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Science Class 9 Maharashtra Board | Menu
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Science and Technology Class 9 Important Questions Chapter 3 Maharashtra Board

Current Electricity


Short Questions


1. What is electric potential?

  • It is the electric level that determines the flow of charge between two points.

2. What is potential difference, and what is its formula?

  • It is the difference in electric potential between two points, calculated as V = W/Q (where V is in volts, W is work in joules, Q is charge in coulombs).

3. How do electrons flow in a circuit?

  • Electrons flow from lower potential (negative) to higher potential (positive).

4. What causes potential difference in a cell?

  • Chemical reactions inside the cell cause the potential difference.

5. What are free electrons?

  • They are weakly bound outermost electrons in a metal that move freely.

6. What is electric current, and what is its formula?

  • It is the flow of electrons through a conductor, calculated as I = Q/t (where I is in amperes, Q is charge in coulombs, t is time in seconds).

7. What is the unit of charge?

  • The unit of charge is Coulomb (C).

8. What does Ohm’s Law state, and what is its formula?

  • It states that current (I) is directly proportional to potential difference (V) if resistance (R) is constant, given by V = I × R.

9. What is the unit of resistance?

  • The unit of resistance is Ohm (Ω).

10. What is resistivity, and what is its formula?

  • It is a material’s property affecting resistance, calculated as ρ = R × A / L (where ρ is resistivity in Ωm, R is resistance in Ω, A is area in m², L is length in m).

11. What is the unit of resistivity?

  • The unit of resistivity is Ohm-meter (Ωm).

12. What are conductors?

  • Conductors are substances with low resistance, like copper, allowing current to flow easily.

13. What are insulators?

  • Insulators are substances with high resistance, like rubber, that block current flow.

14. What is the formula for effective resistance in series?

  • The formula for effective resistance in series is Rs = R1 + R2 + R3 (where Rs is total resistance in Ω).

15. What is the formula for effective resistance in parallel?

  • The formula for effective resistance in parallel is 1/Rp = 1/R1 + 1/R2 + 1/R3 (where Rp is total resistance in Ω).

16. What is the purpose of a fuse wire?

  • It melts to break the circuit if current exceeds a safe limit, protecting appliances.

17. What is the voltage difference between live and neutral wires in India?

  • The voltage difference is about 220 V.

18. Why are appliances connected in parallel at home?

  • They are connected in parallel to get the same 220 V and ensure independent operation.

19. What is the current if 120 C of charge flows in 60 seconds?

  • I = Q/t = 120 C / 60 s = 2 A.

20. What is the resistance if a 12 V potential difference causes a 3 A current?

  • R = V / I = 12 V / 3 A = 4 Ω.

Long Questions


1. What is potential difference, and how is it related to the flow of electrons?

  • Potential difference is the work done per unit charge to move it between two points, calculated as V = W/Q. It drives electrons from the negative terminal (lower potential) to the positive terminal (higher potential) in a circuit, enabling current flow.

2. How does the water flow experiment explain electric potential?

  • In the experiment, water flows from a higher level to a lower level when the clamp is removed (due to gravity), similar to how electric charge flows from higher to lower potential. This analogy shows that potential difference, like height difference, causes flow until the potential becomes equal (V = 0).

3. What are free electrons, and how do they contribute to electric current?

  • Free electrons are loosely bound electrons in a metal’s outer shell that can move freely, as depicted in Figure 3.3. When a potential difference (V) is applied, they flow from negative to positive, creating electric current (I = Q/t), with the current direction conventionally opposite to electron flow.

4. What is Ohm’s Law, and how can you calculate resistance using it with V = 24 V and I = 0.24 A?

  • Ohm’s Law states that V = I × R, where voltage (V) is proportional to current (I) with resistance (R) as a constant. For V = 24 V and I = 0.24 A, resistance is R = V / I = 24 / 0.24 = 100 Ω.

5. How is resistivity related to resistance, and calculate it for a wire with R = 30 Ω, L = 50 cm, and radius = 0.5 mm?

  • Resistivity (ρ) is related to resistance by R = ρL/A, where A = πr²; for R = 30 Ω, L = 0.5 m, and r = 0.5 × 10⁻³ m, A = 3.14 × (0.5 × 10⁻³)² = 7.85 × 10⁻⁷ m². Thus, ρ = R × A / L = 30 × 7.85 × 10⁻⁷ / 0.5 = 4.71 × 10⁻⁵ Ωm.

6. What happens when resistors are connected in series, and calculate the effective resistance for 15 Ω, 3 Ω, and 4 Ω?

  • In series, the same current flows through each resistor, and the effective resistance (Rs) is the sum, Rs = R1 + R2 + R3. For 15 Ω, 3 Ω, and 4 Ω, Rs = 15 + 3 + 4 = 22 Ω, increasing the total resistance.

7. What happens when resistors are connected in parallel, and calculate the effective resistance for 15 Ω, 20 Ω, and 10 Ω?

  • In parallel, the potential difference is the same across each resistor, and 1/Rp = 1/R1 + 1/R2 + 1/R3. For 15 Ω, 20 Ω, and 10 Ω, 1/Rp = 1/15 + 1/20 + 1/10 = 0.0667 + 0.05 + 0.1 = 0.2167, so Rp = 1/0.2167 ≈ 4.615 Ω, which is less than the smallest resistance.

8. How much charge passes through a conductor if a current of 0.4 A flows for 5 minutes?

  •  Charge (Q) is calculated as Q = I × t, where t = 5 × 60 = 300 s and I = 0.4 A. Thus, Q = 0.4 × 300 = 120 C of charge passes through the conductor.

9. Calculate the current flowing through a 1000 Ω bulb with a 230 V potential difference, and the potential difference for the same current in a 500 Ω appliance.

  • Current (I) = V / R = 230 / 1000 = 0.23 A for the bulb; for a 500 Ω appliance with I = 0.23 A, V = I × R = 0.23 × 500 = 115 V. This shows voltage adjusts with resistance for the same current.

10. If a 1 m nichrome wire has a resistance of 6 Ω, what will be its resistance if cut to 70 cm?

  • Resistance (R) is proportional to length, so R₂ = R₁ × (L₂ / L₁), where R₁ = 6 Ω, L₁ = 1 m, L₂ = 0.7 m. Thus, R₂ = 6 × (0.7 / 1) = 4.2 Ω, showing resistance decreases with length reduction.

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