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Define capacitance and state any two factors that affect capacitance.

Physics

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SECTION A

1. Define capacitance and state any two factors that affect capacitance.
Capacitance is the ability of a body to store an electric charge. It is defined as the ratio of the amount of electric charge stored on a conductor to the difference in electric potential.
Two factors that affect capacitance are:

  1. The area of the plates.
  2. The distance between the plates.
  3. The type of dielectric material between the plates.

2. State Ohm's law.
Ohm's Law states that the current flowing through a conductor between two points is directly proportional to the voltage across the two points, and inversely proportional to the resistance between them, provided the temperature and other physical conditions remain constant. Mathematically, it is expressed as V=IRV = IR.

3. Name at least four basic electrical quantities and state their SI units.

  1. Electric Current: Ampere (A)
  2. Voltage (Potential Difference): Volt (V)
  3. Resistance: Ohm (Ω\Omega)
  4. Electric Charge: Coulomb (C)
  5. Electric Power: Watt (W)

4. State three effects of electric current.

  1. Heating effect: Electric current produces heat when it flows through a conductor (e.g., in an electric heater).
  2. Magnetic effect: Electric current produces a magnetic field around the conductor (e.g., in an electromagnet).
  3. Chemical effect: Electric current can cause chemical changes when passed through certain liquids (e.g., in electrolysis).

5. Find a force acting vertically downward on a mass of 200g attached to a wire. (g = 9.81 m/s^2)

  • Given:
    • Mass m=200 gm = 200\, \text{g}
    • Acceleration due to gravity g=9.81 m/s2g = 9.81\, \text{m/s}^2

  1. Convert mass to kilograms.

    m=200 g=2001000 kg=0.2 kgm = 200\, \text{g} = \frac{200}{1000}\, \text{kg} = 0.2\, \text{kg}

  2. Calculate the force (weight).

    The force acting vertically downward is the weight, given by F=mgF = mg.

    F=(0.2 kg)×(9.81 m/s2)F = (0.2\, \text{kg}) \times (9.81\, \text{m/s}^2) F=1.962 NF = 1.962\, \text{N}

    The force acting vertically downward is 1.962 N\boxed{\text{1.962 N}}.

    6. Show that total resistance in series is given by RT=R1+R2+R3R_T = R_1 + R_2 + R_3.

  3. State the properties of a series circuit.

    In a series circuit, the current (II) is the same through all resistors, and the total voltage (VTV_T) is the sum of the individual voltage drops across each resistor.

    VT=V1+V2+V3V_T = V_1 + V_2 + V_3

  4. Apply Ohm's Law to each resistor.

    According to Ohm's Law (V=IRV = IR), the voltage drop across each resistor is:

    V1=IR1V_1 = IR_1 V2=IR2V_2 = IR_2 V3=IR3V_3 = IR_3

  5. Substitute individual voltages into the total voltage equation.

    VT=IR1+IR2+IR3V_T = IR_1 + IR_2 + IR_3

    Factor out the common current II:

    VT=I(R1+R2+R3)V_T = I(R_1 + R_2 + R_3)

  6. Apply Ohm's Law to the entire circuit.

    For the entire circuit, VT=IRTV_T = IR_T, where RTR_T is the total equivalent resistance.

    IRT=I(R1+R2+R3)IR_T = I(R_1 + R_2 + R_3)

    Dividing both sides by II gives:

    RT=R1+R2+R3R_T = R_1 + R_2 + R_3

    Thus, the total resistance in a series circuit is the sum of the individual resistances.

    7. A 100W electric light bulb is connected to a 250V supply. Determine:

    • a) The current flowing in the bulb.

    • b) The resistance of the bulb.

    • Given:

      • Power P=100 WP = 100\, \text{W}
      • Voltage V=250 VV = 250\, \text{V}

    a) The current flowing in the bulb.

  7. Use the power formula P=VIP = VI to find the current.

    I=PVI = \frac{P}{V} I=100 W250 VI = \frac{100\, \text{W}}{250\, \text{V}} I=0.4 AI = 0.4\, \text{A}

    The current flowing in the bulb is 0.4 A\boxed{\text{0.4 A}}.

    b) The resistance of the bulb.

  8. Use Ohm's Law R=V/IR = V/I to find the resistance.

    Using the current calculated in part a):

    R=VIR = \frac{V}{I} R=250 V0.4 AR = \frac{250\, \text{V}}{0.4\, \text{A}} R=625 ΩR = 625\, \Omega

    The resistance of the bulb is 625 Ω\boxed{\text{625 } \Omega}.

    8. a) Differentiate between resistance and conductance.

    • Resistance: Resistance is the opposition to the flow of electric current in a circuit. Its SI unit is the Ohm (Ω\Omega).
    • Conductance: Conductance is a measure of how easily electric current flows through a material. It is the reciprocal of resistance. Its SI unit is the Siemens (S).

    b) Find the total capacitance of 3µF, 6µF, and 12µF capacitors connected in series.

    • Given:
      • C1=3 μFC_1 = 3\, \mu\text{F}
      • C2=6 μFC_2 = 6\, \mu\text{F}
      • C3=12 μFC_3 = 12\, \mu\text{F}

  9. Apply the formula for capacitors in series.

    For capacitors in series, the reciprocal of the total capacitance (CTC_T) is the sum of the reciprocals of individual capacitances:

    1CT=1C1+1C2+1C3\frac{1}{C_T} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} 1CT=13 μF+16 μF+112 μF\frac{1}{C_T} = \frac{1}{3\, \mu\text{F}} + \frac{1}{6\, \mu\text{F}} + \frac{1}{12\, \mu\text{F}}

  10. Find a common denominator and sum the fractions.

    The common denominator for 3, 6, and 12 is 12.

    1CT=412 μF+212 μF+112 μF\frac{1}{C_T} = \frac{4}{12\, \mu\text{F}} + \frac{2}{12\, \mu\text{F}} + \frac{1}{12\, \mu\text{F}} 1CT=4+2+112 μF\frac{1}{C_T} = \frac{4 + 2 + 1}{12\, \mu\text{F}} 1CT=712 μF\frac{1}{C_T} = \frac{7}{12\, \mu\text{F}}

  11. Calculate the total capacitance.

    CT=127 μFC_T = \frac{12}{7}\, \mu\text{F} CT≈1.714 μFC_T \approx 1.714\, \mu\text{F}

    The total capacitance is 1.714 μF\boxed{\text{1.714 } \mu\text{F}}.

    9. Two resistors are connected in series across a 50V supply and a current of 5A flows in the circuit. If one of the resistors is 4Ω. Determine:

    • i) The value of the another resistor.

    • ii) The Pd across the 4Ω resistor.

    • Given:

      • Total voltage VT=50 VV_T = 50\, \text{V}
      • Total current IT=5 AI_T = 5\, \text{A} (In a series circuit, I1=I2=ITI_1 = I_2 = I_T)
      • One resistor R1=4 ΩR_1 = 4\, \Omega

    i) The value of the another resistor.

  12. Calculate the total resistance of the circuit.

    Using Ohm's Law for the entire circuit:

    RT=VTITR_T = \frac{V_T}{I_T} RT=50 V5 AR_T = \frac{50\, \text{V}}{5\, \text{A}} RT=10 ΩR_T = 10\, \Omega

  13. Calculate the value of the second resistor.

    For resistors in series, RT=R1+R2R_T = R_1 + R_2.

    10 Ω=4 Ω+R210\, \Omega = 4\, \Omega + R_2 R2=10 Ω−4 ΩR_2 = 10\, \Omega - 4\, \Omega R2=6 ΩR_2 = 6\, \Omega

    The value of the another resistor is 6 Ω\boxed{\text{6 } \Omega}.

    ii) The Pd across the 4Ω resistor.

  14. Calculate the voltage drop across the 4Ω resistor.

    Using Ohm's Law for the 4Ω resistor (V1=I1R1V_1 = I_1 R_1):

    V1=(5 A)×(4 Ω)V_1 = (5\, \text{A}) \times (4\, \Omega) V1=20 VV_1 = 20\, \text{V}

    The potential difference (Pd) across the 4Ω resistor is 20 V\boxed{\text{20 V}}.

    10. State any two factors affecting resistance of the resistor.

    1. Length of the conductor: Resistance is directly proportional to the length.
    2. Cross-sectional area of the conductor: Resistance is inversely proportional to the cross-sectional area.
    3. Nature of the material: Different materials have different resistivities.
    4. Temperature: For most conductors, resistance increases with temperature.

    SECTION B

    1. (a) Portable machine requires a force 300N to move it. If it moved a distance of 400cm for a time of 200ms. Determine:

    • i) Work done by machine.

    • ii) Average power utilized.

    • Given:

      • Force F=300 NF = 300\, \text{N}
      • Distance d=400 cmd = 400\, \text{cm}
      • Time t=200 mst = 200\, \text{ms}

    i) Work done by machine.

  15. Convert distance to meters.

    d=400 cm=400100 m=4 md = 400\, \text{cm} = \frac{400}{100}\, \text{m} = 4\, \text{m}

  16. Calculate the work done.

    The formula for work done is W=FdW = Fd.

    W=(300 N)×(4 m)W = (300\, \text{N}) \times (4\, \text{m}) W=1200 JW = 1200\, \text{J}

    The work done by the machine is 1200 J\boxed{\text{1200 J}}.

    ii) Average power utilized.

  17. Convert time to seconds.

    t=200 ms=2001000 s=0.2 st = 200\, \text{ms} = \frac{200}{1000}\, \text{s} = 0.2\, \text{s}

  18. Calculate the average power.

    The formula for average power is P=WtP = \frac{W}{t}.

    P=1200 J0.2 sP = \frac{1200\, \text{J}}{0.2\, \text{s}} P=6000 WP = 6000\, \text{W}

    The average power utilized is 6000 W\boxed{\text{6000 W}}.

    1. (b) An electric heater consumes 4.5 MJ when connected to 250V supply for 30 minutes. Determine:

    • i) Power rating of the heater.

    • ii) The current taken from the supply.

    • Given:

      • Energy consumed E=4.5 MJE = 4.5\, \text{MJ}
      • Voltage V=250 VV = 250\, \text{V}
      • Time t=30 minutest = 30\, \text{minutes}

    i) Power rating of the heater.

  19. Convert energy to Joules and time to seconds.

    E=4.5 MJ=4.5×106 JE = 4.5\, \text{MJ} = 4.5 \times 10^6\, \text{J} t=30 minutes=30×60 s=1800 st = 30\, \text{minutes} = 30 \times 60\, \text{s} = 1800\, \text{s}

  20. Calculate the power rating.

    The formula for power is P=EtP = \frac{E}{t}.

    P=4.5×106 J1800 sP = \frac{4.5 \times 10^6\, \text{J}}{1800\, \text{s}} P=2500 WP = 2500\, \text{W}

    The power rating of the heater is 2500 W\boxed{\text{2500 W}}.

    ii) The current taken from the supply.

  21. Use the power formula P=VIP = VI to find the current.

    I=PVI = \frac{P}{V} I=2500 W250 VI = \frac{2500\, \text{W}}{250\, \text{V}} I=10 AI = 10\, \text{A}

    The current taken from the supply is 10 A\boxed{\text{10 A}}.

    1. (c) Five resistors are connected as shown in the diagram below.
    This part cannot be solved as the diagram showing the resistor connections is not visible in the provided image.

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