NCERT Textbook Questions & Answers + Important Extra Questions
Section A: In-Text Questions
What does an electric circuit mean?
Answer
An electric circuit is a continuous, closed path through which electric current can flow. It consists of an energy source (such as a battery or cell), connecting wires, a switch, and one or more electrical devices (resistors, bulbs, etc.). For current to flow, the circuit must be complete — any break in the path stops the flow of current. This is the principle behind how switches work.
Define the unit of current.
Answer
The SI unit of electric current is the ampere (A). One ampere is defined as the flow of one coulomb of charge per second through a conductor.
1 A = 1 C/s
Smaller units: milliampere (mA) = 10⁻³ A, microampere (μA) = 10⁻⁶ A.
Calculate the number of electrons constituting one coulomb of charge.
Answer
Charge on one electron = 1.6 × 10⁻¹⁹ C
Number of electrons in 1 coulomb = 1 / (1.6 × 10⁻¹⁹)
= 6.25 × 10¹⁸ electrons
This enormous number shows that electric current involves the movement of a very large number of electrons.
Name a device that helps to maintain a potential difference across a conductor.
Answer
A battery (or cell) helps maintain a potential difference across a conductor. The chemical energy stored in the battery is continuously converted into electrical energy, which pushes charges through the circuit and maintains the potential difference even as current flows.
What is meant by saying that the potential difference between two points is 1 V?
Answer
A potential difference of 1 volt between two points means that 1 joule of work is done in moving 1 coulomb of charge from one point to the other.
1 V = 1 J/C (1 joule per coulomb)
How much energy is given to each coulomb of charge passing through a 6 V battery?
Answer
V = W/Q → W = V × Q = 6 × 1 = 6 joules
Each coulomb of charge receives 6 joules of energy from the battery.
On what factors does the resistance of a conductor depend?
Answer
The resistance of a conductor depends on:
(i) Length (l): Resistance is directly proportional to the length. R ∝ l (longer wire
⇒ more resistance)
(ii) Area of cross-section (A): Resistance is inversely proportional to the area. R ∝ 1/A (thicker wire
⇒ less resistance)
(iii) Nature of the material: Different materials have different resistivities (ρ). R = ρl/A
(iv) Temperature: Resistance of most metals increases with temperature.
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
Current will flow more easily through a thick wire. A thick wire has a larger cross-sectional area (A), and since resistance R = ρl/A, a larger area means lower resistance. With lower resistance and the same potential difference (V = IR), more current will flow through the thick wire.
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
By Ohm's law: I = V/R
If V becomes V/2 and R remains constant:
I' = (V/2)/R = (1/2)(V/R) = I/2
The current will also reduce to half of its original value.
Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
Answer
Alloys are used because they have:
(i) Higher resistivity than pure metals — they produce more heat for the same current.
(ii) Higher melting point — they can withstand the high temperatures generated during use without melting.
(iii) Do not oxidise (burn) easily at high temperatures, making them durable.
Common alloys used: nichrome (nickel, chromium, manganese, iron) for toasters and irons.
Q11. Use the data in Table 12.2 to answer the following
(a) Which among iron and mercury is a better conductor?
Answer
Iron is a better conductor because its resistivity (10.0 × 10⁻⁸ Ω·m) is lower than that of mercury (94.0 × 10⁻⁸ Ω·m). Lower resistivity means electrons flow more easily.
(b) Which material is the best conductor?
Answer
Silver is the best conductor as it has the lowest resistivity (1.60 × 10⁻⁸ Ω·m). Copper (1.62 × 10⁻⁸ Ω·m) is a very close second, which is why copper is used in wiring (silver being too expensive).
Section B: NCERT Exercise Questions
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 R/R' is:
Answer
(a) 1/25 (b) 1/5 (c) 5 (d) 25
Ans. (d) 25
Each piece has resistance R/5. Five pieces in parallel: 1/R' = 5/(R/5) = 25/R
⇒ R' = R/25
Therefore R/R' = R/(R/25) = 25
Which of the following terms does not represent electrical power in a circuit?
Answer
(a) I²R (b) IR² (c) VI (d) V²/R
Ans. (b) IR². Electrical power P = VI = I²R = V²/R. The term IR² does not represent power in any standard formula.
An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be:
Answer
(a) 100 W (b) 75 W (c) 50 W (d) 25 W
Ans. (d) 25 W
Resistance of bulb: R = V²/P = (220)²/100 = 484 Ω
Power at 110 V: P' = V'²/R = (110)²/484 = 12100/484 = 25 W
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:
Answer
(a) 1:2 (b) 2:1 (c) 1:4 (d) 4:1
Ans. (c) 1:4
Let each wire have resistance R. In series: Rs = 2R. In parallel: Rp = R/2.
Heat H = V²t/R (same V and t).
Hs/Hp = Rp/Rs = (R/2)/(2R) = 1/4
Ratio = 1:4
How is a voltmeter connected in the circuit to measure the potential difference between two points?
Answer
A voltmeter is connected in parallel across the two points between which the potential difference is to be measured. It has a very high resistance so that it draws negligible current from the circuit and does not disturb the circuit's behaviour.
A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10⁻⁸ Ω·m. 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
Diameter = 0.5 mm = 5 × 10⁻⁴ m, radius = 2.5 × 10⁻⁴ m
Area A = πr² = 3.14 × (2.5 × 10⁻⁴)² = 1.964 × 10⁻⁷ m²
R = ρl/A
⇒ l = RA/ρ = 10 × 1.964 × 10⁻⁷ / (1.6 × 10⁻⁸) = 122.7 m
If diameter is doubled, area becomes 4 times. Since R ∝ 1/A, resistance becomes R/4 = 2.5 Ω.
The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below:
Answer
I (amperes): 0.5, 1.0, 2.0, 3.0, 4.0
V (volts): 1.6, 3.4, 6.7, 10.2, 13.2
Plot a graph between V and I and calculate the resistance of that resistor.
Ans. Plotting V (x-axis) vs I (y-axis), we get a nearly straight line passing through the origin, confirming Ohm's law.
Resistance R = V/I
Taking the slope: R = ΔV/ΔI = (13.2 – 1.6)/(4.0 – 0.5) = 11.6/3.5 = 3.31 Ω
(Individual readings give: 3.2, 3.4, 3.35, 3.4, 3.3 — average ≈ 3.3 Ω)
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
V = 12 V, I = 2.5 mA = 2.5 × 10⁻³ A
R = V/I = 12 / (2.5 × 10⁻³) = 4800 Ω = 4.8 kΩ
A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω, and 12 Ω respectively. How much current would flow through the 12 Ω resistor?
Answer
In a series circuit, the same current flows through all resistors.
Total resistance = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 Ω
I = V/R = 9/13.4 = 0.67 A
The same 0.67 A flows through the 12 Ω resistor.
How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?
Answer
Required resistance = V/I = 220/5 = 44 Ω
Let n resistors of 176 Ω be connected in parallel: 1/Rp = n/176
Rp = 176/n = 44
⇒ n = 176/44 = 4 resistors
Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω, (ii) 4 Ω.
Answer
(i) 9 Ω: Connect two 6 Ω resistors in parallel (gives 3 Ω) and then connect this combination in series with the third 6 Ω resistor: 3 + 6 = 9 Ω
(ii) 4 Ω: Connect two 6 Ω resistors in series (gives 12 Ω) and then connect this combination in parallel with the third 6 Ω resistor: (12 × 6)/(12 + 6) = 72/18 = 4 Ω
Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. 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
Current drawn by each bulb: I = P/V = 10/220 = 1/22 A
Number of bulbs = Total current / Current per bulb = 5 / (1/22) = 5 × 22 = 110 lamps
A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?
Answer
(i) Separately: I = V/R = 220/24 = 9.17 A
(ii) In series: Rs = 24 + 24 = 48 Ω
⇒ I = 220/48 = 4.58 A
(iii) In parallel: Rp = (24 × 24)/(24 + 24) = 12 Ω
⇒ I = 220/12 = 18.33 A
Compare the power used in the 2 Ω resistor in each of the following circuits: (i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and (ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.
Answer
(i) Series: Total R = 1 + 2 = 3 Ω, I = 6/3 = 2 A
Power in 2 Ω = I²R = 2² × 2 = 8 W
(ii) Parallel: V across 2 Ω = 4 V (same in parallel)
Power in 2 Ω = V²/R = 4²/2 = 8 W
Power used in both cases is equal (8 W).
Two lamps, one rated 100 W at 220 V, and the other 60 W 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
I₁ = P₁/V = 100/220 = 5/11 A
I₂ = P₂/V = 60/220 = 3/11 A
Total current I = I₁ + I₂ = 5/11 + 3/11 = 8/11 = 0.727 A
Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?
Answer
TV: E = Pt = 250 × 3600 = 900,000 J = 900 kJ
Toaster: E = Pt = 1200 × 600 = 720,000 J = 720 kJ
The 250 W TV set uses more energy (900 kJ vs 720 kJ).
An electric heater of resistance 8 Ω draws 15 A from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater.
Answer
Rate of heat = Power = I²R = (15)² × 8 = 225 × 8 = 1800 W = 1.8 kW
Explain the following: (a) Why is the tungsten used almost exclusively for filament of electric lamps? (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? (c) Why is the series arrangement not used for domestic circuits? (d) How does the resistance of a wire vary with its area of cross-section? (e) Why are copper and aluminium wires usually employed for electricity transmission?
Answer
(a) Tungsten has a very high melting point (3380°C), so it can be heated to white-hot temperatures without melting. It also has high resistivity, making it glow brightly when current passes through it.
(b) Alloys have higher resistivity than pure metals (producing more heat), a higher melting point (withstanding high temperatures), and do not oxidise easily at high temperatures.
(c) In series, if one device fails, the entire circuit breaks and all devices stop working. Also, all devices share the same current, and the voltage is divided — devices don't get their rated voltage. Parallel arrangement is used so each device gets the full voltage, works independently, and can be switched on/off individually.
(d) Resistance is inversely proportional to the area of cross-section: R = ρl/A. Doubling the area halves the resistance.
(e) Copper and aluminium have very low resistivity (1.62 × 10⁻⁸ and 2.63 × 10⁻⁸ Ω·m respectively), meaning minimal energy is lost as heat during transmission. They are also relatively inexpensive and ductile (can be drawn into thin wires easily).
Section C: Important Extra Questions
State Ohm's Law and its limitations.
Answer
Ohm's Law: At constant temperature, the electric current flowing through a conductor is directly proportional to the potential difference applied across its ends.
V = IR (where V = potential difference, I = current, R = resistance)
Limitations
(i) It does not apply to non-linear devices like diodes, LEDs, and transistors.
(ii) It does not hold true if the physical conditions (especially temperature) of the conductor change.
(iii) It does not apply to gaseous conductors.
What is the difference between resistance and resistivity?
Answer
Resistance (R): Opposition offered by a conductor to the flow of current. Depends on length, area, material, and temperature. Unit: ohm (Ω). Changes with dimensions of the conductor.
Resistivity (ρ): Resistance of a conductor of unit length and unit cross-sectional area. Depends only on material and temperature, NOT on dimensions. Unit: Ω·m. It is a characteristic property of a material — ρ = RA/l.
Derive the formula for equivalent resistance when resistors are connected (a) in series and (b) in parallel.
Answer
(a) Series: Same current flows through all resistors. Total voltage = sum of individual voltages. V = V₁ + V₂ + V₃ = IR₁ + IR₂ + IR₃ = I(R₁ + R₂ + R₃)
⇒ Rs = R₁ + R₂ + R₃
(b) Parallel: Same voltage across all resistors. Total current = sum of individual currents. I = I₁ + I₂ + I₃ = V/R₁ + V/R₂ + V/R₃ = V(1/R₁ + 1/R₂ + 1/R₃)
⇒ 1/Rp = 1/R₁ + 1/R₂ + 1/R₃
What is electric power? Derive P = I²R.
Answer
Electric power is the rate at which electrical energy is consumed or dissipated in a circuit. Unit: watt (W).
Work done W = VIt (energy consumed)
⇒ Power P = W/t = VI
From Ohm's law, V = IR: P = (IR)I = I²R
Also, I = V/R: P = V(V/R) = V²/R
1 kWh (kilowatt-hour) = 3.6 × 10⁶ J = 1 unit of electricity.
What is Joule's law of heating? Give its applications.
Answer
Joule's law: The heat produced in a resistor is directly proportional to (i) the square of the current (H ∝ I²), (ii) the resistance (H ∝ R), and (iii) the time (H ∝ t).
H = I²Rt
Applications
(i) Electric heating devices — toasters, irons, heaters, ovens
(ii) Electric bulbs — tungsten filament heats up and glows
(iii) Electric fuse — a thin wire that melts when excess current flows, protecting the circuit