MP Board Class 10 Science Chapter 11: Electricity –…
MP Board Class 10 Science Chapter 11: Electricity (विद्युत) Important Questions — Electricity is a high-weightage chapter in the MP Board Class 10 Science exam, carrying 8–10 marks. This chapter covers electric current, Ohm’s law, resistance, resistors in series and parallel, heating effect of electric current, and electric power. The board exam frequently asks numerical problems, circuit-based questions, and conceptual application questions from this chapter. Here we have compiled the most important questions covering MCQs, short answers, long answers, and previous year questions to help you score full marks in the 2027 board exam.
📑 Table of Contents
- 1. Electric Current and Potential Difference
- 2. Ohm’s Law and Resistance
- 3. Combination of Resistors
- 4. Heating Effect and Electric Power
- 5. Numerical Problems
- 6. Multiple Choice Questions (1 Mark Each)
- 7. Very Short & Short Answer Questions (2–4 Marks)
- 8. Long Answer Questions (5–6 Marks)
- Previous Year Questions (2018–2026)
- Frequently Asked Questions
⚡ 1. Electric Current and Potential Difference
Q1. What is electric current? Define its SI unit.
Answer: Electric current is defined as the rate of flow of electric charge through a cross-section of a conductor. It is given by I = Q/t, where Q is the charge flowing in time t. The SI unit of electric current is the Ampere (A). One ampere is the current flowing when one coulomb of charge passes through a conductor in one second.
Q2. What is potential difference? Explain with an example.
Answer: Potential difference (V) between two points in an electric circuit is the amount of work done in moving a unit positive charge from one point to the other. It is given by V = W/Q, where W is work done and Q is charge. Its SI unit is Volt (V). One volt is the potential difference when one joule of work is done to move one coulomb of charge. Example: A 1.5 V cell provides 1.5 joules of energy per coulomb of charge flowing through the circuit.
💡 Key Formula: I = Q/t V = W/Q V = IR
Q3. Draw a schematic diagram of an electric circuit comprising a cell, a bulb, an ammeter and a voltmeter.
Answer: A simple electric circuit consists of a cell (source), a bulb (load), an ammeter connected in series to measure current, and a voltmeter connected in parallel across the bulb to measure potential difference. The circuit is shown as: Cell (+) → Ammeter → Bulb → Cell (-), with Voltmeter connected across the bulb. In a schematic diagram:
– Cell is represented by two parallel lines (longer = positive, shorter = negative)
– Bulb is represented by a cross inside a circle ×
– Ammeter is represented by A in a circle
– Voltmeter is represented by V in a circle
🔌 2. Ohm’s Law and Resistance
Q4. State and explain Ohm’s law. Give its mathematical expression.
Answer: Ohm’s law states that at constant temperature, the current flowing through a conductor is directly proportional to the potential difference applied across its ends. Mathematically, V ∝ I, or V = IR, where R is the resistance of the conductor. The V–I graph for an ohmic conductor is a straight line passing through the origin. SI unit of resistance is Ohm (Ω). One ohm is the resistance of a conductor when 1 V potential difference produces 1 A of current.
📌 Exam Tip: Ohm’s law is NOT applicable to semiconductors, diodes, and transistors — these are non-ohmic conductors. Always mention “at constant temperature” in the statement.
Q5. What factors affect the resistance of a conductor?
Answer: The resistance of a conductor depends on the following factors:
(a) Length (l): Resistance is directly proportional to length (R ∝ l). Longer wires have higher resistance.
(b) Cross-sectional area (A): Resistance is inversely proportional to area (R ∝ 1/A). Thicker wires have lower resistance.
(c) Nature of material: Different materials have different resistivities (ρ). Copper has low resistivity, while nichrome has high resistivity.
(d) Temperature: For metals, resistance increases with temperature; for semiconductors, resistance decreases with temperature.
The relationship is given by: R = ρl/A, where ρ is the resistivity of the material.
Q6. Define resistivity. Give its SI unit and write the resistivity range of insulators.
Answer: Resistivity (ρ) is an intrinsic property of a material that quantifies how strongly it resists the flow of electric current. It is numerically equal to the resistance of a conductor of unit length and unit cross-sectional area. SI unit: Ohm-metre (Ω m). Insulators have very high resistivity, typically in the range of 10¹² to 10¹⁷ Ω m. In comparison, conductors like copper have resistivity of about 1.7 × 10⁻⁸ Ω m.
🔗 3. Combination of Resistors
Q7. What is meant by series combination of resistors? Derive the expression for equivalent resistance.
Answer: In a series combination, resistors are connected end-to-end so that the same current flows through each resistor. The total potential difference across the combination is the sum of individual potential differences.
V = V₁ + V₂ + V₃
Using Ohm’s law: IRₛ = IR₁ + IR₂ + IR₃
Therefore: Rₛ = R₁ + R₂ + R₃
The equivalent resistance in series is always greater than the largest individual resistance. In a series circuit, if one component fails, the entire circuit breaks.
Q8. Derive the formula for equivalent resistance in a parallel combination of resistors.
Answer: In a parallel combination, resistors are connected such that one end of all resistors is connected to a common point and the other end to another common point. The potential difference across each resistor is the same, while the total current is the sum of currents through each branch.
I = I₁ + I₂ + I₃
Using Ohm’s law: V/Rₚ = V/R₁ + V/R₂ + V/R₃
Therefore: 1/Rₚ = 1/R₁ + 1/R₂ + 1/R₃
The equivalent resistance in parallel is always less than the smallest individual resistance. In a parallel circuit, if one component fails, current still flows through other branches.
🔥 4. Heating Effect and Electric Power
Q9. State and derive Joule’s law of heating.
Answer: Joule’s law of heating states that the heat produced in a resistor is directly proportional to:
(1) The square of current flowing through it (H ∝ I²)
(2) The resistance of the conductor (H ∝ R)
(3) The time for which current flows (H ∝ t)
Mathematically: H = I²Rt
Derivation: When current I flows through a resistor R for time t, work done by the source W = VIt. From Ohm’s law, V = IR. Therefore, W = (IR)It = I²Rt. This entire work is converted into heat energy, so H = I²Rt joules. The heating effect is used in electric heaters, irons, toasters, and incandescent bulbs.
Q10. Define electric power. Derive its formula and list its SI and commercial units.
Answer: Electric power is the rate at which electrical work is done or energy is consumed. Mathematically: P = W/t = VI. From Ohm’s law: P = I²R = V²/R.
SI unit: Watt (W) — one watt is the power consumed when 1 A flows at 1 V potential difference.
Commercial unit: Kilowatt-hour (kWh). 1 kWh = 3.6 × 10⁶ J. The electricity bill is calculated in kWh.
Relation: Energy (kWh) = Power (kW) × Time (h). For example, a 100 W bulb used for 10 hours consumes 100 × 10 = 1000 Wh = 1 kWh.
💡 Key Formulas: P = VI = I²R = V²/R H = I²Rt = Pt 1 kWh = 3.6 × 10⁶ J
Q11. Why is tungsten used as the filament in incandescent bulbs?
Answer: Tungsten is used as the filament in incandescent bulbs because:
(1) It has a very high melting point (3422°C), allowing it to glow white-hot without melting.
(2) It has high resistivity, producing sufficient heat when current flows through it.
(3) It can be drawn into thin wires easily (ductility).
However, incandescent bulbs are inefficient — only about 2–5% of the energy is converted to light, and the rest is lost as heat. This is why they are being phased out in favour of LEDs.
🧮 5. Numerical Problems
Q12. A current of 0.5 A flows through a bulb for 2 minutes. Find the amount of electric charge that flows through the circuit.
Answer:
Given: I = 0.5 A, t = 2 minutes = 2 × 60 = 120 s
Using formula Q = I × t = 0.5 × 120 = 60 C
Thus, 60 coulombs of charge flows through the circuit.
Q13. A wire of resistance 5 Ω is bent in the form of a closed circle. Find the resistance between two points at the ends of any diameter.
Answer:
When a 5 Ω wire is bent into a circle, the total resistance of the wire remains 5 Ω. Between the ends of a diameter, the wire is divided into two equal halves, each of resistance 5/2 = 2.5 Ω. These two halves are in parallel between the two points.
1/Rₚ = 1/2.5 + 1/2.5 = 2/2.5 = 0.8
Rₚ = 1/0.8 = 1.25 Ω
Q14. An electric iron of resistance 50 Ω draws a current of 4 A. Calculate the heat developed in 30 seconds.
Answer:
Given: R = 50 Ω, I = 4 A, t = 30 s
Using Joule’s law: H = I²Rt = 4² × 50 × 30
H = 16 × 50 × 30 = 24,000 J = 24 kJ
Q15. Two resistors of 6 Ω and 3 Ω are connected in parallel. Find the total current flowing in the circuit if the potential difference across the combination is 12 V.
Answer:
R₁ = 6 Ω, R₂ = 3 Ω, V = 12 V
1/Rₚ = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2
Rₚ = 2 Ω
Total current: I = V/Rₚ = 12/2 = 6 A
Q16. A 100 W bulb is used for 8 hours daily. Calculate the energy consumed in 30 days in kWh.
Answer:
Power = 100 W = 0.1 kW, Time per day = 8 hours, Days = 30
Total time = 8 × 30 = 240 hours
Energy = Power × Time = 0.1 × 240 = 24 kWh
If the rate is ₹6 per kWh, the cost = 24 × 6 = ₹144
📝 6. Multiple Choice Questions (1 Mark Each)
The MP Board exam includes 2–3 MCQs from Electricity. Practice these top expected questions.
- The SI unit of electric current is:
(a) Volt (b) Ampere ✓ (c) Watt (d) Ohm - According to Ohm’s law, which of the following is correct?
(a) V ∝ I² (b) V ∝ 1/I (c) V ∝ I ✓ (d) V ∝ R² - The resistance of a conductor depends on its:
(a) Length only (b) Area of cross-section only (c) Material only (d) Length, area, and material ✓ - If three resistors of 2 Ω, 3 Ω, and 5 Ω are connected in series, the equivalent resistance is:
(a) 10 Ω ✓ (b) 1 Ω (c) 0.97 Ω (d) 30 Ω - The heating effect of electric current is given by:
(a) H = IRt (b) H = I²Rt ✓ (c) H = VIt² (d) H = VRt - Which of the following is the commercial unit of electric energy?
(a) Joule (b) Watt (c) Kilowatt-hour ✓ (d) Ampere-hour - A wire of resistivity ρ is stretched to double its length. Its new resistivity will be:
(a) ρ/2 (b) 2ρ (c) 4ρ (d) ρ ✓ - In a parallel circuit, the potential difference across each resistor is:
(a) Different (b) Zero (c) Same ✓ (d) Proportional to current - An electric bulb of 60 W is used for 5 hours. The energy consumed in kWh is:
(a) 0.3 kWh ✓ (b) 3 kWh (c) 12 kWh (d) 300 kWh - Tungsten is used as the filament of bulbs because it has:
(a) Low resistivity (b) Low melting point (c) High melting point ✓ (d) High conductivity
✏️ 7. Very Short & Short Answer Questions (2–4 Marks)
-
Define one ampere of current.
One ampere is the current flowing through a conductor when one coulomb of charge passes through its cross-section in one second. (2 marks) -
What is the difference between a voltmeter and an ammeter? How are they connected in a circuit?
A voltmeter measures potential difference and is connected in parallel across the component. An ammeter measures current and is connected in series with the component. A voltmeter has high resistance, while an ammeter has very low resistance. (2 marks) -
What does an electric circuit mean? Name the components of a basic electric circuit.
An electric circuit is a closed path through which electric current flows. The basic components are: (1) a cell or battery (source of energy), (2) connecting wires, (3) a switch, and (4) a load (bulb, resistor, or appliance). (2 marks) -
Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
Alloys are preferred because: (1) They have higher resistivity than pure metals, generating more heat. (2) They do not oxidize or burn easily at high temperatures. (3) Tungsten is an exception used in bulbs because it has a very high melting point. Nichrome (alloy of Ni, Cr, Mn, Fe) is commonly used in heating appliances. (3 marks) -
Explain why the resistance of a conductor increases with an increase in temperature.
When temperature increases, the atoms and ions in the conductor vibrate more vigorously. This increases the frequency of collisions between free electrons and the vibrating atoms, obstructing the flow of electrons. Hence, the resistance increases. For semiconductors, however, resistance decreases with temperature because more electrons become free to conduct. (3 marks) -
Two wires of equal length, one of copper and the other of nichrome, have the same cross-sectional area. Which one has more resistance? Why?
The nichrome wire has much higher resistance because resistivity of nichrome (~100 × 10⁻⁸ Ω m) is about 60 times that of copper (~1.7 × 10⁻⁸ Ω m). Since R = ρl/A and l and A are same for both, the wire with higher resistivity (nichrome) will have higher resistance. (3 marks) -
How does the use of a fuse wire protect an electric circuit?
A fuse wire is made of a material with low melting point (like tin-lead alloy) and is connected in series with the circuit. When current exceeds the safe limit, the fuse wire heats up and melts (blows), breaking the circuit. This prevents damage to appliances and reduces fire risk. A fuse is always the thinnest wire in the circuit so it blows first. (3 marks) -
Why is parallel circuit used for household electrical wiring and not series circuit?
Parallel circuits are used because: (1) Each appliance gets the same voltage (220 V). (2) Appliances can be operated independently — switching off one does not affect others. (3) If one appliance fails, others continue working. (4) Different appliances with different power ratings can be used simultaneously. In series, the voltage divides and if one appliance fails, all stop working. (4 marks) -
What is electrical resistivity of a material? What is its unit? On what factors does the resistance of a conductor depend?
Resistivity (ρ) is the intrinsic property of a material that quantifies its opposition to current flow. Its SI unit is ohm-metre (Ω m). Factors affecting resistance: (1) Length — R ∝ l, (2) Cross-sectional area — R ∝ 1/A, (3) Material — different ρ values, (4) Temperature — R increases with temperature for metals. R = ρl/A. (4 marks)
📄 8. Long Answer Questions (5–6 Marks)
Q1. State Ohm’s law. Derive the expression for the equivalent resistance of three resistors connected in series. Also, give the characteristics of series combination.
Answer:
Ohm’s law: At constant temperature, the current flowing through a conductor is directly proportional to the potential difference across its ends. V ∝ I, or V = IR.
Series combination: Three resistors R₁, R₂, R₃ are connected in series. Same current I flows through all. Let V₁, V₂, V₃ be the potential differences across each resistor.
V = V₁ + V₂ + V₃
V₁ = IR₁, V₂ = IR₂, V₃ = IR₃
V = I(R₁ + R₂ + R₃)
If Rₛ is equivalent resistance: V = IRₛ
Therefore: Rₛ = R₁ + R₂ + R₃
Characteristics:
(1) Same current flows through all resistors.
(2) Voltage divides across resistors: V₁ : V₂ : V₃ = R₁ : R₂ : R₃
(3) Equivalent resistance is greater than the largest individual resistance.
(4) If one resistor fails, the entire circuit is broken. (5 marks)
Q2. Derive the expression for the equivalent resistance of three resistors connected in parallel. Compare series and parallel combinations. Why is parallel wiring used in homes?
Answer:
Parallel combination: Three resistors R₁, R₂, R₃ are connected in parallel. The potential difference V across each resistor is the same. The total current I splits as I₁, I₂, I₃.
I = I₁ + I₂ + I₃
I₁ = V/R₁, I₂ = V/R₂, I₃ = V/R₃
I = V(1/R₁ + 1/R₂ + 1/R₃)
If Rₚ is equivalent resistance: I = V/Rₚ
Therefore: 1/Rₚ = 1/R₁ + 1/R₂ + 1/R₃
Why parallel wiring in homes: (1) All appliances receive the same 220 V supply voltage. (2) Each appliance can be switched on/off independently. (3) If one appliance fails, others continue to work. (4) Different appliances with different power ratings can be used simultaneously without affecting each other’s performance. (5) The total resistance decreases, allowing more current from the main supply.
For two resistors: Rₚ = (R₁R₂)/(R₁ + R₂). For n equal resistors of R each: Rₚ = R/n. (6 marks)
Q3. State Joule’s law of heating. Derive the formula H = I²Rt. Explain some practical applications of the heating effect of electric current. An electric heater of resistance 100 Ω draws 5 A current. Calculate the heat produced in 10 minutes.
Answer:
Joule’s law: The heat produced in a conductor is directly proportional to (i) the square of current (I²), (ii) the resistance (R), and (iii) the time (t). H = I²Rt.
Derivation: Work done by source: W = VIt
From Ohm’s law: V = IR
W = (IR) × I × t = I²Rt
This work is converted to heat: H = I²Rt
Applications: (1) Electric iron — nichrome coils produce heat for ironing. (2) Electric toaster — heating elements glow red-hot to toast bread. (3) Electric water heater/geyser — immersion heaters warm water. (4) Electric bulb — tungsten filament glows to produce light (though only 2–5% is light). (5) Electric fuse — thin wire melts to break circuit during overloading.
Numerical: R = 100 Ω, I = 5 A, t = 10 min = 600 s
H = I²Rt = 5² × 100 × 600 = 25 × 100 × 600 = 1,500,000 J = 1.5 × 10⁶ J = 1500 kJ (6 marks)
📋 Previous Year Questions (2018–2026)
Electricity questions appear regularly in MP Board exams. Here are the questions from the past 8 years:
❓ Frequently Asked Questions
Q1. What is the difference between current and voltage?
Current is the rate of flow of charge (measured in amperes), while voltage is the electrical pressure or potential difference that causes current to flow (measured in volts). Current is analogous to water flow, and voltage is analogous to water pressure.
Q2. What is the SI unit of resistance?
The SI unit of resistance is the ohm (Ω). One ohm is the resistance of a conductor when 1 volt of potential difference produces 1 ampere of current through it (1 Ω = 1 V/A).
Q3. What happens to the resistance of a wire if its length is doubled?
Since R ∝ l (resistance is directly proportional to length), if the length is doubled, the resistance also doubles. However, if the wire is stretched, its cross-sectional area decreases too, so the actual increase is more due to the combined effect from R = ρl/A.
Q4. Why is household wiring done in parallel?
Household wiring is done in parallel so that each appliance receives the full 220 V supply, can be switched on/off independently, and the failure of one appliance does not affect others. Parallel also allows drawing higher total current when multiple appliances run simultaneously.
Q5. What is a fuse? How does it work?
A fuse is a safety device consisting of a thin wire made of tin-lead alloy with low melting point. It is connected in series with the circuit. When excessive current flows, the fuse wire heats up and melts, breaking the circuit and protecting appliances from damage.
Q6. What is the heating effect of electric current? Give two examples.
When electric current flows through a conductor, it produces heat due to the resistance offered by the conductor. This is the heating effect of current (Joule heating). Examples: (1) Electric iron uses nichrome coils that get hot. (2) Electric toaster uses heating elements that glow red-hot to toast bread.
Q7. How can we calculate the cost of electricity consumption?
Cost = Power (kW) × Time (hours) × Rate (₹/kWh). For example, a 100 W bulb used for 10 hours consumes 1 kWh. If the rate is ₹6/kWh, the cost is ₹6 per day. Energy in kWh is multiplied by the per-unit rate shown on electricity bills.
Q8. What is the resistance of an ideal ammeter and an ideal voltmeter?
An ideal ammeter has zero resistance (so it does not reduce the current when connected in series). An ideal voltmeter has infinite resistance (so it draws no current from the circuit when connected in parallel). In practice, ammeters have very low resistance and voltmeters have very high resistance.
Q9. What happens to the brightness of bulbs connected in series if one bulb fuses?
If bulbs are connected in series and one bulb fuses (burns out), the circuit breaks and all bulbs stop glowing because there is no closed path for current to flow. This is a major disadvantage of series circuits, which is why household wiring uses parallel connections.
Q10. Define 1 kWh. How is it related to joule?
One kilowatt-hour (1 kWh) is the electrical energy consumed by a 1 kW appliance used for 1 hour. 1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J. This is the unit used by electricity companies to measure and bill household consumption.
Q11. What is overloading in an electric circuit?
Overloading occurs when too many electrical appliances are connected to a single socket or circuit, drawing current higher than the rated capacity of the wiring. This causes excessive heating and can lead to fires. Fuses and MCBs (Miniature Circuit Breakers) are used to protect against overloading.
Q12. Why is nichrome used in heating elements and not copper?
Nichrome has high resistivity (~100 × 10⁻⁸ Ω m) compared to copper (~1.7 × 10⁻⁸ Ω m), so it produces more heat for the same current. Nichrome also has a high melting point and does not oxidize easily at high temperatures. Copper, being a good conductor, produces very little heat and would not work effectively in heating elements.