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Task 2: i. Simplify the formula for the current I to its most basic form. ii. Draw a circuit diagram showing how to connect the four batteries in parallel to increase capacity while maintaining 12V. i

Physics

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Here's the solution for Task 2:

Task 2, Part i:

Part i· 3 steps

Simplify the formula for the current I to its most basic form.

  1. Write down the given formula and values.

    The formula is I=k×Sm/SnI = k \times S^m / S^n.
    Given constants are k=5k=5, m=3m=3, n=2n=2.

  2. Substitute the given values into the formula.

    I=5×S3/S2I = 5 \times S^3 / S^2
  3. Apply the law of indices axay=ax−y\frac{a^x}{a^y} = a^{x-y}.

    I=5×S(3−2)I = 5 \times S^{(3-2)} I=5×S1I = 5 \times S^1 I=5SI = 5S

    The simplified formula for current II is I=5S\boxed{I = \text{5S}}.

    Task 2, Part ii:

Part ii

Draw a circuit diagram showing how to connect the four batteries in parallel to increase capacity while maintaining 12V.

To connect four batteries in parallel, all positive terminals are connected together, and all negative terminals are connected together. This maintains the voltage (e.g., 12V if each battery is 12V) while increasing the total current capacity (Ah).

A circuit diagram would show:

  • Four battery symbols arranged side-by-side.
  • A common line connecting all the positive terminals of the batteries.
  • Another common line connecting all the negative terminals of the batteries.
  • The output terminals would be taken from these two common lines.

Task 2, Part iii:

Part iii· 3 steps

Calculate the inductive reactance XL.

  1. Write down the formula for inductive reactance and given values.

    The formula for inductive reactance is XL=2πfLX_L = 2 \pi f L.
    Given values are f=50Hzf = 50Hz, L=0.1HL = 0.1H, and π=3.14\pi = 3.14.

  2. Substitute the values into the formula.

    XL=2×3.14×50Hz×0.1HX_L = 2 \times 3.14 \times 50Hz \times 0.1H
  3. Calculate the result.

    XL=314×0.1 ΩX_L = 314 \times 0.1 \, \Omega XL=31.4 ΩX_L = 31.4 \, \Omega

    The inductive reactance XLX_L is 31.4 \Omega\boxed{\text{31.4 \Omega}}.

    Task 2, Part iv:

Part iv· 3 steps

Calculate the capacitive reactance XC.

  1. Write down the formula for capacitive reactance and given values.

    The formula for capacitive reactance is XC=12πfCX_C = \frac{1}{2 \pi f C}.
    Given values are f=50Hzf = 50Hz, C=100μFC = 100\mu F, and π=3.14\pi = 3.14.
    Convert capacitance to Farads: C=100μF=100×10−6F=1×10−4FC = 100\mu F = 100 \times 10^{-6} F = 1 \times 10^{-4} F.

  2. Substitute the values into the formula.

    XC=12×3.14×50Hz×(1×10−4F)X_C = \frac{1}{2 \times 3.14 \times 50Hz \times (1 \times 10^{-4} F)}
  3. Calculate the result.

    XC=1314×10−4 ΩX_C = \frac{1}{314 \times 10^{-4} \, \Omega} XC=10.0314 ΩX_C = \frac{1}{0.0314} \, \Omega XC≈31.847 ΩX_C \approx 31.847 \, \Omega

    The capacitive reactance XCX_C is 31.85 \Omega\boxed{\text{31.85 \Omega}}.

    Task 2, Part v:

Part v· 4 steps

Calculate the total impedance Z.

  1. Write down the formula for total impedance in a series RC circuit and given values.

    The formula for impedance in a series RC circuit is Z=R2+XC2Z = \sqrt{R^2 + X_C^2}.
    Given values are R=10ΩR = 10\Omega and XC=15ΩX_C = 15\Omega.

  2. Substitute the values into the formula.

    Z=(10Ω)2+(15Ω)2Z = \sqrt{(10\Omega)^2 + (15\Omega)^2}
  3. Calculate the squares.

    Z=100Ω2+225Ω2Z = \sqrt{100\Omega^2 + 225\Omega^2}
  4. Sum the squared values and take the square root.

    Z=325Ω2Z = \sqrt{325\Omega^2} Z≈18.0277 ΩZ \approx 18.0277 \, \Omega

    The total impedance ZZ is 18.03 \Omega\boxed{\text{18.03 \Omega}}.

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