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Explain the concepts of electric charge and electric fields, and solve a circuit problem involving parallel resistors.

Physics

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9. (a)

(i) A charged material is a material that has an excess or deficiency of electrons, resulting in a net positive or negative electric charge.

(ii) To identify the charges on the two spheres (one positively charged, one negatively charged) using a positively charged rod:

  • Bring the positively charged rod close to sphere A without touching it.
  • If sphere A is repelled by the rod, then sphere A is positively charged (like charges repel).
  • If sphere A is attracted to the rod, then sphere A is negatively charged (opposite charges attract).
  • Once the charge of sphere A is identified, the charge of sphere B must be the opposite, as one is positive and the other is negative. Repulsion is the definitive test for identifying the same charge.

(iii) The electric field pattern around a positively charged sphere:

  • Electric field lines originate from the surface of the sphere and point radially outwards.
  • The lines are closer together near the sphere, indicating a stronger field.
  • The lines do not cross each other.

9. (b)

(i) Calculate the combined resistance of the resistors.

  1. Calculate the equivalent resistance of the two 6Ω6 \Omega bulbs in parallel.

    1Rp=16Ω+16Ω=26Ω=13Ω\frac{1}{R_p} = \frac{1}{6 \Omega} + \frac{1}{6 \Omega} = \frac{2}{6 \Omega} = \frac{1}{3 \Omega} Rp=3ΩR_p = 3 \Omega
  2. Calculate the total resistance by adding the parallel equivalent to the series 6Ω6 \Omega bulb.

    Rtotal=Rp+6Ω=3Ω+6Ω=9ΩR_{total} = R_p + 6 \Omega = 3 \Omega + 6 \Omega = 9 \Omega

    The combined resistance of the resistors is 9Ω\boxed{\text{}9 \Omega\text{}}.

    (ii) Calculate the current flowing through the circuit.

  3. Use Ohm's Law, I=VRtotalI = \frac{V}{R_{total}}.

    I=24 V9Ω=83 AI = \frac{24 \text{ V}}{9 \Omega} = \frac{8}{3} \text{ A} I≈2.67 AI \approx 2.67 \text{ A}

    The current flowing through the circuit is 2.67 A\boxed{\text{}2.67 \text{ A}\text{}}.

    (iii) Calculate the total power dissipated by the bulbs.

  4. Use the formula P=VIP = V I.

    P=24 V×83 A=8×8 W=64 WP = 24 \text{ V} \times \frac{8}{3} \text{ A} = 8 \times 8 \text{ W} = 64 \text{ W}

    The total power dissipated by the bulbs is 64 W\boxed{\text{}64 \text{ W}\text{}}.

    (iv) State two advantages of connecting bulbs in parallel.

    • If one bulb fails or burns out, the other bulbs in the parallel circuit will continue to operate.
    • Each bulb receives the full supply voltage, allowing them to operate at their intended brightness (assuming they are rated for that voltage).

    9. (c)

    (i) Compare the resistances of the two wires.

  5. State the formula for resistance.

    The resistance of a wire is given by R=ρLAR = \rho \frac{L}{A}, where ρ\rho is resistivity, LL is length, and AA is cross-sectional area.

  6. Compare the resistances.

    For the first wire, R1=10Ω=ρLAR_1 = 10 \Omega = \rho \frac{L}{A}.
    For the second wire, it is twice as long (L2=2LL_2 = 2L) and has the same thickness (A2=AA_2 = A) and material (ρ\rho).

    R2=ρ2LA=2(ρLA)=2R1R_2 = \rho \frac{2L}{A} = 2 \left( \rho \frac{L}{A} \right) = 2 R_1 R2=2×10Ω=20ΩR_2 = 2 \times 10 \Omega = 20 \Omega

    The second wire has a resistance of 20Ω\boxed{\text{}20 \Omega\text{}}, which is twice the resistance of the first wire.

    (ii) State and explain one other factor which affects the resistance of the wire.

    • Factor: Temperature.
    • Explanation: For most metallic conductors, resistance increases with increasing temperature. This is because higher temperatures cause the atoms within the conductor to vibrate more vigorously, leading to more frequent collisions with the free electrons. These increased collisions impede the flow of electrons, thus increasing the resistance to current.

    9. (d)

    (i) A magnet is a material or object that produces a magnetic field. This magnetic field is invisible but can attract or repel other magnetic materials and exert a force on moving electric charges.

    (ii) Describe test that you would carry out using the magnet M to identify the other bars.

    • Bring one pole of the known bar magnet M close to one end of bar X.
    • If bar X is repelled by magnet M, then bar X is a magnet (as only like poles of magnets repel each other).
    • If bar X is attracted to magnet M, it could be either a magnet (opposite poles attract) or a magnetic material (which is always attracted to a magnet).
    • To confirm, bring the same pole of magnet M close to the other end of bar X. If it is a magnet, it will now be repelled (if the first end was attracted) or attracted (if the first end was repelled). If it is a magnetic material, it will still be attracted.
    • Repeat the process for bar Y. The bar that shows repulsion is the magnet, and the other bar is the magnetic material. Repulsion is the definitive test for magnetism.

    (iii) Draw the magnetic field pattern around the bar magnet.
    (Imagine a bar magnet with a North pole on the left and a

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