Section 1: Basic Concepts of Electric Current
Q1. What is electric current?
Answer: Electric current is the flow of electric charge through a conductor. It is measured as the amount of charge flowing through a cross-section of a conductor per unit time.
Q2. Define the unit of electric current.
Answer: The SI unit of electric current is Ampere (A). One ampere is defined as one coulomb of charge flowing through a conductor in one second.
Q3. Write the mathematical expression for electric current.
Answer: I = Q/t Where:
- I = Electric current (Ampere)
- Q = Electric charge (Coulomb)
- t = Time (second)
Q4. What is the direction of conventional current?
Answer: Conventional current flows from positive terminal to negative terminal of a cell, which is opposite to the direction of electron flow.
Section 2: Electric Potential and Potential Difference
Q5. What is electric potential?
Answer: Electric potential at a point is the work done per unit positive charge to bring it from infinity to that point.
Q6. Define potential difference.
Answer: Potential difference between two points is the work done per unit positive charge to move it from one point to another.
Q7. What is the unit of potential difference?
Answer: The SI unit of potential difference is Volt (V). One volt is the potential difference when one joule of work is done to move one coulomb of charge.
Q8. Write the relationship between work, charge, and potential difference.
Answer: V = W/Q Where:
- V = Potential difference (Volt)
- W = Work done (Joule)
- Q = Electric charge (Coulomb)
Section 3: Ohm's Law
Q9. State Ohm's Law.
Answer: Ohm's Law states that the current flowing through a conductor is directly proportional to the potential difference across its ends, provided the temperature and other physical conditions remain constant.
Q10. Write the mathematical expression of Ohm's Law.
Answer: V = IR or I = V/R or R = V/I Where:
- V = Potential difference (Volt)
- I = Current (Ampere)
- R = Resistance (Ohm)
Q11. What is resistance?
Answer: Resistance is the property of a conductor that opposes the flow of electric current through it. It is measured in Ohms (Ω).
Q12. Draw the V-I graph for an ohmic conductor.
Answer: The V-I graph for an ohmic conductor is a straight line passing through the origin, showing that voltage is directly proportional to current.
Section 4: Factors Affecting Resistance
Q13. What factors affect the resistance of a conductor?
Answer: The resistance of a conductor depends on:
- Length of the conductor (R ∝ l)
- Cross-sectional area (R ∝ 1/A)
- Nature of material (resistivity ρ)
- Temperature
Q14. Write the formula relating resistance with length, area, and resistivity.
Answer: R = ρl/A Where:
- R = Resistance (Ω)
- ρ = Resistivity (Ω⋅m)
- l = Length (m)
- A = Cross-sectional area (m²)
Q15. What is resistivity?
Answer: Resistivity is an intrinsic property of a material that determines its resistance. It is independent of the dimensions of the conductor and depends only on the nature of the material and temperature.
Section 5: Resistors in Series and Parallel
Q16. How are resistors connected in series?
Answer: In series connection, resistors are connected end to end in a single path. The same current flows through all resistors, but the voltage is divided among them.
Q17. Write the formula for equivalent resistance in series.
Answer: Rs = R₁ + R₂ + R₃ + ... The total resistance is the sum of individual resistances.
Q18. How are resistors connected in parallel?
Answer: In parallel connection, all resistors are connected between the same two points. The voltage across each resistor is the same, but current is divided among them.
Q19. Write the formula for equivalent resistance in parallel.
Answer: 1/Rp = 1/R₁ + 1/R₂ + 1/R₃ + ... The reciprocal of total resistance equals the sum of reciprocals of individual resistances.
Q20. Compare series and parallel connections.
Answer: Series Connection:
- Same current through all components
- Voltage is divided
- Total resistance increases
- If one component fails, the entire circuit stops working
Parallel Connection:
- Same voltage across all components
- Current is divided
- Total resistance decreases
- If one component fails, others continue to work
Section 6: Electric Power
Q21. What is electric power?
Answer: Electric power is the rate of consumption of electrical energy. It is the electrical energy consumed per unit time.
Q22. Write the formulas for electric power.
Answer:
- P = VI (Power = Voltage × Current)
- P = I²R (Using Ohm's Law)
- P = V²/R (Using Ohm's Law)
Q23. What is the unit of electric power?
Answer: The SI unit of electric power is Watt (W). One watt is equal to one joule per second.
Q24. What is the commercial unit of electrical energy?
Answer: The commercial unit of electrical energy is kilowatt-hour (kWh). 1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J
Section 7: Heating Effect of Electric Current
Q25. State Joule's Law of heating.
Answer: Joule's Law states that the heat produced in a conductor when current flows through it is directly proportional to:
- Square of the current (H ∝ I²)
- Resistance of the conductor (H ∝ R)
- Time for which current flows (H ∝ t)
Q26. Write the mathematical expression for Joule's heating.
Answer: H = I²Rt Where:
- H = Heat produced (Joule)
- I = Current (Ampere)
- R = Resistance (Ohm)
- t = Time (second)
Q27. Give applications of heating effect of current.
Answer: Applications include:
- Electric heater
- Electric iron
- Electric bulb
- Electric fuse
- Electric stove
Section 8: Electric Fuse
Q28. What is an electric fuse?
Answer: An electric fuse is a safety device that protects electrical circuits from damage due to overcurrent. It melts when excessive current flows through it, breaking the circuit.
Q29. How does a fuse work?
Answer: A fuse contains a thin wire made of low melting point material. When current exceeds the safe limit, the wire heats up due to Joule heating and melts, breaking the circuit and preventing damage to appliances.
Q30. What material is used in fuse wire?
Answer: Fuse wire is made of materials with low melting point and high resistivity, such as tin-lead alloy or copper.
Section 9: Numerical Problems
Q31. Calculate the current when 240 C of charge flows through a conductor in 2 minutes.
Answer: Given: Q = 240 C, t = 2 min = 120 s I = Q/t = 240/120 = 2 A
Q32. Find the resistance of a conductor if a current of 0.5 A flows through it when a potential difference of 12 V is applied.
Answer: Given: I = 0.5 A, V = 12 V Using Ohm's Law: R = V/I = 12/0.5 = 24 Ω
Q33. Calculate the equivalent resistance when resistors of 4 Ω and 12 Ω are connected in parallel.
Answer: Given: R₁ = 4 Ω, R₂ = 12 Ω 1/Rp = 1/R₁ + 1/R₂ = 1/4 + 1/12 = 3/12 + 1/12 = 4/12 = 1/3 Therefore, Rp = 3 Ω
Q34. Find the power consumed by a 100 Ω resistor when 2 A current flows through it.
Answer: Given: R = 100 Ω, I = 2 A P = I²R = (2)² × 100 = 4 × 100 = 400 W
Q35. Calculate the heat produced in a 10 Ω resistor in 5 minutes when 3 A current flows through it.
Answer: Given: R = 10 Ω, I = 3 A, t = 5 min = 300 s H = I²Rt = (3)² × 10 × 300 = 9 × 10 × 300 = 27,000 J = 27 kJ
Important Formulas Summary
- Current: I = Q/t
- Ohm's Law: V = IR
- Resistance: R = ρl/A
- Series Resistance: Rs = R₁ + R₂ + R₃ + ...
- Parallel Resistance: 1/Rp = 1/R₁ + 1/R₂ + 1/R₃ + ...
- Electric Power: P = VI = I²R = V²/R
- Joule's Heating: H = I²Rt = VIt = V²t/R
- Energy: E = Pt = VIt