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These are complete exercise solutions for Samacheer Kalvi 9th Science Chapter 4 — Electric Charge and Electric Current. Solve each question yourself first, then verify below. For numerical problems, always show the formula, substitution, calculation, and final answer with correct units. If you get stuck on a question, review the Key Concepts table first — most answers follow directly from the definitions and formulae listed there.
Key Concepts (Quick Recap)
| Term | Definition |
|---|
| Electric charge (Q) | Fundamental property of matter; Q = It. SI unit: coulomb (C). Two types: positive and negative. |
| Electric current (I) | Rate of flow of charge through a conductor; I = Q/t. SI unit: ampere (A). |
| Potential difference (V) | Work done per unit charge to move charge between two points; V = W/Q. SI unit: volt (V). |
| Ohm's law | At constant temperature, current is directly proportional to potential difference: V = IR. |
| Resistance (R) | Opposition a conductor offers to current flow; R = V/I. SI unit: ohm (Ω). |
| Resistivity (ρ) | Material property relating resistance to dimensions; R = ρL/A. SI unit: ohm-metre (Ω·m). Depends on material and temperature, not on dimensions. |
| Series circuit | Components connected end-to-end; same current flows through all, voltages add up. R_total = R₁ + R₂ + ... |
| Parallel circuit | Components connected across same two points; same voltage across all, currents add up. 1/R_total = 1/R₁ + 1/R₂ + ... |
| Electric power (P) | Rate at which electrical energy is consumed; P = VI = I²R = V²/R. SI unit: watt (W). |
| Heating effect (Joule's law) | Heat produced when current flows through a resistor: H = I²Rt. Unit: joule (J). |
| Fuse | Safety device with thin wire of low melting point; melts and breaks circuit during overload or short circuit. |
| Ammeter | Instrument measuring current; always connected in series. Has low resistance. |
| Voltmeter | Instrument measuring potential difference; always connected in parallel. Has high resistance. |
| kWh | Unit of electrical energy; 1 kWh = 3.6 × 10⁶ J. Used in electricity bills. |
I. Choose the Correct Answer
Q1. Which of the following is a conductor? (a) Glass (b) Rubber (c) Copper ✓ (d) Plastic
Answer(c) ✓ — Copper is a metal and allows electric current to flow easily.
Q2. The unit of electric current is: (a) volt (b) watt (c) ohm (d) ampere ✓
Answer(d) ✓ — Current is measured in ampere (A).
Q3. According to Ohm's law: (a) V = IR ✓ (b) V = I/R (c) V = R/I (d) I = VR
Answer(a) ✓ — Ohm's law: V = IR, where V is voltage, I is current, R is resistance.
Q4. The resistance of a conductor is directly proportional to its: (a) area (b) length ✓ (c) temperature (d) volume
Answer(b) ✓ — R = ρL/A; resistance increases with length.
Q5. In a series circuit, the quantity that remains the same through all components is: (a) voltage (b) current ✓ (c) resistance (d) power
Answer(b) ✓ — In series, the same current flows through all components.
Q6. In a parallel circuit, the quantity that remains the same across all components is: (a) current (b) resistance (c) voltage ✓ (d) power
Answer(c) ✓ — In parallel, the same voltage appears across all components.
Q7. The resistivity of a conductor depends on: (a) its length (b) its cross-sectional area (c) its material ✓ (d) its volume
Answer(c) ✓ — Resistivity (ρ) is a material property. It depends on the nature of the material and temperature, not on dimensions like length or area.
Q8. Which instrument is always connected in series in a circuit? (a) Voltmeter (b) Ammeter ✓ (c) Galvanometer (d) Rheostat
Answer(b) ✓ — An ammeter measures current and must be connected in series so that the full circuit current passes through it. It has very low resistance to avoid affecting the circuit.
Q9. Electric power is given by: (a) P = IR (b) P = V/I (c) P = I²R ✓ (d) P = R/V
Answer(c) ✓ — Power can be expressed as P = VI = I²R = V²/R. Option (c) I²R is one valid form.
II. Fill in the Blanks
1. The direction of conventional current is from positive to negative terminal. 2. The SI unit of resistance is ohm (Ω). 3. The instrument used to measure current is ammeter. 4. The instrument used to measure potential difference is voltmeter. 5. 1 kWh = 3.6 × 10⁶ joules. 6. The SI unit of electric charge is coulomb (C). 7. In a series circuit, the total resistance is greater than the largest individual resistance. 8. The relationship between resistance, resistivity, length, and area is given by R = ρL/A. 9. A fuse wire is made of a material with a low melting point.
III. True or False
1. Silver is the best conductor of electricity.
AnswerTrue ✓ (Silver has the lowest resistivity among metals.)
2. A conductor offers more resistance than an insulator.
AnswerFalse — Correction: A conductor offers less resistance; an insulator offers very high resistance.
3. In a parallel circuit, if one bulb fuses, the others continue to glow.
AnswerTrue ✓
4. Resistivity depends on the dimensions of the conductor.
AnswerFalse — Correction: Resistivity (ρ) depends on the material and temperature, not on length or cross-sectional area.
5. An ammeter is always connected in parallel.
AnswerFalse — Correction: An ammeter is always connected in series. A voltmeter is connected in parallel.
6. The equivalent resistance of parallel resistors is always less than the smallest individual resistance.
AnswerTrue ✓ (In parallel, 1/R_total is greater than 1/R_smallest, so R_total < R_smallest.)
7. When a wire is stretched to double its length without changing volume, its resistance becomes four times.
AnswerTrue ✓ (Length doubles and area halves, so R' = ρ(2L)/(A/2) = 4R.)
8. An electric bulb rated 100 W, 220 V draws more current than a bulb rated 60 W, 220 V.
AnswerTrue ✓ (I = P/V; at the same voltage, higher power means higher current.)
IV. Match the Following
| Column I | Column II |
|---|
| Ammeter | (d) Current |
| Voltmeter | (e) Potential difference |
| Ohm | (a) Resistance |
| Coulomb | (c) Charge |
| Watt | (b) Power |
| Fuse | (g) Safety device |
| Resistivity | (f) Ohm-metre |
| Kilowatt-hour | (h) Electrical energy |
V. Answer Briefly (2-mark)
Q1. Define electric current.
Answer
Electric current is the rate of flow of electric charge through a conductor. It is measured by the amount of charge passing through a cross-section of the conductor per unit time, expressed as I = Q/t. The SI unit of current is the ampere (A); one ampere equals one coulomb of charge flowing per second.
Q2. Define potential difference.
Answer
Potential difference between two points in a circuit is the work done to move a unit positive charge from one point to the other. It is given by V = W/Q, where W is work in joules and Q is charge in coulombs. The SI unit is the volt (V); one volt equals one joule per coulomb.
Q3. Define resistance.
Answer
Resistance is the opposition a conductor offers to the flow of electric current through it. It depends on the material, length, cross-sectional area, and temperature, and is given by R = V/I. The SI unit is the ohm (Ω); one ohm equals one volt per ampere.
Q4. What is the heating effect of electric current?
Answer
When current flows through a conductor, the collisions between free electrons and atoms convert electrical energy into heat energy. The heat produced is given by Joule's law: H = I²Rt, where I is current, R is resistance, and t is time. Common applications include electric heaters, irons, toasters, and fuses.
Q5. State Ohm's law.
Answer
Ohm's law states that at constant temperature, the current flowing through a conductor is directly proportional to the potential difference across its ends. Mathematically, V = IR, where V is voltage, I is current, and R is resistance (a constant for the given conductor). The V-I graph for an ohmic conductor is a straight line passing through the origin.
Q6. What is resistivity?
Answer
Resistivity (ρ) is the resistance of a conductor of unit length and unit cross-sectional area. From R = ρL/A, resistivity determines how strongly a material opposes current flow. Its SI unit is ohm-metre (Ω·m). Resistivity depends only on the material and temperature — not on the dimensions of the conductor.
Q7. Differentiate an ammeter from a voltmeter.
Answer
An ammeter measures electric current and is always connected in series; it has very low resistance so it does not significantly alter the circuit current. A voltmeter measures potential difference and is always connected in parallel; it has very high resistance so it draws minimal current from the circuit.
VI. Answer in Detail (5-mark)
Q1. State Ohm's law and describe its experimental verification.
Answer
Ohm's law states that at constant temperature, the current flowing through a conductor is directly proportional to the potential difference across its ends: V ∝ I, or V = IR, where R is the constant of proportionality called resistance.
Experimental verification: 1. Connect a resistor, ammeter (in series), voltmeter (in parallel), battery, and rheostat to form a complete circuit. 2. Close the circuit and adjust the rheostat to set a particular voltage. Record the voltmeter reading (V) and the ammeter reading (I). 3. Vary the rheostat to change the voltage in steps and note the corresponding current for at least five readings. 4. Calculate the ratio V/I for each reading — it should be approximately constant, confirming R is constant. 5. Plot a V-I graph; it should be a straight line through the origin, confirming V ∝ I.
Precautions: All connections should be tight, the conductor should not heat up during the experiment, and readings should be taken only after the ammeter pointer stabilizes.
Q2. Differentiate series and parallel circuits in detail.
Answer
| Property | Series Circuit | Parallel Circuit |
|---|
| Connection | Components connected end-to-end | Components connected across same two points |
| Current | Same through all components | Different through each branch |
| Voltage | Different across each component | Same across all components |
| Total resistance | R = R₁ + R₂ + R₃ (adds up) | 1/R = 1/R₁ + 1/R₂ + 1/R₃ (decreases) |
| Equivalent resistance | Greater than the largest individual R | Less than the smallest individual R |
| If one component fails | Entire circuit breaks | Others continue to work |
| Example use | Decorative series lights, old-style tree lights | Household wiring, parallel-connected appliances |
In household wiring, appliances are connected in parallel so each receives the full 220 V supply and can be switched on or off independently. Series circuits are rarely used for domestic power because if one appliance fails, all others stop working.
Q3. What are the factors affecting the resistance of a conductor? Explain each.
Answer
The resistance of a conductor depends on four factors:
1. Length (L): R ∝ L — Resistance is directly proportional to length. A longer wire offers more resistance because electrons must travel a greater distance and undergo more collisions. Doubling the length doubles the resistance.
2. Cross-sectional area (A): R ∝ 1/A — Resistance is inversely proportional to the cross-sectional area. A thicker wire has more space for electrons to flow, reducing collisions and hence resistance. Doubling the area halves the resistance.
3. Material (resistivity ρ): Different materials have different resistivities. Copper and aluminium have low resistivity (good conductors), while rubber and glass have very high resistivity (insulators). The combined relation is R = ρL/A.
4. Temperature: For metallic conductors, resistance increases with temperature. As temperature rises, atoms vibrate more vigorously, increasing collisions with free electrons and hindering current flow.
Q4. Explain the domestic electric circuit with a diagram.
Answer
Domestic wiring uses a parallel circuit so that each appliance operates independently:
1. Supply lines: Two wires enter the house — the live (phase) wire at 220 V and the neutral wire at 0 V. A third earth wire provides safety. 2. Main fuse and switch: Connected on the live wire before distribution. The main fuse protects against overload; the main switch allows disconnecting the entire supply. 3. Electric meter: Records energy consumption in kWh for billing. 4. Distribution board: Splits supply into separate circuits for lighting (5 A), heating (15 A), and power (15 A or higher), each with its own fuse or circuit breaker. 5. Parallel connection: All appliances are connected in parallel across the live and neutral wires. Each gets the full 220 V and can be operated independently — if one appliance fails, others are unaffected. 6. Earthing: The earth wire connects the metallic body of appliances to the ground. If a fault causes the body to become live, current flows to earth instead of through a person, preventing electric shock.
Q5. Explain the working of an electric fuse.
Answer
A fuse is a safety device designed to protect electrical circuits from damage caused by excessive current. It consists of a thin wire or strip made of an alloy with a low melting point (typically tin-lead alloy), enclosed in a ceramic or glass body.
Working principle: The fuse wire is connected in series with the live wire of the circuit. During normal operation, the current flows through the fuse without issue. When excessive current flows — due to a short circuit (live wire touching neutral) or an overload (too many appliances on one circuit) — the fuse wire heats up rapidly because H = I²Rt. Since the wire has low melting point, it melts and breaks the circuit, cutting off the current supply and protecting appliances and wiring from damage.
Important points: The fuse rating must be slightly above the normal operating current. Common ratings are 5 A for lighting circuits and 15 A for power circuits. Once a fuse blows, it must be replaced — it cannot be reused. Using a fuse wire with a higher rating than recommended is dangerous as it will not melt in time.
VII. Numerical Problems
Q1. A current of 2 A flows through a resistor of 10 Ω. Find the potential difference across it.
Answer
V = IR = 2 × 10 = 20 V
Q2. Three resistors of 2 Ω, 4 Ω, and 6 Ω are connected in series. Find total resistance.
Answer
R = R₁ + R₂ + R₃ = 2 + 4 + 6 = 12 Ω
Q3. Three resistors of 2 Ω, 4 Ω, and 6 Ω are connected in parallel. Find total resistance.
Answer
1/R = 1/2 + 1/4 + 1/6 = 6/12 + 3/12 + 2/12 = 11/12 R = 12/11 = 1.09 Ω
Q4. A wire of resistance 20 Ω is stretched to double its length. Find the new resistance.
Answer
When stretched to 2L, area becomes A/2 (volume = AL = constant). R = ρL/A; R' = ρ(2L)/(A/2) = 4ρL/A = 4R = 4 × 20 = 80 Ω
Q5. An electric bulb is rated 100 W at 220 V. Find the resistance and current.
Answer
P = V²/R → R = V²/P = 220²/100 = 48400/100 = 484 Ω I = P/V = 100/220 = 0.45 A
Q6. Calculate the cost of running a 2 kW heater for 3 hours at Rs. 5 per unit.
Answer
Energy = P × t = 2 × 3 = 6 kWh Cost = 6 × 5 = Rs. 30
Common Mistakes to Avoid
1. ❌ Mixing up series and parallel formulae — Using R = R₁ + R₂ for parallel, or 1/R = 1/R₁ + 1/R₂ for series. ✅ Series adds directly: R = R₁ + R₂. Parallel uses reciprocals: 1/R = 1/R₁ + 1/R₂.
2. ❌ Forgetting to convert units — Substituting mA directly as A or kW as W. ✅ Always convert: 1 mA = 10⁻³ A, 1 kW = 10³ W, 1 kV = 10³ V.
3. ❌ Connecting ammeter in parallel or voltmeter in series. ✅ Ammeter always in series (low resistance); voltmeter always in parallel (high resistance).
4. ❌ Confusing resistivity (ρ) with resistance (R). ✅ Resistivity is a material property (depends on material and temperature only). Resistance also depends on length and area: R = ρL/A.
5. ❌ Using wrong formula for heating effect — H = IRt or H = V²Rt. ✅ Correct formula: H = I²Rt (Joule's law). Also H = VIt = V²t/R.
6. ❌ Adding parallel resistors directly — R_total = R₁ + R₂ for two parallel resistors. ✅ For two resistors in parallel: R_total = R₁R₂/(R₁ + R₂). The result is always less than the smaller resistor.
7. ❌ Confusing kWh (energy) with kW (power). ✅ kW is power (rate of energy use). kWh is energy = Power × Time. Electricity bills charge for kWh.
Exam Checklist
- [ ] Ohm's law V = IR memorized and verified
- [ ] Series and parallel resistance formulae practiced
- [ ] Factors affecting resistance (L, A, ρ, T) memorized
- [ ] Heating effect H = I²Rt and applications
- [ ] Domestic circuit wiring understood
- [ ] Numerical problems on V = IR, R = ρL/A, P = VI practiced
- [ ] Ammeter (series) and voltmeter (parallel) connections memorized
- [ ] Fuse working and safety understood
- [ ] Unit conversions mastered (mA → A, kW → W, kWh → J)
- [ ] Resistivity vs resistance distinction clear
- [ ] Parallel resistance shortcut for two resistors: R = R₁R₂/(R₁ + R₂)
- [ ] Power formulae P = VI = I²R = V²/R all memorized
Internal Links
- [Chapter 4 Guide — Electric Charge and Electric Current](/school/tamil-nadu-state-board/class/9/science/electric-charge-and-electric-current)
- [All Class 9 Science Solutions](/school/tamil-nadu-state-board/class/9/science/solutions)
- [Class 9 Science Subject Guide](/school/tamil-nadu-state-board/class/9/science)
