb; a resistor of resistance R (zigzag line); a variable resistance or rheostat (zigzag line with an arrow); an ammeter (circle with the letter A); and a voltmeter (circle with the letter V)]

CROSSWORD

Across
1. Rate of flow of charge: CURRENT
2. Ratio of potential difference to the current: RESISTANCE
6. Rate of consumption of electrical energy: POWER
7. SI unit of power: WATT

Down
1. Closed path for flow of electricity: CIRCUIT
3. SI unit of electric current: AMPERE
4. SI unit of charge: COULOMB
5. SI unit of potential difference: VOLT

QUESTION BANK

VERY SHORT ANSWER TYPE

1. Calculate the amount of charge flowing in a wire if it draws a current of 2A in 10 minutes. [1 Mark]

Answer:
1. Current I = 2 A, and time t = 10 minutes = 10 × 60 = 600 s.
2. Charge Q = I × t = 2 × 600.
3. Q = 1200 C.

Teacher's Note:
a) Convert minutes into seconds before using Q = It.

2. What happens to resistance of a conductor if area of cross-section is doubled? [1 Mark]

Answer:
1. Resistance R is inversely proportional to the area of cross-section A, that is, R ∝ 1/A.
2. When the area is doubled, the resistance becomes half of its original value.

Teacher's Note:
a) A thicker wire offers less resistance to the flow of current, because more area is available for the charge to flow.

3. Which device helps to maintain a potential difference across a conductor? [1 Mark]

Answer: A cell or a battery.

Teacher's Note:
a) A cell or battery converts chemical energy into electrical energy and keeps the potential difference across the conductor.

SHORT ANSWER TYPE

1. Draw a circuit diagram having the following components a. Bulb b. A two cell battery c. Ammeter d. A closed key [2 Marks]

Answer:
1. Connect the two cells in series to form the battery.
2. Connect the bulb, the ammeter and the closed key in series with the battery, in one closed loop, as shown in Figure 12.1.

[Figure: A single closed loop with a two-cell battery, a bulb, an ammeter (A) and a closed plug key, all connected in series, drawn as in Figure 12.1]

Teacher's Note:
a) Use the standard symbols from Table 12.2 for the cell, bulb, ammeter and key.
b) The ammeter must be connected in series with the bulb to measure the current through it.

2. Why are heating elements made of alloys rather than metals? [2 Marks]

Answer:
1. Alloys have high resistivity, so they produce more heat for the same current.
2. Alloys do not oxidise easily at high temperatures, and they have a high melting point, so they last longer.

Teacher's Note:
a) Nichrome is a common alloy used for heating elements.
b) Pure metals oxidise quickly at high temperatures and melt more easily.

3. What do we mean when we say that potential difference between two points is 1 volt? [2 Marks]

Answer:
1. When the potential difference between two points is 1 volt, 1 joule of work is done in moving 1 coulomb of charge from one point to the other.
2. This is the definition of the volt, that is, 1 volt = 1 joule / 1 coulomb.

Teacher's Note:
a) Potential difference is the work done per unit charge.
b) The volt is named after Alessandro Volta.

4. If three resistors of 6Ω, 9Ω and 21Ω are connected in series to a 12V battery, find a) The total resistance of the circuit. b) The current flowing through the circuit. c) The potential difference across the 21 Ω resistor. [3 Marks]

Answer:
1. Total resistance R = 6 + 9 + 21 = 36 Ω.
2. Current I = V / R = 12 / 36 = 0.33 A (approximately).
3. Potential difference across 21 Ω = I × R = 12 × 21 / 36 = 7 V.
4. So a) 36 Ω, b) 0.33 A, c) 7 V.

Teacher's Note:
a) The printed key gives 6.93 V for part (c); the correct value is 7 V, because the printed value comes from rounding the current to 0.33 A before multiplying.
b) In a series circuit, the same current flows through each resistor.
c) The largest resistor takes the largest share of the battery voltage.

5. What are the advantages of connecting electrical devices in parallel with the battery rather than in series? [3 Marks]

Answer:
1. The current required by each device is different, which is possible only in parallel.
2. If one device fails, the others can still work.
3. The total resistance of the circuit is decreased.

Teacher's Note:
a) In parallel, each device gets the full battery voltage.
b) In series, one failed device breaks the whole circuit.
c) Adding resistors in parallel always lowers the equivalent resistance.

VALUE BASED QUESTION

Rohan had gone to Sohan's house for studying. Sohan was worried about the high electricity bill and was wondering how to reduce it. Rohan suggested using LED bulbs instead of fluorescent bulbs as it consumes less electric energy and switch to solar power. After two months when they met again Sohan was very happy as the number of units consumed and the electric bill had reduced drastically.

1. What is electric power? What is its SI unit?

Answer:
Electric power is the rate of consumption of electric energy. Its SI unit is the watt.

Teacher's Note:
a) Power is energy used per second.
b) A 60 W bulb uses 60 joules every second.

2. What is the commercial unit used to measure electric energy?

Answer: The commercial unit of electric energy is the kilowatt hour (kWh).

Teacher's Note:
a) One kilowatt hour is one unit of electricity on the electricity bill.
b) 1 kWh is equal to 3.6 × 10⁶ joules.

3. What values were shown by Rohan?

Answer: Rohan showed caring, awareness and conservation of energy.

Teacher's Note:
a) He cared for his friend's financial problem and offered a practical solution.
b) LED bulbs and solar power save energy and protect the environment.

LONG ANSWER TYPE

1. a) What is the function of fuse in an electric circuit? b) What would be the rating of the fuse for an electric kettle which is operated at 220V and consumes 500 W power? c) How is the SI unit of electric energy related to its commercial unit? [5 Marks]

Answer:
a) A fuse protects the circuit from excess current. When the current exceeds the safe limit, the fuse wire melts and breaks the circuit.
b) Current drawn by the kettle, I = P / V = 500 / 220 = 2.27 A. The fuse should be rated just above this value, so a 3 A fuse is used.
c) The SI unit of electric energy is the joule, and the commercial unit is the kilowatt hour. 1 kWh = 3.6 × 10⁶ J.

Teacher's Note:
a) A fuse must be connected in series with the appliance it protects.
b) The printed answer gives 2.2 A; the more exact value is 2.27 A, and the fuse rating is still 3 A.
c) The energy used by a 1 kW appliance in one hour is one unit.
d) A fuse that is too high in rating does not protect the appliance, so the rating must be chosen carefully.

2. a) State Ohms law. Derive the relation and give graphical representation for it. b) An electric oven rated at 500W is connected to a 220V line and used for 2 hours daily. Calculate the cost of electric energy per month at the rate of Rs.5 per KWh. [5 Marks]

Answer:
a) Ohm's law states that, at constant temperature, the current through a conductor is directly proportional to the potential difference across its ends.
1. Let V be the potential difference and I the current, so that V ∝ I.
2. Therefore V = IR, where R is a constant called the resistance.
3. Graph: a plot of V against I is a straight line passing through the origin, and its slope gives the resistance R.
b) 1. Power = 500 W = 0.5 kW, and time per day = 2 h.
2. Energy consumed per day = 0.5 kW × 2 h = 1 kWh.
3. Energy consumed in 30 days = 30 kWh.
4. Cost = 30 × Rs. 5 = Rs. 150.

Teacher's Note:
a) The graph is a straight line only at constant temperature.
b) A 500 W oven running for 2 hours uses exactly one unit per day.
c) Check the units: kilowatt multiplied by hour gives kWh.
d) Monthly cost is the daily energy multiplied by the number of days and the rate per unit.

Question. 10,000 alpha-particles per minute are passing through a straight tube of radius r.
The resulting electric current is approximately
(a) 0.5 × 10^–16 A
(b) 2 × 10¹² A
(c) 0.5 × 10¹² A
(d) 2 × 10^–12 A
Answer. A

Question. Two metallic wires A and B are connected in series. Wire A has length l and radius r, while wire B has length 2l and radius 2r. If both the wires are of same material then find the ratio of the total resistance of series combination to the resistance of the wire A.
(a) 3/4
(b) 3/2
(c) 6/2
(d) 6/5
Answer. B

Question. The resistivity does not change if
(a) the material is changed
(b) the temperature is changed
(c) the shape of the resistor is changed
(d) both material and temperature are changed.
Answer. C

Question. A cylindrical conductor of length l and uniform area of cross section A has resistance R.
Another conductor of length 2l and resistance R of the same material has area of cross-section
(a) A/2
(b) 3A/2
(c) 2A
(d) 3A
Answer. C

Question. Consider a simple circuit containing a battery and three identical incandescent bulbs A, B and C.Bulb A is wired in parallel with bulb B and this combination is wired in series with bulb C. What would happen to the brightness of the other two bulbs if bulb A were to burn out?
(a) Only bulb B would get brighter.
(b) Both A and B would get brighter.
(c) Bulb B would get brighter and bulb C would get dimmer.
(d) There would be no change in the brightness of either bulb B or bulb C
Answer. C

Question. In a metallic conductor, electric current thought to be due to the movement of
(a) ions
(b) amperes
(c) electrons
(d) protons
Answer. C

Question. 1 volt = ..............
(a) 1 joule/coulomb
(b) 1 coulomb/joule
(c) 1 joule/coulomb2
(d) 1 joule coulomb
Answer. A

Question. If a wire of resistance R is melted and recast to half its length, the new resistance of the wire will be
(a) R/4
(b) R/2
(c) R
(d) 2R
Answer. A

Question. A piece of aluminium of finite length is drawn or stretched such that to reduce its diameter to one fourth its original value, its resistance will become
(a) 256 times
(b) four times
(c) eight times
(d) sixteen times
Answer. A

Question. Choices for the correct combination of elements from column-I and column-II are given as options (a), (b), (c) and (d) out of which one is correct?
     Column-I        Column-II
(P) Current          1. ohm
(Q) Potential        2. ampere
(R) Resistance     3. ohm m
(S) Resistivity      4. volt
(a) P-2, Q-4, R-1, S-3
(b) P-3, Q-4, R-2, S-1
(c) P-4, Q-3, R-1, S-2
(d) P-2, Q-1, R-4, S-3
Answer. A

Question. A multimeter is used to measure
(a) current only
(b) resistance only
(c) voltage only
(d) current, resistance and voltage.
Answer. D

Question. There are three copper wires of lengths and cross-sectional areas (L, A), (2 L, A/2), (L/2, 2 A). Resistivity is
(a) minimum of wire of cross-sectional area A/2
(b) minimum of wire of cross-sectional area A
(c) minimum of wire of cross-sectional area 2A
(d) same in all the three cases.
Answer. D

Question. In an electrical circuit, three incandescent bulbs A, B and C of rating 40 W, 60 W and 100 W respectively are connected in parallel to an electric source. Which of the following is likely to happen regarding their brightness?
(a) Brightness of all the bulbs will be the same
(b) Brightness of bulb A will be the maximum
(c) Brightness of bulb B will be more than that of A
(d) Brightness of bulb C will be less than that of B
Answer. C

Question. Masses of three wires of copper are in the ratio 1 : 3 : 5 and their lengths are in the ratio of 5 : 3 : 1. The ratio of their electrical resistances are
(a) 1 : 3 : 5
(b) 5 : 3 : 1
(c) 1 : 15 : 125
(d) 125 : 15 : 1
Answer. D

Question. The amount of heat energy produced in 5 minutes by an electric heater rated at 1000 W is
(a) 2 × 105 J
(b) 3 × 105 J
(c) 4 × 105 J
(d) 300 J
Answer. B

CBSE Class 10 Physics Chapter 12 Electricity Case Based MCQs 

Case I : Read the passage given below and answer the following questions.

The heating effect of current is obtained by transformation of electrical energy in heat energy. Just as mechanical energy used to overcome friction is converted into heat, in the same way, electrical energy is converted into heat energy when an electric current flows through a resistance wire. The heat produced in a conductor, when a current flows through it is found to depend directly on (a) strength of current (b) resistance of the conductor (c) time for which the current flows.

The mathematical expression is given by
H = I2Rt.
The electrical fuse, electrical heater, electric iron, electric geyser etc. all are based on the heating effect of current.

Question. What are the properties of heating element?
(a) High resistance, high melting point
(b) Low resistance, high melting point
(c) High resistance, low melting point
(d) Low resistance, low melting point.
Answer. A

Question. What are the properties of electric fuse?
(a) Low resistance, low melting point
(b) High resistance, high melting point.
(c) High resistance, low melting point
(d) Low resistance, high melting point
Answer. C

Question. When the current is doubled in a heating device and time is halved, the heat energy produced is
(a) doubled
(b) halved
(c) four times
(d) one fourth times
Answer. A

Question. A fuse wire melts at 5 A. It is is desired that the fuse wire of same material melt at 10 A. The new radius of the wire is
(a) 4 times
(b) 2 times
(c) 1/2 times
(d) 1/4 times
Answer. B

Question. When a current of 0.5 A passes through a conductor for 5 min and the resistance of conductor is 10 W, the amount of heat produced is
(a) 250 J
(b) 5000 J
(c) 750 J
(d) 1000 J
Answer. C

CBSE Class 10 Physics Chapter 12 Electricity Assertion & Reasoning Based MCQs

For question numbers 46-55, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
(a) Both assertion and reason are true, and reason is correct explanation of the assertion.
(b) Both assertion and reason are true, but reason is not the correct explanation of the assertion.
(c) Assertion is true, but reason is false.
(d) Assertion is false, but reason is true.

Question. Assertion : It is advantageous to transmit electric power at high voltage.
Reason : High voltage implies high current. 
Answer. C

Question. Assertion : Good conductors of heat are also good conductors of electricity and vice versa.
Reason : Mainly electrons are responsible for conduction.
Answer. A

Question. Assertion : If 10 bulbs are connected in series and one bulb fused, then the remaining 9 bulbs will not work.
Reason : Bulb of higher wattage will give less bright light.
Answer. B

Question. Assertion : The 200 W bulbs glows with more brightness than 100 W bulbs.
Reason : A 100 watt bulb has more resistance than a 200 W bulb.
Answer. A

Question. Assertion : A voltmeter and ammeter can be used together to measure resistance and power.
Reason : Power and resistance can be calculated from voltage and current.
Answer. A